gkeswani92 Most pythonic solution I could come up with :)
marksheet = [] for _ in range(0,int(input())): marksheet.append([input(), float(input())]) second_highest = sorted(list(set([marks for name, marks in marksheet])))[1] print('\n'.join([a for a,b in sorted(marksheet) if b == second_highest]))
omalsa04 I like your use of join. Btw, you could use list expansion for the read as well if you wanted:
n = int(input()) marksheet = [[input(), float(input())] for _ in range(n)]gkeswani92 That's cool. Reduces the code bya further one line :)
utsav_krishnan [deleted]
pankajpal Why we are trying to cram the code in 2-3 lines. Is this what pythonic is?
If you are writing a code more than 50 lines and there is bug you will have hell of headache in debugging it.
One great personaility said once:
Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it.
mathPythonCoder I feel like people prefer "to code in fewer lines" instead "to write beautiful and clean code" here.
sim_astro I also agree with you and here is reason for code should be clear and simple to read.
Think someone (may be you) after few months/years may need to change/debug it, then you might have forget the cleaverness done in past :/
Be practicle and try to write clear (might not be in fewer lines or cleaverest) but working and easy read (by someone else) and in future easy to maintaine.
kurian_benoy Truely agree with u,but his is elegant coding
venkatraju_1284 ridiculously rich and cool
itsbruce I don't disagree with the principle but in this specific case there's absolutely nothing wrong with using a list comprehension instead of a for loop. The comprehension is doing a simple job elegantly.
If this were production code, I'd use a generator rather than a comprehension, to avoid loading the entire data set into memory unnecessarily, but that's not a problem with the tiny data set here.
ic3balandran Java dev here learning python. I don't think pythonic means cram code in as fewest lines. In fact pythonic just means to use python features to your advantages and create CLEAR, CONCISE, MAINTANABLE CODE. And also follow the "standard" ways of doing something in the python community.
The key word here is MAINTANABLE. I've seen many python challenges were people strive for less lines and even one liners and in some occassions they even fail to do the assignment for the sake of a one liner.
xq9125 True
meet_rahulkumar8 Writing a code which machine can understand is easy. Writing a code which human can understand easily is difficult.
jerodg Once you get the hang of comprehensions you will use nothing else. They are very easy to read/understand/debug. It's no different than when you learned how to use a loop for the first time.
GoodBOi3000 This is python if your code is not small consice and small , it kind of defeats the purpose of coding in python
ManiChilzz python is type of snake. It is large is size. It is non venomous.
SHEERSENDU_GHOSH [deleted]SHEERSENDU_GHOSH r u mad or what? first learn english then come to comment on other's post.
generalom1234 lol...
pipa0979 omalsa04 snippet is simple yet elegent, This is an industry standard code, why are you guys complaining about this ?
pankajpal I beleive it will bloat your memory in case list is huge thats first.
Second the code will be hard to debug.
itsbruce This challenge requires the whole list to be loaded into memory. Unless you are saving each student record to a file as you read it, there is no other option. And we're told there are never more than 5 students in the input.
I use generators rather than comprehensions wherever it is possible and reasonable, but this isn't one of those situations. If you are doing this with nested lists (a dict would be more efficient), the whole list has to be traversersed at least twice, depending on your sorting strategy.
Hard to debug? Comprehensions have become core to Python. They beat hand-crafted for loops for several reasons:
- More expressive (if not abused)
- Less fragile and error prone
- Can often easily be turned into generators for memory efficiency.
Learn them. Modern Python is full of them. Which of these is clearer?
validGroup = all(validUser(user) for user in group)
for user in group: if not validUser(user): validGroup = False break else validGroup = True
nick3499 [deleted]
TayyabKhan [deleted]ManiChilzz your answer is shit
tincumagic You are wrong, for this chalenge you only need to traverse the list once. You will keep two lists: 1. list of name found for the second position 2. list of names found for the currently min value. In case you find a another min value, the second position list of names becomes the min value list of names.
captainkangaroo1 +1 for tincumagic
This version only stores 2 lists. If the input scores are taken from a normal distribution, the 2 lists will be extremely short and will use virtually no memory. Runs in linear time.
lowest=[] secondLowest=[] for _ in range(int(raw_input())): name=raw_input() score=float(raw_input()) if len(lowest)==0: #initialize lowest list lowest=[ [name,score] ] elif score < lowest[0][1]: secondLowest=lowest lowest=[ [name,score] ] elif score == lowest[0][1]: lowest.append( [name,score] ) elif len(secondLowest)==0: secondLowest=[ [name,score] ] elif score < secondLowest[0][1]: secondLowest=[ [name,score] ] elif score == secondLowest[0][1]: secondLowest.append( [name,score] ) names=[item[0] for item in secondLowest] for name in sorted(names): print name
itsbruce -1 for ugly and verbose code. That is a mess.
lnie1 Hi, Bruce,
People are using different benchmarks.
your code is beautiful because it is easy to read and friendly to starters especially for python starters
tincumagic's code is beautiful because it runs in linear time. If you want to work for big IT companies, like Goolge, Facebook as a professional developer, please note this is the answer that they want. Sometimes, verbose code can be extremely effecient
itsbruce You can run in linear time without ugliness. Google and Facebook value maintainable code and have extensive coding standards devoted to that.
Start with the clean code, move to the fast code as necessary. The result will still look better than that.
The Hackerrank challenges actually offer good training in this approach. Start with the clean solution, then see which tests - if any - it fails, then see how much you have to contort the code to make it meet the challenge. Aim for the least distortion necessary, because the clean code is easier to maintain. Which will oten matter even in these challenges, when your optimisation clears one difficult test but still fails the next.
There's never an excuse for starting with something that much of a mess.
loki29464 God thats some fugly code.
itsbruce You mean like this?
scores = {} top2 = [] def logScore(score): global top2 xs = [x for x in top2 if x < score] + [score] + [x for x in top2 if x > score] top2 = xs if len(xs) < 3 else xs[0:2] for _ in range(int(raw_input())): name = raw_input() score = float(raw_input()) logScore(score) scores[score] = scores.get(score,[]) + [name] for name in sorted(scores[top2[1]]): print name
I think you may have missed the point of what I was commenting on (the solution techniques presented in this thread, not on how I did it).
captainkangaroo1 Hey itsbruce, I think you missed the point from tincumagic. If you have a gazillion names in the input, your code will store all gazillion names in your "scores = {}" dict. (And BTW your code doesn't use nested lists, which is the subject of this assignment.) My solution (that you find "ugly and verbose") only stores 2 lists. If the input scores are taken from a normal distribution, my 2 lists will be extremely short and my code will use virtually no memory.
itsbruce [deleted]itsbruce "If you have a gazillion names in the input"
The challenge explicitly states there will never be more than 5. If the number were open ended, I'd optimise for that. As it is, there's no point. Pointless optimisation is a waste of time, in real work even more so than in toy challenges. Your solution does more work than mine and doesn't have to because there's no gain here.
"And BTW your code doesn't use nested lists, which is the subject of this assignment.)"
A fixed limit of 5 is also part of the assignment but you pointlessly ignored that.
Nested lists are the wrong solution for this challenge. That's typical of the author, most of whose challenges are poor quality and poor teaching material. Fortunately, there's zero requirement to use nested lists.
I find your solution ugly and verbose because all the logic is tied up in one long if/else statement. If I wanted to optimise for saving space (not necessary with a guaranteed limit of 5), i'd simply adjust the code so that when a value is pruned from top2, its names are pruned from dict and all future entries for that score are ignored. And because my code isn't a long chain of if/else, that's an easy change to make.
captainkangaroo1 show me
itsbruce scores = {} top2 = [] def logScore(score): global top2 xs = [x for x in top2 if x < score] + [score] + [x for x in top2 if x > score] if len(xs) > 2: scores.pop(xs[2], None) xs.pop() top2 = xs return score <= xs[-1] for _ in range(int(raw_input())): name = raw_input() score = float(raw_input()) if logScore(score): scores[score] = scores.get(score,[]) + [name] for name in sorted(scores[top2[1]]): print name
tincumagic -1 for incomprehensible code for the old school programmers like me. Maybe some Python experts find it readable.
itsbruce Truly it is indecipherable.
After all, a for loop that iterates over the input and feeds each parsed number to a function called "logScore". Deeply mysterious.
IF logScore returns true, the corresponding name is added to the dictionary entry for that score, otherwise not. How could that possibly be related to the challenge? The Rosetta Stone to translate this esoteric concept has not been carved.
Sheesh. It's mostly simple Python and a couple of core library functions that have obvious names. There are, to be fair, one or two things that will not be obvious to somebody completely unfamiliar with Python. The underscore is the variable equivalent of /dev/null - here, we're iterating over a range but don't actually care about the individual numbers in the range. Using -1 to index a sequence gives you the last element. List comprehensions used to filter a list (although you'd think the "< score" and "> score" might be a clue). But somebody told me this was a Python tutorial, which might mean that people with some interest in learning the language might try it.
I never had much time for the definition of "old school programmer" that says "Don't learn anything that doesn't look like C and complain if anybody else uses it". Have you considered actually learning the language?
k11h12anh I think everyone mentally processes code differently so some people might find one way is better than the other. I'm sure some will like itsbruce's code (verbose in a different way as in too many for loops) but for me I find tincumagic/captainkangaroo1's code cleaner and easier to read.
unknowndoor Hey Bruce!
I am new to programming and have started with python. I like your code because i can comprehend it, but i am having difficulty understanding some parts of it. If you could help clarifying the concepts it'd be great:
I dont understand this:
xs = [x for x in top2 if x < score] + [score] + [x for x in top2 if x > score]
First you mention x < score and then x > score, you consider both the cases (!) and how can you concat score in between. I'm like totally lost on this line. What is (xs[2], None) in:
scores.pop(xs[2], None)
What does (score,[]) + [name] do in:
scores[score] = scores.get(score,[]) + [name]
Thanks in advance!
itsbruce Hi. OK, first line...
[x for x in top2 if x < score]
This is a list comprehension. It creates a new list based on the contents of another sequence (top2). In this case, it returns a list containing all the elements in top2 which are less than score (preserving the original order). The other comprehension returns a list with all the elements which are greater than score. Meanwhile, [score] is a list containing the single value score.
So what I do in that line is create three lists - everything less than score, [score], everything bigger than score - and concatenate them into one list. What it does is take a list which may already contain scores sorted in order and adds a new item in the correct position within the list.
By doing it this way, I don't have to check to see if the list is empty or contains only one item or contains two items. I don't have to create the different strategies to cope with those 3 different possibilities. This simple strategy works in all of those situations. For example, if top2 is empty, the two list comprehensions will return empty results and two empty lists will be concatenated onto a list containing only score, which is what I would want to do if top2 were empty. If score is lower than any of the current scores in top2, then the first list comprehension will be empty and the second one will contain everything from top2; an empty list will have score added to the end of it and then the contents of top2 added to the end of that.
Oh, and I also don't have to deal with the situation where the new score has the same value as one of the scores already in top2. can you see why?
So this one line always does the right thing, no matter how many values are already in top2 and no matter how the new score value compares to them. So it works much better than a long sequence of if/else statements. It's also much clearer in its intent than a long chain of if/else - if you understand list comprehensions.
unknowndoor First, two thumbs up for for the detailed explanation, thank you! Now it makes so much sense to me! This is such a nifty way to deal with the problem since the order of the entries are maintained and you know where what is.
I understand what will happen when top2 is empty, that number of elements is irrelevant since you iterate over all the entries in top2 with score and you maintain the order of entries in top2, wether the score is higher or lower than top2, but i'm sure why it'd not add two entries of same value because the first comprehension will give an empty list, score will be score and the third comprehension will give an empty list so it should append score to top2 otherwise the first time the code is run, it wouldn't work in that case too?
itsbruce Imagine the current content of top2 is [1, 2] and logScore is passed a value of 2. The first list comprehension returns all the scores in top2 which are less than 2: [1]. The second list comprehension returns all the scores in top2 which are greater than 2: [], the empty list. So the result is
[1] + [2] + []
which means that the contents of top2 don't actually change. logScore replaces the contents of top2 with the result of that line of code, it doesn't add it to the end of top2.
itsbruce On to the second line you had an issue with:
scores.pop(xs[2], None)
So I want top2 to contain only the lowest two scores (yes, I should probably have named it bottom2), but I've just added a new value to the list, so there may now be 3 scores in there. So I check and if there are 3 then I remove the last one (which will be the highest of the three and not the one I want to keep) by doing this:
xs.pop()
pop() is a function that removes the last item from a list and returns that value. But I ignore it because I don't care what it was.
But before I do that, I also remove from the scores dictionary the key and the list of names associated with the score that I'm discarding. I know it can't be the second lowest score, so I don't need to keep that list of names around. Deleting it from the dict saves memory.
Now, there is a del function that removes a key and its value from a dictionary. But there's a chance that there may not be any entry in the dictionary for this score; if score was bigger than the existing values in the list, then it's been added to the end and is about to be removed from it without there ever having been an entry for it created in. So you have to check if the key exists in the dictionary before deleting it - if you use del to try and remove a key that doesn't exist, it raises an error.
Or you can use the dictionary version of the pop() function, which takes a key as input. Like the list version of pop(), it removes an element from the collection and returns its value. If you also provide a default value (as the optional second argument), it won't complain if the key doesn't exist. I don't care what that value is, so I provide a default value of None.
So that line removes an entry from the scores dictionary if it is there; otherwise it does nothing.
unknowndoor Just to clarify here,
scores.pop(xs[2], None)
Means: Python pop key position 2 from the list xs and return value None
itsbruce No, it means
- Find the third score in the list xs (which will be the highest score of the three, the one we don't want to keep)
- Delete that score and its list of names from the scores dictionary and return the value of that list. If there is no such entry in the dictionary, return None.
scores.pop is a method belonging to the dictionary class. It modifies the dictionary it belongs to, it doesn't modify the values passed to it. It always needs one parameter which is the key to fetch/remove from the dictionary. It optionally takes a second value, which is a default value to use if there is no such key in the dictionary.
I don't do anything with the returned value, I ignore it, so it could return anything and it wouldn't matter. I'm only providing a default value to stop pop from complaining if there is no such key in the dictionary. I use the pop method rather than the del function because pop can be made not to complain; I know there will be occasions where there is no such key and I'd rather use a function that doesn't mind, rather than have to make that check myself. I provide None as the default value to make it clearer that I don't care about the result (and to not waste the program's time by having it create and store a value that will never be used).
itsbruce OK, third and final line:
scores[score] = scores.get(score,[]) + [name]
So if logScore(score) returned true, score is one of the two lowest scores we've seen so far, which means we want to add the name to the list of names associated with that score (if it isn't one of the lowest two, we don't care). Which is what the line above does.
So we want to fetch the list of names associated with score in the scores dictionary. But what if this is the first time we've seen this score? In that case, there will be no entry - trying to fetch it will raise an error. However, if we provide a default value, get() will return that default value if score is not a key in the scores dictionary.
So that line fetches the existing list from the dictionary or returns an empty list if there is no entry for score in the dictionary. Then we add name to the end of that list and put the resulting list into the dictionary as the list associated with score.
unknowndoor Here i get a bit confused when you mention "if logScore(score) returned true, score is one of the two lowest scores we've seen so far" > i thought that the job of logscore is to only get second heighest score every time it iterates and this line adds the identified second heigest scores to the dictionary scores with their names **+ [name]
And thanks again!
itsbruce According to the problem description, it's the names of the students who got the second lowest score we're looking for. My code certainly finds the lowest two scores and since my code passes the tests, that seems to be what's wanted. Can't remember precisely what I was thinking when I called the variable top2 - the two most significant and/or the two which would be at the top if you sorted the list in descending order, I suppose.
You're wrong about what logScore does. Its responsibility is to add the score to the list of the two lowest scores if it is one of the two lowest scores we've seen so far. It returns True if the score is currently one of the two lowest, False otherwise.
The program checks the result; if the returned value is True, then score is currently one of the two lowest scores (lowest or second lowest) and we should keep a record of the name associated with it. But recording the name isn't logScore's responsibility.
To be honest, logScore probably has either too much responsibility or not enough; either a separate function should manage both the addition of names to the dictionary and their removal if a score drops out of the lowest-two-scores list, or logScore should completely manage both the top2 list and the scores dictionary. As it is, it's a little messy. My excuse was that it was a quick change made to my code at somebody else's request.
matthewzalex Hi itsbruce. I read your explanations and understood the code. I was wondering, if i use the same comprehension to sort the names, will the code be better?
I modified the if statement,
if logScore(score): scores[score] = scores.get(score,[]) + [name]to
if logScore(score): temp = scores.get(score,[]) scores[score] = [x for x in temp if x < name] + [name] + [x for x in temp if x >name]and avoided sorted() function in the end. full code:
scores = {} top2 = [] def logScore(score): global top2 xs = [x for x in top2 if x < score] + [score] + [x for x in top2 if x > score] if len(xs) > 2: scores.pop(xs[2], None) xs.pop() top2 = xs return score <= xs[-1] for _ in range(int(raw_input())): name = raw_input() score = float(raw_input()) if logScore(score): temp = scores.get(score,[]) scores[score] = [x for x in temp if x < name] + [name] + [x for x in temp if x >name] for name in scores[top2[1]]: print nameitsbruce I applaud your perception, but the first objection is that you've created a significantly less efficient solution. You sort the list of students with the grade you've just seen if that score is currently in the lowest two scores. Imagine if the first few students in the input get the highest score and there are more than 2 different scores. Your code will diligently sort that list, only to throw it away later. That's a waste of time.
Aside from that, reusing the technique from logScore on the list of names is likely to be significantly less efficient than Python's sorted function.
What I did in the logScore function is an insertion sort - one of the easiest sort functions to implement but one of the least efficient on data sets of any significant size. It isn't even the most efficient implementation of insertion sort - an efficient implementation would compare the next value to insert to the value just inserted, remembering the position of the value just inserted, so that the next iteration can work forward or backward from that point rather than starting from the beginning each time. That doesn't matter in logScore, because the data set never grows beyond 3 (being cropped back to a maximum of 2 elements after the sort).
As it happens, we're told there will never be more than 4 students with the second-lowest grade, so the difference in efficiency is unlikely to matter, but a) I don't see the point in dropping the core sort function for a more-than-usually-inefficient insertion sort here and b) I prefer any piece of code to do only one thing unless there's a clear and needed performance gain from doing more than one thing at once. Trying to do two things at once is also why you missed the fact that your proposed solution is inefficient.
matthewzalex Thank you for the clear explanation. Now i see what i did there. I never thought of insertion sort that way too! Good to have you commenting on what I did.
danielgcwik thanks bruce this worked for me in python 2
sim_astro Its just a challenge with title "Nested List" and not about cleaverest solution.
nath_ajit01 [deleted]
ericali nicely done!!!
pashnidgunde marksheet = [[input(), float(input())] for _ in range(int(input()))]
codeharrier I would be very leery of the order in which those input() calls are being done. And given the way Python tends to upend its language standards with any given revision, I'd be very, very, very leery of relying on it to never change.
itsbruce Counting on the order of items in a dictionary would be an unwise thing, which some languages deliberately randomise just to discourage the habit. Counting on the order of evaluation of items in a sequence is extremely reasonable for Python, a strictly-evaluated, imperative language.
Comprehensions and generators have long been central to idiomatic Python. The usage you're complaining about is widespread. Making the evaluation of sequences non-deterministic a) isn't something GvR is ever likely to consider given his history and expressed opinions and b) would be a code-breaking change for so much production code that the complaints would make the Python 3 hoo-ha seem a mild disagreement.
codeharrier I dunno. Someone who'd change print from a statement to a function is not going to make anyone trust the stability of the platform.
itsbruce You mean a minor change between one major version and the next (2.x to 3.x) is reason to believe that sequences aren't going to be evaluated in order from one point release to the next? Python is curated conservatively compared to most dynamic languages. I'm no great fan of Python's design but there's no evidence GvR is batshit crazy.
codeharrier A "minor" change could alter anything internal that isn't written down in a standard. And when it does affect order of execution, they'll shrug their shoulders and walk away.
And as for GvR, that change in print, which could have been averted simply by calling the new function pr() or something, will probably induce on the order of millions of man-hours of rework, re-review, and re-verification over time, not to mention all the arguments like this one. If that doesn't ring up as batshit crazy, the bar is set way too high.
itsbruce You can argue over whether it was the best move, but it was highlighted in advance and 2.x is still supported. Nobody's code was in danger. The problem has been one of persuading everybody in the community to move, not people's code crashing.
You're criticising somebody's perfectly good, entirely idiomatic code. By your argument, all Python code is vulnerable, not just that code. Unreasonable singling out that decent solution.
ManiChilzz don't you have any other work?
satadhi hi can u kindly tell me how th loop is working in your code ! i am new to this language and i am pretty confused my now. thanks for the help
yash1th he is using list comprehensions, for more
https://www.youtube.com/watch?v=3dt4OGnU5sM
visit that link
bryon_nicoson this is a helpful, well-presented introduction to python list comprehensions. thank you for posting it :)
bitb3ast I tried this method, but I get the following error. Do, I need to import some module?
Traceback (most recent call last): File "solution.py", line 4, in <module> marksheet = [[input(), float(input())] for _ in range(n)] File "<string>", line 1, in <module> NameError: name 'Harry' is not definedsteeeenos I was having the same problem. The way i fixed it is by using raw_input() for the name instead of input(). Hope this helps
styx97 This maybe a matter of your python version. In python 2 , input takes only integer inputs and raw_input takes strings, ints and floats. In python 3, there is only one input() function which is practically same as raw_input in Python2.
Long story short, the code above is in Python3, so suit yourself.
rakshanihamad change your compiler to python 3 if you are using python 2 or vice versa
Hope it will help :)
akulkapoor marksheet = [[input(), float(input())] for _ in range(int(input()))]
Reduce one more line :-)
shivamgoel977 will u please tell me why we use[1] at the end second heighest=sorted(())[1]
tarunchand280211 because we want the second highest value of the list. if you do [0], you will get the highest which the problem has not asked.
loinger_adiel Should be second lowest grade ;)
beckwithdylan [deleted]mail2karthk07 i did it in one line, it works
marksheet = [[input(), float(input())] for _ in range(int(input()))]
karthiky7cs1212 Is there a need to convert a set to list , Also what is the result of sorted(marksheet) will it sort based on Names or marks?
jerodg or simply:
marksheet = [[input(), float(input())] for _ in range(int(input()))]
alan_cruz_tj marksheet = [[input(), float(input())] for _ in range(int(input()))]
Shivan111 what does [marks for name,marks in marksheet] mean can you explain that to me please ! ?
latived This is called "list comprehension" -- an awesome way to create lists. He takes only "marks" for every (name, marks) in marksheet, where marksheet is a nested list.
Example:
In [682]: marksheet
Out[682]:
[['Harry', 37.21],
['Berry', 37.21],
['Tina', 37.2],
['Akriti', 41.0],
['Harsh', 39.0]]
In [683]: [marks for name, marks in marksheet]
Out[683]: [37.21, 37.21, 37.2, 41.0, 39.0]karan_youcan2438 Thanks a lot.. that was useful or me :)
sumitmcc i see that u explained well, but i'm still not clear. would you mind elaborating it for me? i am not clear on "marks for name, marks in marksheet" part..
niwot "list comprehension" is in the form : [ A B C]
Part A defines the element(s) put into the new list. Part B is a set of <"for" element(s) "in" collection> statements. Part C contains optional "if" statements that filter each collection element from consideration. In this case it looks like...
A B C [marks for name, marks in marksheet] there is no C [new2 for new1, new2 in parent]
The tricky part is between "for" and "in" of Part B.
B slices the parents child lists horizontaly into two new lists (name and marks), iteratively assigning name=marksheet [0] and marks=marksheet [1]
Part A says "im only interested in keeping the marks list." We throw away the new list called "name."
I suppose if the childlists each had N elements you could say, "for new1, new2, new3,,,newN in ParentList" and be able to manipulate N new lists in your part A.
Hope helpful
niwot The answer in the Editorial tab
a = [[raw_input(), float(raw_input())] for i in xrange(int(raw_input()))]
shows something about the order in which comprehensions are executed. "B" (in comment above) happens first, so xrange gets N. I suppose if there was a raw_input() call in C (the optional "if" statement), that woud happen next (left to right?). Finally the A bit is executed left to right.
I'm just making assumptions here. Please correct me if its wrong.
BhagyeshPanchal great explanation thank u
NseEtim i thought the set() function removes repitition, if thats true then set(marks for name, marks in marksheet) would not return those scores that are of the same value someone should please clearify
kevinbrewer04 Correct, that line of code is sorting all the unique marks (scores), and grabbing the second from the bottom (
[1]). It's slightly confusing because he named the variablesecond_highest, when it's really the second lowest score. The reason for usingsetis to guard against the case where more than one student is tied for the lowest score.
harshavardhanm03 What if we want to find the duplicate names? lets say I have [['A',12],['A',24],['B',22]]
How can I find the unique set based on the first element. Please let me know
achieng91 Hey nice solution man, but 1 issue though. When the no. of students is 2 and they have the same score,
second_highest = sorted(list(set([marks for name, marks in marksheet])))[1]this line of code result in IndexError as the size of list is 1.
Esser420 I just started learning python and this commetn just helped me understand it a lot better!
students = sorted([[raw_input(), float(input())] for _ in range(int(input()))])
secndLow = sorted(list(set(grade for [_, grade] in students)))[1]
print ("\n".join( name for [name, grade] in students if grade == secndLow))
manasveemittal If somebody wants to know reason of 2 different sorts on output
students
[['Dheeraj', 39.0], ['Pallavi', 36.0], ['Prashant', 32.0], ['Shivam', 40.0]]
secndLow
36.0
so basically first sort(on students) sorts by name in case there are repeated names for a the given second lowest grade
second sort just sorts the scores, and set is used for unique values of scores to be sorted
Correct me if wrong
sharath14187 what if there are two names with same scores
Shriswissfed Thank you for posting and explaining code...learnt two new things......!!!!!!
th3ac3 Couldn't sorted(list(set([marks for name, marks in marksheet])))[1] be simplified down to sorted([marks for name, marks in marksheet])[1]
kenzcross This works the same for the initial tests however what happens when you have duplicates of the lowest value (ie 62 62 73 87 or 62 62 62 62 73). How can you determine what arbitrary index will hold the lowest value?
TL;DR Set removes the duplicates.
cbrit Why is it necessary to first turn the comprehension into a set and then back into a list? EDIT: I think I found out that it's because it removes duplicates. Thanks.
Velvet why have you used "set" in the 4th line ?
.....list(set(..)) ?
What is the purpose of set here ? Using list(..) instead of list(set(..)) should be fine I think. Where am I wrong ?
geekysalsero What if there are two top marks ? Then the second highest would be [2] ?
sushilyadav It doesn't matter because the set() function will remove all the duplicates as said by @aswathm78 and list provided by sorted() function will only contain unique marks. So, sorted(...)[1] will always contain 2nd lowest marks.
Spraynard_K [deleted]
dillonredding [deleted]newkid Great idea to use the set.
deepu_ac21 Could you explain the last line to me "[a for a,b in sorted(marksheet) if b == second_highest]))"
Bonsoirelliot what does this mean[1]?
newmocart great solution!!!
a bit improved on py2
arr = [[raw_input(), float(raw_input())] for _ in range(input())] c = sorted(set([b for a,b in arr]))[1] print '\n'.join(sorted([a for a,b in arr if b == c]))
prakashyadav008 c = sorted(set([b for a,b in arr]))[1]
what does [1]in this line mean
tao_zhang get the second element in the sorted collection
Max_Vignesh It gets the second element from the array.
Thanks & Regards
tao_zhang Interesting, even without creating the list before sorting, after sorting
print(type(sorted(set([b for a, b in sheet]))))
result:
class 'list'
omavinashkalal can you explain me following code : arr = [[raw_input(), float(raw_input())] for _ in range(input())] c = sorted(set([b for a,b in arr]))[1]
NseEtim that is list comprehension the varriable "arr" takes two inputs, but the second input is placed inside the float function, so the input is converted to a floating point value,which means it cant take strings,then "for _ in range(input())", the "_" can be replaced with anything, its just that if you are writing code where you wont need to use that particular variable for anything, you could call it _, its just a community rule of thum thingy,so the third input is taken at the point of the range, its just a way of reducing the amount of code one has to write, so instead of doing something like,
a = raw_input() b = float(raw_input()) d = input() arr = [a, b] for _ in range(d) :# you can use i or any variable name in place of '_' c = sorted(set([b for a, b in arr]))[1]# the 1 in braces means the order in which the sorting should be done, if you use -1 it will be in reverse order
you just use the short version you have there
starcoder thanks gkeswani92 for this solution bt since i am just new in python world so can you plz explain each line of code
satadhi [marks for name, marks in marksheet] how this is working ? can you kindly tell me, i am new to this language .
amit_badve what if bottom 2 students have same marks? we will need a loop to find out second min person from sorted list.
tao_zhang [deleted]yash1th is there any reason why you created a list in the statement?
sorted(list(set....)
as even if you leave it as a set, it will create a set and then sort it, right?
rchidb Here is a variance of your implementation that sorts the the nested list using its 2nd element directly to avoid using Set(). It may save some time in the best case scenario. Comments welcomed.
N = int(raw_input()) students = [] for i in xrange(N): grade = [raw_input(), float(raw_input())] students.append(grade) students = sorted(students, key=lambda x: x[1]) second_highest = students[0][1] for name, grade in students: if grade > second_highest: second_highest = grade break print('\n'.join([name for name, grade in sorted(students) if grade == second_highest]))
neil_a_ramsay Very similar to mine and much more legible than the obfuscated skinny code that the experts are posting above!
kurian_benoy for name, grade in students: if grade > second_highest: second_highest = grade break
I am new to python. Can you please explain what does grade> SECOND_highest means?
enilaydagdemir [deleted]mitkatwala95 students = sorted(students, key=lambda x: x[1])
Can someone please explain me how does this line work? Bit confused on the lambda x: x[1] thing!
valibraimi lambda x:x[1]
Is an anonymous function that will select the second element from lists that are processed in the nested list. also it is equivalent to
def select_second_element(x): return x[1]
XRohan Great thinking. salute to you sir.
Krrish_bro [deleted]
enilaydagdemir Can you explain what this "join" part of the code does?
landonkoo0207 This is not my code, but I can comment on that.
It is to print the results as strings in separate lines.
Without it, the results would be like ['Harry', 'Berry'].
'\n' part is to print each name in a new line. This is just the way of python to print a list as strings :)
kurian_benoy second_highest = sorted(list(set([marks for name, marks in marksheet])))[1]
What is meaning of [1] at end of snippet?
enilaydagdemir I think it means taking the second last element from the list.
generalom1234 Not second last but it determines the order of sorting which is to be in done for every 2nd element... I guess...
Rick356 It will display the element at 1 position
vikas_p_manav [deleted]
paritosh17 can please please explain me this " second_highest = sorted(list(set([marks for name, marks in marksheet])))[1] " ??
jonobjornstad excellent solution once again!
SebastianNielsen in the line were you define "second_highest": Why do you say: "list( blah blah.. )" The list comprehension will always return a list anyway, or am I wrong?
ameybh to use the sort
wuestenblume An alternative to using join is the unpacking operator and setting sep='\n' for print.
eyshikaengineer Can you please explain how your code takes second highest value, also what is the use of set()
Lucasjuv [deleted]pshaikh sorted(list(set([marks for name, marks in marksheet])))[1]
In above statement how does this work?
set([marks for name, marks in marksheet])
How is "marks for name" and "marks in marksheet" interpreted?
[deleted] Wow! Elegant.
oksuzgonulh Great solution. Because this is what happens if it is not pythonic at all!
if __name__ == '__main__': final = [] n = int(input()) for _ in range(n): name = input() score = float(input()) final.append([score,name]) final.sort() k = 0 for x in range(n): if final[x][0] > final[0][0]: k = x break for x in range(n): if final[k][0] == final[x][0]: print(final[x][1])
I have k = 0 for the case all students have the same grade, so the solution is all of them.
ankitathummar final = [] n = int(raw_input()) for i in range(n): score = raw_input() name = float(raw_input()) l = [name,score] final.append(l) final.sort() needed = final[1] c = final.count(needed[0]) if(c == 1): print needed[1] else: for i in range(n): if(final[i][0] == needed[0]): print final[i][1]
I have use same logic as your code. You said it in not pythonic. But this is what logic came into my mind as I am new to python. Though it is failing for TC 2 & 3. What should I do for more cool work as everyone have done for this statement problem.
hivex201 I think there is something wrong with the compiler. I made some changes to your code to run the rest of the test cases and made some changes:
final = [] n = int(input()) for _ in range(n): name = input() score = float(input()) l = [name, score] final.append(l) final.sort() need = final[1] for i in range(n): if(final[i][1] == need[1]): print(final[i][0])
When I ran it on a offline IDLE compiler all outputs are correct but when I ran it on HackerRank, it shows scores instead of names.
sdovejero [deleted]
saransh_arora123 Can you please explain me how are the last two lines working ?
second_highest = sorted(list(set([marks for name, marks in marksheet])))[1] print('\n'.join([a for a,b in sorted(marksheet) if b == second_highest]))
ameybh will you please explain the join part to me?
nandi_abir [marks for name, marks in marksheet] how does this give me a list of marks .... name is not defined separately ... what does it do???
kevinbrewer04 marksheet is a 2D array, something like
[["Sally", 20.3], ["John", 19.5], ["Susan", 21.2]].So
name, markseffectively destructures each element in the 2D array. For every element,namegets bound to the first item, e.g."Sally", andmarksgets bound to the second item, e.g.20.3. So we iterate through the 2D array, and for each element, we return only the second item in the subarray, and we end up with a 1D array like this:[20.3, 19.5, 21.2].The code is slightly misleading because as the list comprehension iterates over the 2D array
marksheet, it's only looking at 1markat a time. So perhaps, it should have been written[mark for name, mark in marksheet].Does that make sense?
ottawabiju Thanks, how to get the name list (remove sorting), instead of mark pls.
arunangshubsws Who told you that this code is pythonic?? read PEP 8 and PEP 20 again! I dont care if you cram your code into 5 or 4 lines but please refrain from saying that cramming codes is pythonic! It goes directly against the core principles of python programming!
sristi_gupta92 second_highest = sorted(list(set([marks for name, marks in marksheet])))[1]
can you explain this line , i didn't understand why a list has to be created to be sorted, we can place a collection in sorted method.
seejoy [deleted]bubun71 what is that [1]
vishalit10 sorted(list(set([marks for name, marks in marksheet]))) it will provide a list, and of that list object at index 1 will be carried and valuue transfer to second_highest. index start with zero
vishalit10 [marks for name, marks in marksheet] how it will work??
itsbruce The most pythonic thing would be not to use a sorted set when a dict or defaultdict is a significantly better choice.
- Create defaultdict(list) grades
- Read in each name and grade, grades[grade].append(name)
- Sort the keys and take the second lowest.
- Sort the list at that key and print each element
I'm not inclined to use an inappropriate solution when nothing compels me to.
humichina Based on your code, I got this solution. Learned a lot. (py3)
student_marks = [[input(), float(input())] for _ in range(int(input()))] snd_mark = sorted(list(set([mark for stu, mark in student_marks])))[1] print(*sorted([stu for stu, mark in student_marks if mark == snd_mark]), sep = '\n')
RahulBhusare Can you please explain me this code?
bONSPARROW even more compact. is it pythonic?
students = [[input(),float(input())] for _ in range(int(input()))]
print(*(sorted([x[0] for x in students if x[1] == sorted({x[1] for x in students})[1]])),sep='\n')
itsbruce Pythonic aimes for the ideal mix of the explicit and the expressive, so no. Your compactness is at the expense of efficiency as well as clear meaning - you sort the list, then traverse the unsorted list. You traverse the entire list twice. You do more sorting than you need, because you're sorting alphabetically the names of students whom you will discard. That can be improved significantly.
I've said elsewhere that using nested lists for this is an inappropriate solution (dicts much better). But if you want to use nested lists then
- Sort the list by score only.
- Drop from the front of the list all elements with the same score as the first element
- take all the elements from the new sequence with the same score as the first element of the new sequence, discarding the scores and ignoring the rest of the new sequence
- Sort what you have left.
If you want to be efficient and compact to the point of unreadability, here's how your code could be changed to do that...
from itertools import dropwhile, takewhile, tee students = ([input(),float(input())] for _ in range(int(input()))) print(*(x for s1 in [sorted(students,key=lambda y: y[1])] for lowest in [s1[0][1]] for (s2, s2b) in [tee(dropwhile(lambda y: y[1] == lowest, s1))] for next_lowest in [next(s2b)[1]]for x in sorted(y[0] for y in takewhile(lambda y: y[1] == next_lowest, s2))), sep="\n")
For larger data sets with a spread of scores, that's going to be much more efficient than your solution (partly because I use generators and iterators, partly because I do as little traversal as possible). But neither of them are pythonic. Line noise is not pythonic.
petersonpeter elegant!
nath_ajit01 Nice. You could also unpack the list like this when printing in python3
print(*[a for a,b in sorted(marksheet) if b == second_highest], sep="\n")
iamalokit what's the use of set() in the above code. Please explain. ThankYou
nath_ajit01 a set is an iterable with unique elements. So when you do something like,
list(set(some_list))
You get a list with just unique elements
sycramore I am curious about taking position 1 in second_highest. I would prefer something like counting how often the minimum arises and then take the count + 1. That might help if you have more than one person with the lowest grade.
nihl01 Short code does not equals efficient code.
In this code there's a loop of O(n) to read the input, then another loop of O(n) to get the marks, then set() (I'm not sure but I think) is O(n), the sorted marks is O(nlgn) (here n being the size of the set), then another sorted() by names O(nlgn) and a O(n) loop to get the names with the second highest and print them.
A solution: In the same loop that reads the input you could find out the 2nd lowest.
And instead of using a list to store the data use a dict of lists, where the key is the mark, and append to it the names.
After reading all the input you know the 2nd lowest mark, asking the dict for the data of that key will get you the list of names, sort them and print them.
MatricksDecoder You can actually be more pythonic and get rid of the for loop completely and only use list comprehensions all the way n = int(input()) students = [[input(), float(input())] for i in range(n)] second_lowest_score = sorted(list(set([i[1] for i in students])))[1] print(*sorted([i[0] for i in students if i[1]==second_lowest_score]), sep="\n")
saif_bhatkar Super I followed ur way and tried to perform this same operation on list of dict();
marksheet=[]
for _ in range(0,int(raw_input())): d = dict() d['name'],d['score'] = raw_input(),float(raw_input()) marksheet.append(d)
second_lowest = sorted(list(set([score['score'] for score in marksheet])))[1]
print '\n'.join(sorted([a['name'] for a in marksheet if a['score']== second_lowest]))
DebasishOpu Such a beautiful code <3
tongjin what does [marks for name, marks in marksheet] mean? Any rules for that? Thx
kozuch_tomasz1 Such code pieces are too hard for debbuging and maintaining, here is my edit of your code for minimal readability:
marksheet = [] noStudents = int(input()) for _ in range(noStudents): name = input() score = float(input()) marksheet.extend([[name, score]]) uniqueOrderedMarks = sorted(list(set([marks for _, marks in marksheet]))) second_highest = uniqueOrderedMarks[1] print('\n'.join([name for [name, score] in sorted(marksheet) if score == second_highest]))mrchethan print('\n'.join([a for a,b in sorted(marksheet) if b == second_highest])) can you explain this line pleas?
pradippatel7792 Lets first break whole code into pieces.
students = [[raw_input(), float(raw_input())] for _ in range(input())]
it creates nested lists.
students = [['Harry', 37.21], ['Berry', 37.21], ['Tina', 37.2], ['Akriti', 41.0], ['Harsh', 39.0]]
second_lowest = sorted(set([grade for name, grade in students]))[1]
First it removes repeated values and sort by their values. so sorted list becomes [37.2, 37.21, 39.0, 41.0].
then it copies 2nd element as second lowest value which is 37.21 to compare whether it has more than one student or not.
print '\n'.join(sorted([name for name, grade in students if grade == second_lowest]))
it is called list comprehensions. syntax is : [ expression for item in list if conditional ]
First it compares every values with second lowest value if it is true then it copies new name object which is Harry and Berry in new list then it sorts by name.
'\n'.join prints each object from new list with new line.
For more information, you can refer below link.
atulksingh4 Great work!
bhaskargoswamy_1 kindly explain what is "second_highest = sorted(list(set([marks for name, marks in marksheet])))[1]"[The Bolted and Italic Section] is used for. As I am new to coding, i can't find what's the role of this expression... Thanks in Advance..!!!
shanishani54321 line 3rd of your code shows error
JJJason second_highest = sorted(list(set([marks for name, marks in marksheet])))[1]
You do not need a list() here, because sorted() will return a sorted list.
nano_yabizniz I am a newbie to this but care you able to explain why there is the need to put it in set?
silvinho485 second_highest = sorted(set(marksheet))[1][1]
Looks cleaner for me.
mgnanaprakash Can u explain these two lines
second_highest = sorted(list(set([marks for name, marks in marksheet])))[1] print('\n'.join([a for a,b in sorted(marksheet) if b == second_highest]))
silvinho485 Of course!
second_highest = sorted(list(set([marks for name, marks in marksheet])))[1]
I just need to know what is the the second lowest mark, if you clean the repeated marks and sort, it always be in the second position ([1]).
print('\n'.join([a for a,b in sorted(marksheet) if b == second_highest]))
Reading this part of code, make me see, i must use more specific var names :P . Basicaly you pass an temporary var in [0] and [1] of marksheet it is sorted. And compare if the b value at is equal do second lowest mark, if yes, return and put inside of the "a" in the begining.
PS: Sorry my english
NikhilReddy533 Using [1] doesn't pass all the testcases. For example if the first 2 values in the sorted list can be same, so the second lowest grade can be at index [2].
rishilja [deleted]Suyashpawar24 can you explain second last line in code please i am not getting it.
tkosa You can omit the list since sorted returns a list.
janakiram_katto1 wouldn't the above code sorted() does a sort as per name and not on marks!!!??? rather i think, we can try this: second=sorted([a[k][1] for k in range(len(a))])[1] correct me,if i'm wrong
leslie5 Nice solution! Super late note here, but I don't think you need the list constructor, as sorted() will take any iterable and return a list.
suryabteja Can you please explain me how second_highest is obtained?
nil_mohanty can you explain why did you use marks for name,marks in marksheet? i am new to python so couldnt understand it.
MaxNRG [deleted]MaxNRG sorted(list(set([marks for name, marks in marksheet])))[1]
list() is not necessary, because sorted() function returns list object:
sorted(set([marks for name, marks in marksheet]))[1]
MaxNRG optimized version imho:
grades = sorted([[input(), float(input())] for i in range(int(input()))]) print(*[x for x, y in grades if y == sorted(set([score for name, score in grades]))[1]], sep='\n')
earic20pal Doesn't work for me with this input 3 amit 50.55 brij 40.40 charles 40.40
dpkdeepakpandey second_highest = sorted(list(set([marks for name, marks in marksheet])))[1] print('\n'.join([a for a,b in sorted(marksheet) if b == second_highest]))
I am finding it difficult to understand.Can you please help me in understanding the code. Thanks in advance
nsd13 Hi! i am so noob, can somebody explain me this line? second_highest = sorted(list(set([marks for name, marks in marksheet])))[1]
- Why we must change a type so many times?
- [marks for name, marks in marksheet] - this is called list comprehension? Is it every time in "[ ]" inside expression?
- also, why we need "set" before list?
codeharrier using
set()removes duplicates; usinglist()turns it back into a listbut
sorted()also returns a list even if it's passed a set, so the use oflistis not necessarythe list comprehension is taking
marksheet, which has 2-element lists in it, and just getting one of the elements; it's turning[[name1,mark1],[name2,mark2],...]into[mark1,mark2,...].also, the casting and sorting can be replaced by calling
max:second_highest = max([marks for name, marks in marksheet])
nsd13 thank you! it's clear now
mca2k6_jsr can you please explain below line:
second_highest = sorted(list(set([marks for name, marks in marksheet])))[1]
umapathy Good Stuff., I reached till second_highest mark but i couldn't find the duplicates and get the name of the respective duplicate marks. The solution provided teased my brain.
imilchakov if name == 'main': long = [] numb = [] names = [] for _ in range(int(input())): name = input() score = float(input()) long.append([name, score]) numb.append(score)
numb1 = list(set(numb)) numb1.sort() for el in long: if el[1] == numb1[1]: names.append(el[0]) names.sort() for _ in names: print(_)amithdsilva007 can you explain why [1] is used
nguyenson3822 Can you explain these, : second_highest = sorted(list(set([marks for name, marks in marksheet])))[1] print('\n'.join([a for a,b in sorted(marksheet) if b == second_highest])) Thank you so much!
praveenpdhr [marks for name, marks in marksheet] I do not understand here.. what really happens?
[i for j,j in marksheet] how does all marks get extracted?
ozpy1 I don't think you need the list call. sorted(set([...])) seems to do it.
Limckos Thank you, it works.
vaibhavlabs Hey gkeswani92,
Could you explain to me what you did in the last two lines?
second_highest = sorted(list(set([marks for name, marks in marksheet])))[1] print('\n'.join([a for a,b in sorted(marksheet) if b == second_highest]))
patricklyngrutz Mostly the same approach but chose tuples and explicitly sorting by name.
if name == 'main':
students = [] for _ in range(int(input())): students.append((input(), float(input()))) # Second highest score s2 = sorted(list(set([x[1] for x in students])))[1] # Sort by name students.sort(key = lambda x: x[0]) [print(s) for s,x in students if x == s2]sunfq603 [deleted]vishalbhardwaj51 Can you please explain what you are doing in the last two lines.
P.S: I'm proper newbie to python.
pratik_nalage Can you explain this line
set([marks for name, marks in marksheet])
tusharchatur01 can anyone explain the last print statement
ankush25kumar heyy can someone make it more comprehensible?? i cant understand... can u solve this same code without list comprehension . thanks in advance
janu_s Can u please explain that? :)
f20170605 can you explain this code
itsbruce The guy hasn't visited the site for over two years now.
f20170605 can you explain if you know that method
itsbruce He has a for loop building the nested list called marksheet.
Then "marks for name, marks in marksheet" is a list comprehension. It's extracting the second item - the scores - from marksheet into a new list. He then turns the list into a set (which removes the duplicates), turns it back into a list and then sorts the list. The second item in the list is the second lowest score.
"a for a,b in sorted(marksheet) if b == second_highest" is another list comprehension. He sorts the marksheet (the list he built at the beginning, containing both names and scores); because the name is the first in each pair, this is basically putting the list in alphabetical order He's taking the first item in each pair - the name - if the second item - the score - matches the second lowest.
itsbruce Two simple ways to improve this without abandoning the basic idea.
- Create the set of marks at the same time as creating the marksheet. Just put "marks = set()" at the beginning, then in the for loop add the score to that set as well as to the marksheet. Then there's no need to traverse marksheet to extract the marks and put them in a set - they're already there. ** "second_highest = sorted(scores)[1]"**
- Filter marksheet to extract only the names of students who have the right score and then sort them. This answer does it the wrong way round; why bother sorting names you're going to throw away?
basithnitk second lowest ...not highest btw there could be more than one entry with the lowest and this code cannot handle such a case
itsbruce You're wrong on the second point. He puts the scores into a set before sorting them, so there's only unique values and the second in the sorted output will always be the second lowest..
balasaheb_kiswe excellent coding, simple, readable and easy to understand though efficient.
itsbruce It's quite inefficient.
- It traverses the entire data set twice, the second time to extract the scores (which it already saw in the first traversal, so the second traversal is unnecessary).
- Turning the list of scores into a set and then sorting the result and extracting the second one is significantly less efficient than simply traversing the list once and keeping the lowest two numbers you see, then discarding the lowest once you've reached the end of the list.
- Sorting the entire list of students (despite the fact that most of them probably didn't have the second lowest score and won't be printed out) and then excluding the students with the wrong score is less efficient than excluding the wrong students and sorting only the right students.
- Concatenating all the names and printing the result as one big string is less efficient (both in terms of time taken and memory used) than simply iterating over the names and printing each one in turn.
It is simple but an efficient solution can be just as simple and at least as readable.
thirumalesuthiru second_highest = sorted(list(set([marks for name, marks in marksheet])))[1] can any one explain this line
michael_lowjial1 can u explain the term [marks for name, marks in marksheet]?
fragboy4123 hey ,i am new to python so i did not understand what you did here please explain print('\n'.join([a for a,b in sorted(marksheet) if b == second_highest]))
sibilhacker your code won't work if we have two highest score, since you have considered second element of the sorted list as second highest. some thing like [1,1,1,3,4,5] as of scores. but anyway neat and high class code u have. :D <3
itsbruce You're wrong about that. He turns the list of scores into a set, then turns it back into a list. Turning it into a set removes the duplicates. Turning the set back into a list doesn't restore the duplicates.
sibilhacker You are right! I didn't know the set function do the work of duplicate. so if duplicates removed, what happens to the ones that we want as second worst list of class. it could be more than one. tnx for your reply any way! <3
pavanpurohit95 u r such a beauty man i just love the way you wrote code
shivani_k449 can you please explain your code as im new to python
dikshaagrawal564 can you explain 4th line??
gautiharsh55 set([marks for name, marks in marksheet]) will you please explain me , how this line is working.
aalam0774 list inside set how it work definetly you got error because set is mutable and list is also mutabal that way set can not take list
aalam0774 [deleted]aalam0774 [deleted]zhengyaojun1204 neat and nice!!!!Love it!!!!!
misterudupa Very nice method!
shrinivas_iyeng1 I have a few doubts. Why did you assume that the first element of the list would be the second highest. If two people would have the same lowest scores, this would fail to work. Second, why did you create a list, convert it to a set and then back to list again. You could have directly created a list and then sort that.
itsbruce Turning the list into a set removes all the duplicate elements. Sorting the result puts the lowest element at index 0, the second lowest at index 1.
That's why he converted to a set and back.
The second conversion isn't necessary, in fact, as the sorted function returns a sequence. But the idea is sound and this code will pass the tests.
You could copy and paste the code, try it out with an input where there are two lowest scores - you'll find it works.
sp427661 bro,append() just takes 1 input then how can we take both in marksheet.append([input(), float(input())])
shubham_joshi151 i didnt understand "second_highest = sorted(list(set([marks for name, marks in marksheet])))[1] print('\n'.join([a for a,b in sorted(marksheet) if b == second_highest]))"
this part of code, please can anyone help me to get this
simrand0508 can you explain this: second_highest = sorted(list(set([marks for name, marks in marksheet])))[1]
aalam0774 this is nothing it is list comprehension
here we use set for the removing the dublicate no then again set converted into listafter sorting list we appply list indexing
sbasudha marks for name,marks in marksheet
I am confused because we have not defined 'name' and 'marks'. Can you please explain.
aalam0774 just like [b for a,b in range (marks)] in nested list there is two element first one is name and second one is marks so here we find the second_hig marks so here we write b for marks and a,for name i hope now u undestand
sbasudha ohh..I got it..Thank you very much
generalom1234 no problem my fellow programmer
itsbruce You need to read the Python documentation on comprehensions or a tutorial on them.
When you type something like
[2 * x for x in numbersList]
The "for x" is defining the variable x. This is also what "for" does in for loops - why does that not confuse you but this does?
simrand0508 Can you pleae explain this: [marks for name, marks in marksheet]
maunose [deleted]sashijeevan This isnt scalable although it might be efficient. What if you second lowest or highest is at a 5th position in the list? Your code fails in that case.
Always remember the code is only efficient if it is scalable and ofcourse smaller the better.
itsbruce You're wrong. His code works. First he extracts the scores and puts them in a set, so that any duplicates are removed. Then he sorts the unique scores. Then he takes the second item from the sorted scores. This can only be the second-lowest score, it' can't possibly be anything else.
The code is inefficient, but it covers all the possible inputs (as long as the quarantee about there being a second-lowest score is kept - and Hackerrank tests always keep their promises about the input data).
SHEERSENDU_GHOSH can u pls explain the last 2 lines of code? i couldn't get y did u use that [1] in second_highest. Thank u.
pbhauri Can anyone explain me this part set([marks for name, marks in marksheet])
mahdi420 The simplest and accurate solution:
Whith explaination of each line of code:
@python 2.7
d={} #1 for _ in range(int(raw_input())): #2 Name=raw_input() #3 Grade=float(raw_input()) #4 d[Name]=Grade #5 v=d.values()#6 second=sorted(list(set(v)))[1] #7 second_lowest=[] #8 for key,value in d.items(): #9 if value==second: #10 second_lowest.append(key) #11 second_lowest.sort() #12 for name in second_lowest: #13 print name #14@1) Empty dictionary.
@2) Range for number of students.
@3) Accepting name of the student.
@4) Accepting the grade of the student.
@5) Assigning name as key and grade as value for the dictionary.
@6) Obtaining the values of dictionary(all grades of students.
@7) Remoing duplicte grades by using set data type , changing it to list, sorting in ascending order and taking the second lowest grade.
@8) Declaring an empty list for storing the name of the students who got the second lowest grade.
@9) Going through the name and grade of each students by using items() method of dictionary.
@10) Checking whether the grade is equal to the second lowest grade.
@11) If yes , append it to the second_lowest list.
@12) bSorting the name of students in asceding order
@13) Going through the name of each students who got the second lowes grade.
@14) Printing each name of students in seperate line.
Moazam awsome,really helpful ..thanks
murtaza_98 thks a lot
palpad2002 awesome
prabhjitsingh_t1 Amazing..Thanks
RAM_LODHI nice....
Pulgul Thanks a lot....helped a lot
saifulcseng what's the point using dictionary. This problem is for nested list.
gartman_dale I use dict to store score: [names] pairs, get the appropriate names list using the second key in the sorted keys, then print the unpacked sorted result using sep='\n'.
This takes care of the duplicate grades because the grades are the key.
d = {} #dict for _ in range(int(input())): name = input() score = float(input()) # if the score has names associated, get the list, otherwise empty list score_names = d.get(score, []) # append to list and store list back in dict (just in case it's new) score_names.append(name) d[score] = score_names # get the second of the keys when sorted ascending, use it for the dict to get names, sort + unpack to print print(*sorted(d[sorted(d.keys())[1]]), sep='\n')
chefyunfei Perfect explanation! Feels a little complicated since it uses a dictionary rather than a nested list, but this works neatly!!
vjraghu019 Very nice explaination ! thanks a lot
raj_thakur Thanks for the pretty clear explanation. Cheers !!!!
scintilla My shortest version, while still keeping things readable
n = int(raw_input()) marks = [[raw_input(), float(raw_input())] for i in xrange(n)] scores = sorted({s[1] for s in marks}) result = sorted(s[0] for s in marks if s[1] == scores[1]) print '\n'.join(result)ganuraag31 Suppose there two lowest marks. I mean if marks set is (20,20,24,26) Then second lowest will be 24 but it will check only for 20 which is lowest. Could u please explain
ganuraag31 minv = min(n for n in scores if n!=scores[0])
result = sorted(s[0] for s in marks if s[1] == minv)
ZNevzz Good one ! Readable Too :)
metoinside accidently, I removed the curly brackets from the scores and it failed. May you please describe what makes the difference of: scores = sorted({s[1] for s in marks}) and scores = sorted(s[1] for s in marks)
harshit890 [deleted]
dejzhang Forgive my limited knowledge with Python.. but why do we have to use curly braces, I mean why set it to be a set? When i used list, it just failed on some cases. Please enlighten me! Thanks!
dejzhang I think I found some explanation. Set prevents duplicates...so s[1] is truly the second lowest in case there are people scoring the same lowest
whitewolf09 dude your code doesn't work if there are more than 1 entry for minima... try eliminating minima entries from whole list lmark=[x for x in lmark if x !=y]
wufun This code cannot solve the problem which has two highest marks.
nandhakumarm Kindly @scintilla explain these two line
scores = sorted({s[1] for s in marks})
result = sorted(s[0] for s in marks if s[1] == scores[1])
fuzzeh I used your code within my own understanding of the challenge, I tried to explain this to help myself understand, hopefully this can help others can understand too (and if anyone can let me know if i need to improve any of it!)
I break down the descriptions of what each statement does ABOVE each statement.
studentList= [] for i in range(int(input())): name = input() score = float(input()) studentList.append([name, score]) # for each nest in studentList # sort via the score in ascending # with {} within sorted[] it creates a "set" # output is [37.2, 37.21, 39, 41] - sets remove additonal same numbers scoreSort = sorted({s[1] for s in studentList}) # for each nest in studentList (s) # return a sorted list (sorted by their names (s[0])) of students whos score (s[1]) is equal to s[1] in scoreSort # nameSort = ["Berry", "Harry"] nameSort = sorted(s[0] for s in studentList if s[1] == scoreSort[1]) #.join(nameSort) takes each element in the nameSort list # so ["Berry", "Harry"] # "\n" seperates using a newline BETWEEN each element # output is "Berry\nHarry" print('\n'.join(nameSort))
thekitchensink List comprehensions are so great!
if __name__ == '__main__': a= [] for _ in range(int(input())): name = input() score = float(input()) a.append([score, name]) a.sort() b = [i for i in a if i[0] != a[0][0]] c = [j for j in b if j[0] == b[0][0]] c.sort(key=lambda x: x[1]) for i in range(len(c)): print(c[i][1])
Explanation
list a is simply a list of all the input data. It is sorted so that the lowest score is in the first index.
list b uses a list comprehension to remove all instances of the lowest score from list a. The lowest remaining score in the resulting list (found in the first index since it will be the first value taken from the already-sorted list a) is the second-lowest score overall.
list c then uses another list comprehension to trim down list b to only those scores that are equal to that second-lowest overall score.
list c is now a nested list containing the scores and names of the students that will make up the output. It is sorted alphabetically according to student name by using a lambda expression which indicates to use the contents of the second index of each object to be sorted ('name' in [score, name]) as the key for sorting.
Finally the names from list c are printed to satisfy the problem.
dsrg1 Very nice code and explanation. Good pointers for newbies
_harsha c.sort(key=lambda x: x[1])
Kindly explain this line
thekitchensink The sort() built-in function is being
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