Print Function
SemperPeritus Solution in a single line (Python 3):
print(*range(1, int(input())+1), sep='')dark_passengerMO Can you please explain what the '*' operator is for?
Sammieo I would also like to know this.
Rusher Look for "*identifier" under https://docs.python.org/3/reference/expressions.html#calls
sanjna_srivatsa I think it means "all". That's how we know it in SQL
daturaman [deleted]
SemperPeritus For short operator '*' allows you to pass an array. Example:
user = [name, password, captcha]log_in(*user)sopan94 please can u explain in detail..??
JBallin did you see my explanation below? give it an upvote if helpful. @RahulDamineni took a crack at it too. What about it do you not understand?
--
In this example given by @SemperPeritus, he's providing a function
log_inthat takesname, password, captchaas arguments.In general, you pass in arguments like so:
function(arg1,arg2,arg3)
But, if you put them in a list such as:
function([arg1,arg2,arg3])
then it's like you're only passing in 1 argument (the list) and you'll get an error saying
functionwas expecting 3 arguments and only got 1.On the other hand if you were to include
*it would "unpack" the list (imagine the list is a suitcase and the elements are pieces of clothing you want to put on the bed to look at individually. you can't do that without unpacking the suitcase, right?) so that it would still work:function(*[arg1,arg2,arg3])
becomes
function(arg1,arg2,arg3)
So
*userchanges from a list:user = [name, password, captcha]
to arguments:
*user = name, password, captcha
So now
log_inwill work without errors because it was passed 3 arguments instead of 1!
Want to learn more about a related topic in Python called
zip? Check out my blog post: medium.com/@PrintSupWorld/unzipping-python-2-340701d300a6yidingyang1027 Great interpretation!
satyakibose98 It was helpful, but can you please explain this for print(*range(1, int(input())+1), sep='')?
PS1985 n = int(input()) *range (1, n+1) = range is from 1 to n, exclusive of n. so if your input is 3 then it prints 1 2. So input to the range function is 1 to n+1
if you don't give sep='' in print statement then each value is seperated by a space
sksaravananp Thank you for the explanation. But why we need to use * range() instead of range()
vishnu_b All non-keyword arguments are converted to strings like str() does and written to the stream, separated by sep and followed by end. From printfunction doc.
So if we just give range(1, n) python3 will just print out range(1, n) since str(range(1, n)) is range(1, n).
Also read @JBallin's explanation below.
thomasokonkwo91 wow... Nice one. Thanks man!
thomasokonkwo91 wow... Nice one. Thanks man!
smeshram258 thanks it was very much helpful
researchers902 Superb explanation. So this will work for lists only or dictionary too ?
vishnu_b Python will unpack just the keys.
albaqawi fantastic explaination... for any new comer that needs more clarity make sure you know that you need to use n instead of int(input())
like this print(*range(1,n+1),sep='')
now as Explained by JBallen the range now is 1 to n+1 arrguments passed to the print function and the seperator is no space
Happy coding! :)
0strnz0 sexy AF..
s0rav Now this is the thing! Deserves an upvote!
greggouveia Awesome explanation @JBallin! Thanks for sharing it on such a good willing way. I'll be following youo on medium =) Keep up the good work
nksinghal41 great explanation
vdeverdu That's more or less like the spread operator '...' in javascript, right?
jaswanthsai121 noo
willyhhsu thank you. Please, accept this gif as a thanks
trinadhkoya it removes the end characters like [,{,],}
JBallin I've never come across this explanation and I'm pretty sure it's overly simplistic and incorrect. Please provide references if there is any truth to this. Thanks!
RahulDamineni It is called unpacking operator.
Meaning, if you have a function which takes multiple input arguments, say func(a,b,c); instead of passing them seperately, you can pack them in a list, args = [a,b,c]; and pass this list to your function, func(*args)
And How does your function know that the received argument (args) is a packed list? It'll when you supply unpack argument '*'
Note, operator '**', when used on a dictionary, returns its keys.
Source: https://docs.python.org/2/tutorial/controlflow.html#unpacking-argument-lists
JBallin *ExplanationPython's Print Function defines the input as
*objects, not asiterablelike it does for Map.An example of an iterable is a list because we can keep stepping through the elements one at a time.
printinstead needs multiple arguments for eachobjectyou want to print.*is a way to "unpack" a list so that it becomes arguments instead of a list.f(*[1,2,3])converts tof(1,2,3).Example using
print(from future in py2):print([1,2,3]) #[1, 2, 3] print(*[1,2,3]) #1 2 3 print(1,2,3) #1 2 3
The last two exhibit the desired result, you just need to set
sep=''to get123(w/o spaces).digitalchemy Nice explanation! Just a little remark: Saying that *[1,2,3] converts to (1,2,3) is a bit misleading, becouse it suggests that it converts the list to a tuple, which for print() function beahves the same as the list.
JBallin @digitalchemy great point! I clarified that it applies to arguments inside a function call (see "f()")
sri_sam229 it's for optional varied length arguments to be recieved
g_jaswanth unpacking list i mean it converts [1,2,3] to 1,2,3 that sep = '' replaces , with blank
7dec1997 Its a multiplication
vishalit10 print (vishal*5) it will print vishal five times.
rajatjog1294 it containts the series of values:
Ex:
a = [1,2,3,4,5,6,7,8,9] x, *y = a print(x) print(y)
then x will be 1 and y will be 2,3, ...,9
lalit_chandora this is known as the "multiplication operator".
alaniane The context determines what type of operator it is. Many languages overload symbols to mean more than just one operator. For example in C:
int *ptr = &y; *ptr = 5; z = y * x;
* in the first instance declares a pointer variable
* in the second instance is called a dereference operator
* in the third instance is a multiplication operator.So, from what I can tell about the comments above in Python: What type of operator '*' is depends on its context.
*[1,2,3] the * is not the same as 2 * 3 where it is the multiplication operator.
rosiecakes cool, never heard of the * identifier before, thanks.
local_bantai What would be an alternative to this in python 2>
surajdeshpande print(*range(1, int(raw_input())+1), sep='')
mubarakbsc123 Thanks a lot.
JBallin print(*range(1,input()+1),sep='')JBallin if this isn't working it's because you forgot to import print_function from future:
from __future__ import print_function print(*range(1,input()+1),sep='')
mikee531 what is sep=??
JoJo_wen it is abbreviations of "separator"
JoJo_wen I think "xrange" is better than "range",there are some difference between them.
hikinthru The question was about Python 2. Importing future essentially sets the program to Python 3. Which means it doesn't actually answer the question.
JBallin they instruct us to import from future for python 2
yuanster From https://docs.python.org/2.7/library/functions.html?highlight=print#print"
"Note This function is not normally available as a built-in since the name print is recognized as the print statement. To disable the statement and use the print() function, use this future statement at the top of your module:
from __future__ import print_function
"New in version 2.6"
(Interestingly, this is now inserted in the template when you select Python 2.) Have to say that I'm a bit surprised that Python didn't offer something like sprintf.
christokleu Thanks, better than my solution
print (''.join([str(i) for i in range(1,int(input())+1)]))
Reading through comments now, this fails the following criteria:
Without using any string methods
hacker_gf [deleted]
SANJAY5797 but i didn't get it worked in python 2 version
JBallin you have to import the
print_functionandinputvs.raw_inputworks differently in py2from __future__ import print_function print(*range(1,input()+1),sep='')
Sani1997 why the * for? new to python.. so.asked..
omavinashkalal - Explanation Python's Print Function defines the input as *objects, not as iterable like it does for Map. An example of an iterable is a list because we can keep stepping through the elements one at a time. print instead needs multiple arguments for each object you want to print. * is a way to "unpack" a list so that it becomes arguments instead of a list. *[1,2,3] converts to (1,2,3). Example using print (from future in py2): print([1,2,3])
[1, 2, 3]
print(*[1,2,3])
1 2 3
print(1,2,3)
1 2 3
The last two exhibit the desired result, you just need to set sep='' to get 123 (w/o spaces). Source: JBallin
SANJAY5797 but i didn't get it worked in python 2 version
swapneil_basutk1 I just did and this worked
from __future__ import print_function print(1,2,3,sep='')
SANJAY5797 but i didn't get it worked in python 2 version
danielbarreto557 I ran this and it worked. My question is how does adding the asterisk to "range" make it work. im supposing that range(arg) is returned as a list so adding the asterisk unpacks the list correct?
JBallin That's correct, it unpacks the list so that they are multiple arguments (which
print_functionfromfuturein py2 (or justprintin py3) accepts) to print the range in succession instead of just printing alist.I explained it in depth, above, in a couple different comments.
vikas_kr or you can do like this
[print(i, end='') for i in range(1,int(input())+1)]
SANJAY5797 thanq
john_hollwitz Thanks any recommendations on how to do this in Python 2?
harshit_thakore doesnt get sep='' it reduce it by storing it?
ajaymania12 what is the meaning of "sep''" which is last argument of your code ?
masquerade0097 expalining in simple terms sep='' insert 0 space between the list items output.For example, sep='' output = 123 sep=' ' output = 1 2 3 sep='1 ' output = 11 21 31
that is sep will insert (whatever you enter in it) between the succesive elemnts of the list passed as first argument
STMohd getting error in *range ,showing it as invalid syntax
hijamesgill One line, but less clear. I'd rather read something simple and easy to scan, like:
for x in range(1,n+1): print(x,end='')
kumarparth123 not working
kumarparth123 please help
jaffarjawed thanks bro, it works in python3
shaikotrio This one seems more clear to understant.
abhinavbahl51 what does end do here?
madhu4mobile Is there any similar way in python 2 too? For sure * operator is not working in python 2.
satyakibose98 it is showing error: Sorry: IndentationError: expected an indented block (solution.py, line 2)
lrodlima this for me looks like more the packing and unpacking way to receive parameters
https://hangar.runway7.net/python/packing-unpacking-arguments
Norse I get the following error, when trying this example: Traceback (most recent call last): File "solution.py", line 5, in print(*range(1, int(input())+1), sep='') EOFError: EOF when reading a line
vaisakhsrinivas This is what worked for me.
for i in range (1,n+1): print(i, end='')
ic3balandran I never understand why people are obsessed with one liners or short answers. Beginners should strive to write understanble code, that makes sense in your mind. I'll leave link here List Comprehensions but a solution with a for loop is only two lines:
for i in range(1, n + 1): print(i, end="")
septimo I totally agree with you, your answer is understandable and also short for beginners.
www_shreyanshda1 what is sep
dimithras Python print function sep is a parameter for separator
dimithras Both python version using generator
print(*[n+1 for n in range(input())], sep='')
sai4ever the output was a list in this case.can you explain me why? and moreover in the question the output should not be a list. thank you.
bad_coder_lol if name == 'main': n = int(input()) for i in range(n): print(i + 1, end='')
#this might be the easiest way to solve using python 3shampa1 The solution is following: n=int(input()) print(*range(1, n+1), sep='')
Here, if we don't use *range & sep='' then the answar will be respectively range(1, 4) & 1 2 3.*range function which takes multiple input arguments & sep='' function call makes sure that there is no space between the values.
emrepallamut Tried this:
for a in range (1,n+1) print(str(a))
obviously i got wrong because it printed out vertical.
then this:
print(*range(1, n+1), sep='')
Much easier to understand.
mstevenbowen Does this method print a string as opposed to an integer?
chiragckarkera For input =3,
print(range(1, int(input())+1), sep='')
O/p
range(1,3)
If we put *
print(*range(1, 3), sep='')
O/p
123
ss_sariana hey dude! what function is used in python 2 for sep='' or end=''
mix123 My answer, I prefer clear and simple logic ;-)
from __future__ import print_function N = int(raw_input()) for i in range(1, N+1): print(i, end="")a_yashrocky But the challenge is to write it in one line
hijamesgill Nope. The challenge is to print the output on one line.
mehulj5775 hi, can u tell me why have you used end="" in your code. thank you
gosar95 Without 'end="" ', every number would be on next line. for example:
input:4
Output:
1
2
3
4
Expected Output: 1234
mehulj5775 thank u so much :)
swasti29 Thanks! I was missing the 'end=''' function, and this helped.
JBallin try
N = input()
Supreme_One for i in range(1, N+1)
how does this line of code work? why does it not go past N? if N=3 then it should print 12345.... what am i missing here?
JBallin that's just how
rangeworks. play around with it. It doesn't include the last number.print range(3) #(0,1,2) print range(1,3) #(1,2)
Supreme_One i understand that, but, range(start, stop) so if you have n=3 and range is range(1, n+1) wouldnt that mean that n=1 then n=2 etc, because of the n+1 wouldnt it just keep going past 3 to 4, 5, 6, 7....infinite loop?
nevermind i figured it out writing this post, lol. thanks!
Supreme_One i was thinking of n+=1 which would make n=4, 5, 6 and so on. the range function being used is range(1, n+1) making it range(2, n+1)....and so on. correct me if im wrong.
JBallin rangeproduces alistand it doesn't change once it's produced.I made a mistake above, it
prints out alist([]instead of()) becauserangeis just a way to make alist:print range(3) #[0,1,2] type(range(3)) #<type> 'list' range(3) == [0,1,2] #True
n = 3 print range(1,n+1) #[1,2,3]
Next the code goes through the
list(range(1,4)=[1,2,3]), element by element, and prints each number. It first prints 1, then 2, then 3. It doesn't go to the next line bcend=''and insteadprints them next to each other.Make sense?
JBallin It calculates
rangeonce, and keeps referring back to the samelist.
sagar10451 in for loop that we use i is incremented and not the range. range is fixed it gives order to for loop i to go till itd range is given and i start from lower value to higher in higher value of range. hope this helps!!
MakoShark This is the pythonic way of doing it. Keep it simple and explicit. Nice.
satyakibose98 bt it is giving indentation error.
Bhagya_Gaikwad end="" doesn't work in Python 2. What else can I do?
TytoCorvus Apparently it does if you import the print_function method like so:
from future import print_function
w_merc I think without going to google, and not having previous Python experience, its impossible to do this task. The tutorial is very poor.
mcclintock85 I agree. As a very green python student, I've never encountered 'end= '. I wouldn't even know how to search for it from just this problem.
saurabhmithe How about this approach? It passes all the test cases.
from __future__ import print_function if __name__ == '__main__': n = int(raw_input()) for i in range (1,n+1): print(i, end="")
The comma(,) after i supresses the new line and the end is the delimiter that will be used while printing (can be replaced with space for printing numbers with a single space between)
S_Sambit_Rout it's good, we can do by this way also if name == 'main': n = int(input()) i=1 while i<=n: print(i,end="") i+=1
marinskiy Here is Python solution from my HackerrankPractice repository:
for i in range(1, int(input()) + 1): print(i, sep='')
Feel free to ask if you have any questions :)
Zinngg if name == 'main': n = int(input())
for i in range (0,n): print(i+1,end = '')This is my code. How can i make it look more professional?
dhaval4friends this is neat
it_pladinc You can check my one-line code:
print ("".join([str(num) for num in range(1,n+1)]))
-iterating each number in the range(1,n+1), turning it into a string object and connected using "".join string method.
stancsmith The problem states that our solution must be done "Without using any string methods"
xuliying201000 from future import print_function
for i in range(1, int(raw_input())+1): print (i, sep='', end='')
JBallin you can just use
input()instead ofraw_inputso that you don't have to convert it usingint
dineshsonachalam A Better solution
n = int(input()) for i in range(1,n+1): print(i,end="")
lerdsouza from future import print_function if name == 'main': n = int(raw_input()) for i in range(n + 1): print i
This the code I am trying to run but it throws an invalid syntax error on line 5. The same code when I run on the compiler it works.
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