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  1. Prepare
  2. Python
  3. Itertools
  4. Compress the String!
  5. Discussions

Compress the String!

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  • alex_mourousias
     + 26 comments

    Looks pretty clean to me (Python 3)

    from itertools import groupby
print(*[(len(list(c)), int(k)) for k, c in groupby(input())])

  • mike006322
     + 3 comments

    "In this task, we would like for you to appreciate the usefulness of the groupby() function of itertools . To read more about this function, Check this out ."

    This is a very bad introduction to "groupby()" function. This omits virtually every aspect of the function that one would need to understand it.

    At the very least the problem description could include an example.

    I had trouble undetstanding how to write a list comprehension over the output of the groupby() function.

  • [deleted] + 2 comments

    I have absolutely no clue what's going on here. The challenge gives no explanation of what this function does, and the docs hardly help at all. I don't know anything about how to use this function.

  • johnoharaa
     + 3 comments
    from itertools import groupby
for k, g in groupby(input):
    print("({}, {})".format(len(list(g)), k), end=" ")

  • jEeTu_pYthOn
     + 7 comments
    a=input()
s=''
for i in range(0,len(a)):
    if i!=0:
        if a[i]==a[i-1]:  
            continue
    p=0
    for j in range(i,len(a)):
        if a[i]==a[j]:
            p+=1
        else:
            break
    s+='('+str(p)+', '+a[i]+')'+' '
print(s)

    my python3 approach without groupby and passes all the test cases but more lenghty as compare to groupby

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