Better way to do this dict comprehesion

Chris Angelico rosuav at gmail.com
Tue Mar 7 21:37:23 EST 2017


On Wed, Mar 8, 2017 at 1:28 PM, Sayth Renshaw <flebber.crue at gmail.com> wrote:
> I have got this dictionary comprehension and it works but how can I do it better?
>
> from collections import Counter
>
> def find_it(seq):
>      counts = dict(Counter(seq))
>      a = [(k, v) for k,v in counts.items() if v % 3 == 0]
>      return a[0][0]
>
> test_seq = [20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5]
>
> so this returns 5 which is great and the point of the problem I was doing.
>
> Can also do it like this
> def find_it(seq):
>      counts = dict(Counter(seq))
>      a = [(k) for k,v in counts.items() if v % 3 == 0]
>      return a[0]
>
> But the given problem states there will always only be one number appearing an odd number of times given that is there a neater way to get the answer?

Take a step back for a moment. Are you trying to find something that
appears an odd number of times, or a number of times that counts by
three? First figure that out, THEN see if there's a better way to do
what you're doing.

To find an unpaired number in linear time with minimal space, try
stepping through the list and either adding to a set or removing from
it. At the end, your set should contain exactly one element. I'll let
you write the actual code :)

ChrisA


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