On Wed, Mar 8, 2017 at 1:28 PM, Sayth Renshaw <flebber.c...@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 -- https://mail.python.org/mailman/listinfo/python-list
