Hi
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?
Thanks
Sayth
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