On 2017-03-08 02:37, Chris Angelico wrote:
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 :)
Using Counter seems like a simpler way to me. I wouldn't bother about
other ways unless that way wasn't "good enough" for some reason.
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