CODEWITHSUNDEEP
pythondictionary
Bryan
Bryan

Reputation: 6149

finding and returning names and counts of duplicate values in Python list

I have a list of values that denote the number of times a piece of regex matches in a string. From this, I want to find the numbers that appear more than once, and their count. For example, for [2, 2, 2, 0, 2, 1, 3, 3] I want {2:4,3:2} as output if it's in a dict or [[2,4],[3,2]] if it's in a list of lists. I'm looking for the fastest, most concise way to do this. Right now, I do it via the following code, but think it's way to verbose to be optimal. 

numWinners=[2, 2, 2, 0, 2, 1]
tieCount={x:numWinners.count(x) for x in numWinners}
ties=dict()
for key, value in tieCount.items():
    if value>1:
        ties[key]=value

print ties
{2: 4, 3: 2}

A list or dict output isn't really an issue for me - again, whatever is fastest and concise.

Upvotes: 2

Views: 1512

Answers (3)

meghamind
meghamind

Reputation: 96

Try using collections.defaultdict

import collections
ties = collections.defaultdict(lambda:0)
for num in numWinners:
     ties[num] = ties[num]+1


 for key,value in ties.iteritems():
     print key, value

Upvotes: 0

dansalmo
dansalmo

Reputation: 11694

You can use a dict comprehension to create a histogram:

>>> ns=[2, 2, 2, 0, 2, 1, 3, 3]
>>> {x: ns.count(x) for x in set(ns) if ns.count(x) > 1}
{2: 4, 3: 2}

Upvotes: 2

DSM
DSM

Reputation: 353499

I'd combine collections.Counter with a dictionary comprehension to select the duplicates:

>>> from collections import Counter
>>> numWinners = [2, 2, 2, 0, 2, 1, 3, 3]
>>> counts = Counter(numWinners)
>>> {k: v for k,v in counts.items() if v > 1}
{2: 4, 3: 2}

Upvotes: 5

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