How to get the most common element from a list in python

Question

I'm having a list as shown here. a=[1936,2401,2916,4761,9216,9216,9604,9801] I want to get the value which have more duplicates. In here it is '9216' how can i get this value? Thanks

Answer 1

You can use collections.Counter for this:

from collections import Counter

a = [1936, 2401, 2916, 4761, 9216, 9216, 9604, 9801] 

c = Counter(a)

print(c.most_common(1)) # the one most common element... 2 would mean the 2 most common
[(9216, 2)] # a set containing the element, and it's count in 'a'

From the docs:

Answer 2

https://docs.python.org/2.7/library/collections.html#counter

from collections import Counter Counter(a).most_common(1)

Answer 3

Here is another one not using counter

a=[1936,2401,2916,4761,9216,9216,9604,9801]
frequency = {}
for element in a:
    frequency[element] = frequency.get(element, 0) + 1
# create a list of keys and sort the list
# all words are lower case already
keyList = frequency.keys()
keyList.sort()
print "Frequency of each word in the word list (sorted):"
for keyElement in keyList:
    print "%-10s %d" % (keyElement, frequency[keyElement])

Answer 4

There's two standard library ways to do this:

statistics.mode :

from statistics import mode
most_common = mode([3, 2, 2, 2, 1])  # 2
most_common = mode([3, 2])  # StatisticsError: no unique mode

collections.Counter.most_common :

from collections import Counter
most_common, count = Counter([3, 2, 2, 2, 1]).most_common(1)[0]  # 2, 3
most_common, count = Counter([3, 2]).most_common(1)[0]  # 3, 1

Both are identical in terms of performance, but the first raises an exception when there is no unique most common element and the second returns the frequency as well.