How to find most common elements of a list?

user_434180 0
Viewed 164679

Given the following list

['Jellicle', 'Cats', 'are', 'black', 'and', 'white,', 'Jellicle', 'Cats', 
 'are', 'rather', 'small;', 'Jellicle', 'Cats', 'are', 'merry', 'and', 
 'bright,', 'And', 'pleasant', 'to', 'hear', 'when', 'they', 'caterwaul.', 
 'Jellicle', 'Cats', 'have', 'cheerful', 'faces,', 'Jellicle', 'Cats', 
 'have', 'bright', 'black', 'eyes;', 'They', 'like', 'to', 'practise', 
 'their', 'airs', 'and', 'graces', 'And', 'wait', 'for', 'the', 'Jellicle', 
 'Moon', 'to', 'rise.', '']

I am trying to count how many times each word appears and display the top 3.

However I am only looking to find the top three that have the first letter capitalized and ignore all words that do not have the first letter capitalized.

I am sure there is a better way than this, but my idea was to do the following:

  1. put the first word in the list into another list called uniquewords
  2. delete the first word and all its duplicated from the original list
  3. add the new first word into unique words
  4. delete the first word and all its duplicated from original list.
  5. etc...
  6. until the original list is empty....
  7. count how many times each word in uniquewords appears in the original list
  8. find top 3 and print
frequency list python
11 Answers

nltk is convenient for a lot of language processing stuff. It has methods for frequency distribution built in. Something like:

import nltk
fdist = nltk.FreqDist(your_list) # creates a frequency distribution from a list
most_common = fdist.max()    # returns a single element
top_three = fdist.keys()[:3] # returns a list
user_5779380

There's two standard library ways to find the most frequent value in a list:

statistics.mode:

from statistics import mode
most_common = mode([3, 2, 2, 2, 1, 1])  # 2
most_common = mode([3, 2])  # StatisticsError: no unique mode
  • Raises an exception if there's no unique most frequent value
  • Only returns single most frequent value

collections.Counter.most_common:

from collections import Counter
most_common, count = Counter([3, 2, 2, 2, 1, 1]).most_common(1)[0]  # 2, 3
(most_common_1, count_1), (most_common_2, count_2) = Counter([3, 2, 2]).most_common(2)  # (2, 2), (3, 1)
  • Can return multiple most frequent values
  • Returns element count as well

So in the case of the question, the second one would be the right choice. As a side note, both are identical in terms of performance.

user_21323120

I will like to answer this with numpy, great powerful array computation module in python.

Here is code snippet: 

import numpy
a = ['Jellicle', 'Cats', 'are', 'black', 'and', 'white,', 'Jellicle', 'Cats', 
 'are', 'rather', 'small;', 'Jellicle', 'Cats', 'are', 'merry', 'and', 
 'bright,', 'And', 'pleasant', 'to', 'hear', 'when', 'they', 'caterwaul.', 
 'Jellicle', 'Cats', 'have', 'cheerful', 'faces,', 'Jellicle', 'Cats', 
 'have', 'bright', 'black', 'eyes;', 'They', 'like', 'to', 'practise', 
 'their', 'airs', 'and', 'graces', 'And', 'wait', 'for', 'the', 'Jellicle', 
 'Moon', 'to', 'rise.', '']
dict(zip(*numpy.unique(a, return_counts=True)))

Output

{'': 1, 'And': 2, 'Cats': 5, 'Jellicle': 6, 'Moon': 1, 'They': 1, 'airs': 1, 'and': 3, 'are': 3, 'black': 2, 'bright': 1, 'bright,': 1, 'caterwaul.': 1, 'cheerful': 1, 'eyes;': 1, 'faces,': 1, 'for': 1, 'graces': 1, 'have': 2, 'hear': 1, 'like': 1, 'merry': 1, 'pleasant': 1, 'practise': 1, 'rather': 1, 'rise.': 1, 'small;': 1, 'the': 1, 'their': 1, 'they': 1, 'to': 3, 'wait': 1, 'when': 1, 'white,': 1}

Output is in dictionary object in format of (key, value) pairs, where value is count of particular word

This answer is inspire by another answer on stackoverflow, you can view it here

user_86307920
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