Count frequency of words in a list and sort by frequency

use this

from collections import Counter
list1=['apple','egg','apple','banana','egg','apple']
counts = Counter(list1)
print(counts)
# Counter({'apple': 3, 'egg': 2, 'banana': 1})

You can use

from collections import Counter

It supports Python 2.7，read more information here

1.

>>>c = Counter('abracadabra')
>>>c.most_common(3)
[('a', 5), ('r', 2), ('b', 2)]

use dict

>>>d={1:'one', 2:'one', 3:'two'}
>>>c = Counter(d.values())
[('one', 2), ('two', 1)]

But, You have to read the file first, and converted to dict.

2. it's the python docs example,use re and Counter

# Find the ten most common words in Hamlet
>>> import re
>>> words = re.findall(r'\w+', open('hamlet.txt').read().lower())
>>> Counter(words).most_common(10)
[('the', 1143), ('and', 966), ('to', 762), ('of', 669), ('i', 631),
 ('you', 554),  ('a', 546), ('my', 514), ('hamlet', 471), ('in', 451)]

words = file("test.txt", "r").read().split() #read the words into a list.
uniqWords = sorted(set(words)) #remove duplicate words and sort
for word in uniqWords:
    print words.count(word), word

Pandas answer:

import pandas as pd
original_list = ["the", "car", "is", "red", "red", "red", "yes", "it", "is", "is", "is"]
pd.Series(original_list).value_counts()

If you wanted it in ascending order instead, it is as simple as:

pd.Series(original_list).value_counts().sort_values(ascending=True)

Yet another solution with another algorithm without using collections:

def countWords(A):
   dic={}
   for x in A:
       if not x in  dic:        #Python 2.7: if not dic.has_key(x):
          dic[x] = A.count(x)
   return dic

dic = countWords(['apple','egg','apple','banana','egg','apple'])
sorted_items=sorted(dic.items())   # if you want it sorted