Python-计算列表中列表的出现次数（集合，计数器）

Question

I imported collections

import collections

I use it to count the occurence of certain strings in a list

counts = collections.Counter(list)
counts = counts.most_common()

This works perfectly fine. Now my needs changed and I have a nested list and I want to count the occurence of lists in that list:

lists = [['word1', 'word2'], ['word4', 'word6'], ['word1', 'word2']]

I get "lists" from a list of single words:

list = ['word1', 'word2', ...]

The result should look like this

[(('word1', 'word2'), 2), (('word4', 'word6'), 1)]

I hope it is clear what I want.

Answer 1

collection.Counter() works perfectly fine. Below is the example you mentioned:

>>> from collections import Counter
>>> my_list = [('word1', 'word2'), ('word4', 'word6'), ('word1', 'word2')]

>>> Counter(my_list)
Counter({('word1', 'word2'): 2, ('word4', 'word6'): 1})  # dict object

>>> Counter(my_list).most_common()  
[(('word1', 'word2'), 2), (('word4', 'word6'), 1)] 

Answer 2

Use hashing using dict() . This is a simple code illustrating it:

lists = [('word1', 'word2'), ('word4', 'word6'), ('word1', 'word2')]

hashmap = dict()
for item in lists:
    if item in hashmap:
        hashmap[item] += 1
    else:
        hashmap[item] = 1


new_list = []

for key, value in hashmap.iteritems():
    new_list.append(((key), value))

print new_list  # output: [(('word1', 'word2'), 2), (('word4', 'word6'), 1)]

Answer 3

Okay I just manage to answer the question on my own.
The problem was that it was a list of lists.
I needed to have a list of tupples to get Counter working.

I managed to do that while collecting the words from the list of words:

... tuple(words[i:i+2])