比较python中两个列表中的单词

Question

I would appreciate someone's help on this probably simple matter: I have a long list of words in the form ['word', 'another', 'word', 'and', 'yet', 'another'] . I want to compare these words to a list that I specify, thus looking for target words whether they are contained in the first list or not.

I would like to output which of my "search" words are contained in the first list and how many times they appear. I tried something like list(set(a).intersection(set(b))) - but it splits up the words and compares letters instead.

How can I write in a list of words to compare with the existing long list? And how can I output co-occurences and their frequencies? Thank you so much for your time and help.

Answer 1

>>> lst = ['word', 'another', 'word', 'and', 'yet', 'another']
>>> search = ['word', 'and', 'but']
>>> [(w, lst.count(w)) for w in set(lst) if w in search]
[('and', 1), ('word', 2)]

This code basically iterates through the unique elements of lst , and if the element is in the search list, it adds the word, along with the number of occurences, to the resulting list.

Answer 2

Preprocess your list of words with a Counter :

from collections import Counter
a = ['word', 'another', 'word', 'and', 'yet', 'another']
c = Counter(a)
# c == Counter({'word': 2, 'another': 2, 'and': 1, 'yet': 1})

Now you can iterate over your new list of words and check whether they are contained within this Counter-dictionary and the value gives you their number of appearance in the original list:

words = ['word', 'no', 'another']

for w in words:
    print w, c.get(w, 0)

which prints:

word 2
no 0
another 2

or output it in a list:

[(w, c.get(w, 0)) for w in words]
# returns [('word', 2), ('no', 0), ('another', 2)]