compare words in two lists in python

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.

Can you post some code you tried. set(['word', 'another']) evaluates to set(['word', 'another']) and does not split up words to letters. — class stacker
– class stacker, Commented Mar 14, 2013 at 10:52

Volatility · Accepted Answer · 2013-03-14 10:33:54Z

7

>>> 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.

answered Mar 14, 2013 at 10:33

Volatility

32.4k11 gold badges84 silver badges90 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

eumiro · Accepted Answer · 2013-03-14 10:37:12Z

4

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)]

edited Mar 14, 2013 at 10:37

answered Mar 14, 2013 at 10:32

eumiro

214k36 gold badges307 silver badges264 bronze badges

1 Comment

Thom_bund Over a year ago

Thank you very much. Both solutions seem to do fine, but my code allows for input of the sort [('S'), ('t'), ('o'), ('c'), ('k')] when i type Stock in sys.argv(2). How am I able to input more words to a comparable list as i execute the program? and with both your suggested solutions, it compares the letters rather then the entire word conll=open(sys.argv[1],'r') targetword=str(sys.argv[2]) vocab=[] c = Counter(vocab) print c for w in targetword: print w, c.get(w, 0) print [(w, vocab.count(w)) for w in set(vocab) if w in targetword] print targetword

Collectives™ on Stack Overflow

compare words in two lists in python

2 Answers 2

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related