Count how many times a word appears in a text file

Question

def paraula(file,wordtofind):

    f = open(file,"r")
    text = f.read()
    f.close()
    count = 0
    for i in text:
        s = i.index(wordtofind)
        count = count + s
    return count
paraula (file,wordtofind)

What have you tried and what exception did you get? There are at least two exceptions you'll hit with the code as-is that have nothing to do with counting words in a string. — CivFan, Oct 02 '15 at 23:06
`for i in text` will return each character one by one, so all other things equal you'll only find single character words that way. Next, assuming that's resolved, `s = i.index(...)` will return the position in the string the word was found, so incrementing by `s` will skew your count considerably. Also, if the word isn't found, [you'll get a `ValueError` exception](https://docs.python.org/2/library/string.html). This is actually to your advantage, because you don't really care about the index, you just want to know if it was found or not. — CivFan, Oct 02 '15 at 23:17
Finally, [do you care about overlapping matches](http://stackoverflow.com/questions/5616822/python-regex-find-all-overlapping-matches)? — CivFan, Oct 02 '15 at 23:23

score 3 · Answer 1 · edited May 23 '17 at 10:27

3

Why reinvent the wheel?

def word_count(filename, word):
    with open(filename, 'r') as f:
        return f.read().count(word)

edited May 23 '17 at 10:27

Community

1
1

answered Oct 02 '15 at 22:56

CivFan

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score 0 · Accepted Answer · answered Oct 02 '15 at 22:46

def paraula(file,wordtofind):
    f = open(file,"r")
    text = f.read()
    f.close()
    count = 0
    index = text.find(wordtofind) # Returns the index of the first instance of the word
    while index != -1:
        count += 1
        text = text[index+len(wordtofind):] # Cut text starting from after that word
        index = text.find(wordtofind) # Search again
    return count

paraula (file,wordtofind)

Count how many times a word appears in a text file

2 Answers2