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Python Python Collections (Retired) Dictionaries Word Count

Posted by john larson
john larson
16,594 Points

Create a function named word_count() that takes a string. Return a dictionary with each word in the string as the key...

Create a function named word_count() that takes a string. Return a dictionary with each word in the string as the key and the number of times it appears as the value.

  • Bummer! Didn't get the right count for some of the words.
  • I've tested it with several different strings and it appears to work
  • what am I missing?
word_count.py 
# E.g. word_count("I am that I am") gets back a dictionary like:
# {'i': 2, 'am': 2, 'that': 1}
# Lowercase the string to make it easier.
# Using .split() on the sentence will give you a list of words.
# In a for loop of that list, you'll have a word that you can
# check for inclusion in the dict (with "if word in dict"-style syntax).
# Or add it to the dict with something like word_dict[word] = 1.
def word_count(s):
    s = s.lower()
    dct_cnt = {}
    count = 0
    list_s = s.split()
    for word in list_s:
        if word in dct_cnt:
            count +=1
            dct_cnt.update({word: count})
        else:
            count = 1
            dct_cnt.update({word: count})
    return dct_cnt

1 Answer

Cheo R
Cheo R
37,150 Points
Cheo R
Cheo R
37,150 Points

Hello John, you're on the right track.

Your code fails with the test string:

test = "This test string is only a test string"
word_count(test)
{'this': 1, 'is': 1, 'test': 2, 'string': 3, 'a': 1, 'only': 1}

Because it thinks the word string has been seen three times. Here is a chart showing word and count:

WORD----- ----IN DICT---- ----COUNT
this No 1
test no 1
string no 1
this YES 2
is no 1
only No 1
a no 1
test YES 2
string YES 3

The last three words are important. Because a has not been found in the dictionary, count is one. count is still one, so when test is found in the dictionary, count increases by one. Now, before it checks if string is in the dictionary, count is two, so when it finds string, it increases by one.

To fix your problem, you can either get the word's value (since you already know it's in the dictionary) and update it, like:

if word in dct_cnt:
    count = dct_cnt[word]
    count +=1
    dct_cnt.update({word: count})


test = "This test string is only a test string"
word_count(test)
{'this': 1, 'is': 1, 'test': 2, 'string': 2, 'a': 1, 'only': 1}

or combine them into one line:

if word in dct_cnt:
    dct_cnt[word] += 1

test = "This test string is only a test string"
word_count(test)
{'this': 1, 'is': 1, 'test': 2, 'string': 2, 'a': 1, 'only': 1}

I'm sure there are other ways of doing it, but I'm tired.

john larson
john larson
16,594 Points

Cheo, thanks for your response. Nicely presented