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Python Python Collections (Retired) Dictionaries Word Count

Posted by drpac
drpac
3,405 Points

My problem: I know how to do for loop over the list, and how to map list-items to keys - then im stuck....

def word_count(string): string = string.lower().split(" ") word_dict = {} for word in string: word_dict[word] = 1 return word_dict

word_count("I am a stupid man and I am a dumb man") --Result----> {'a': 1, 'i': 1, 'stupid': 1, 'man': 1, 'and': 1, 'am': 1, 'dumb': 1}

Any Suggestions?

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.

2 Answers

Peter Lawless
Peter Lawless
24,404 Points
Peter Lawless
Peter Lawless
24,404 Points

The issue appears to be that your word_dict assigns a value of one, and does not iterate up when a word occurs a second time. This means you need to write in two conditions within your for loop, maybe something like:

for word in string:
    if word not in word_dict:
        word_dict[word] = 1
    else:
        word_dict[word] += 1
drpac
drpac
3,405 Points
drpac
drpac
3,405 Points

Thanks for answer(s). In the end i solved it like this:

def word_count(string):
    string = string.lower().split(" ")
    word_dict = {}
    for word in string:
        word_dict[word] = string.count(word)
    return word_dict

I think that, what was my problem - with your kind of (the suggested) approach - was just how exactly to turn the given list into a dictionary (word_dict)