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Python Python Collections (Retired) Dictionaries Word Count

selcuk baran
selcuk baran
3,526 Points

I am stuck. Please help

def word_count(string): string.lower() x=string.split() for item in x: word_dict[item]=x.count(item) return word_dict

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(string):
    string.lower()
    x=string.split()
    for item in x:
        word_dict[item]=x.count(item)
    return word_dict
Y B
Y B
14,136 Points

Hint: You need to tell split what to split on.

Not true. By default (no args added) str.split() splits on spaces.

Yeah, Nate Sturgeon is right about that.

Try using help on the split function like this:

help(str.split)

By default Python splits the string by spaces if you don't specify arguments.

Just a little note :)

1 Answer

You are very close. As far as I can tell, your only problem is that you are trying to add an key/value pair to a dictionary (inside your for loop) that doesn't yet exist. If you create it before the for loop starts (word_dict = {}) then you can append to it as you are doing. If you add this one line of code your function works perfectly.