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3,526 PointsI 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
# 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
Nate Sturgeon
2,255 PointsNot true. By default (no args added) str.split() splits on spaces.
Alexander Davison
65,469 PointsYeah, 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
Nate Sturgeon
2,255 PointsYou 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.
Y B
14,136 PointsY B
14,136 PointsHint: You need to tell split what to split on.