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Python Python Collections (Retired) Dictionaries Word Count

Mangesh Somvanshi
Mangesh Somvanshi
3,521 Points

whats wrong with my code it work well in my local.

def word_count(x): x = x.lower() items = x.split() countr = dict()

for i in items:
    if not i in countr:
        countr[i] = 1
    else:
        countr[i] += 1
print countr

word_count("I am that I am")

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(x):
    x =  x.lower()
    items = x.split()
    countr = dict()

    for i in items:
        if not i in countr:
            countr[i] = 1
        else:
            countr[i] += 1
    print countr

word_count("I am that I am")

1 Answer

Matthew Rigdon
Matthew Rigdon
8,223 Points

I believe your code works fine. Just make sure to change your last line to return instead of print since we are working with a function.