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Python Python Collections (Retired) Dictionaries Word Count

Sergey Mangov
Sergey Mangov
3,651 Points

what's wrong with my word_count function?

def word_count(string): new_dict = {} new_string=string.lower() new_string_list = string.split() for word in new_sting_list: if word in new_dict: new_dict[word] +=1 else:
new_dict[word] = 1 return new_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):
    new_dict = {}
    new_string=string.lower()
    new_string_list = string.split()
    for word in new_sting_list:
        if word in new_dict: 
            new_dict[word] +=1
        else:   
            new_dict[word] = 1
    return new_dict

1 Answer

Russell Sawyer
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.a{fill-rule:evenodd;}techdegree
Russell Sawyer
Front End Web Development Techdegree Student 15,705 Points
def word_count(string):
    new_dict = {}
    new_string=string.lower()
    new_string_list = string.split()
    for word in new_string_list: # You have a typo.  You need a r in string
        if word in new_dict: 
            new_dict[word] +=1
        else:   
            new_dict[word] = 1
    return new_dict