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Python Python Collections (2016, retired 2019) Dictionaries Word Count

Posted by Hsufeng Lee
Hsufeng Lee
16,440 Points

my code work on my machine but not on the site

E.g. word_count("I do not like it Sam I Am") gets back a dictionary like:

{'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1}

Lowercase the string to make it easier.

function

def word_count(str): target = str.lower().split(" ") result = {} for key in target: if key in result: result[key]+=1 else: result[key]=1 return result;

test

if word_count("I do not like it Sam I Am")=={'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1}: print("Pass!") print(word_count("I do not like it Sam I Am")) else: print("Fail!") print(word_count("I do not like it Sam I Am"))

wordcount.py 
# E.g. word_count("I do not like it Sam I Am") gets back a dictionary like:
# {'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1}
# Lowercase the string to make it easier.

# function
def word_count(str):
    target = str.lower().split(" ")
    result = {}
    for key in target:
        if key in result:
            result[key]+=1
        else:
            result[key]=1
    return result;

# test
if word_count("I do not like it Sam I Am")=={'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1}:
    print("Pass!")
    print(word_count("I do not like it Sam I Am"))
else:
    print("Fail!")
    print(word_count("I do not like it Sam I Am"))

1 Answer

Alexander Davison
Alexander Davison
65,469 Points
Alexander Davison
Alexander Davison
65,469 Points

You need to split on all whitespace, not just spaces.

Thus you should replace .split(" ") with simply .split().

If you pass no arguments to .split(), it—by default—would split on any whitespace.

Happy coding! :zap: ~Alex