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

Posted by Buck Randall
Buck Randall
1,714 Points

Not sure what I'm missing here . . . the outcome is correct but "Bummer - check lower and remove whitespaces"

def word_count(s): word_list=s.lower().split(' ') word_dict={} for word in word_list: if word in word_dict: word_dict[word]+=1 else: word_dict[word]=1 return word_dict

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.

def word_count(s):
    word_list=s.lower().split(' ')
    word_dict={}
    for word in word_list:
        if word in word_dict:
            word_dict[word]+=1
        else:
            word_dict[word]=1
    return word_dict

print(word_count('I do not like it Sam I Am'))

1 Answer

james south
seal-mask
.a{fill-rule:evenodd;}techdegree seal-36
james south
Front End Web Development Techdegree Graduate 33,271 Points
james south
.a{fill-rule:evenodd;}techdegree seal-36
james south
Front End Web Development Techdegree Graduate 33,271 Points

try split() without any argument to check for more whitespace characters than just space, i think this has worked for other people asking about this one.