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

David Gray
David Gray
1,531 Points
Posted by David Gray
David Gray
1,531 Points

I get the expected value when I run my function locally, however the keys are in a different order, which is ok

def word_count(phrase): frequencies = {} lower_phrase = phrase.lower() words = lower_phrase.split(" ") for x in words: if x in frequencies.keys(): frequencies[x] += 1 else: frequencies[x] = 1 return(frequencies)

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(phrase):
    frequencies = {}
    lower_phrase = phrase.lower()
    words = lower_phrase.split(" ")
    for x in words:
        if x in frequencies.keys():
            frequencies[x] += 1
        else:
            frequencies[x] = 1
    return(frequencies)

1 Answer

Steven Parker
Steven Parker
230,995 Points
Steven Parker
Steven Parker
230,995 Points

The challenge error message says "Be sure you're lowercasing the string and splitting on all whitespace!".

To split on "all whitespace" the argument to "split" should be left empty. Specifying a space causes it to split only on literal spaces. You probably didn't notice it in outside testing because you probably just used single spaces as separators in the data.