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eodell
26,386 PointsWord_Count function. My code works but isn't accepted..
Hello. I am trying to pass the word_count challenge and in my personal python (v2.7) editor I am running:
def word_count(string):
dictionary = {}
words = string.lower().split(" ")
for word in words:
key = word
count = 0
for word in words:
if word == key:
count += 1
dictionary[key] = count
return dictionary
and getting the correct results. I return a dictionary with each key being the lowercase version of the words (found by splitting the argument on " ") and an accurate count of the frequency as an integer for the value of each key. Am I missing a step or is something wrong the code? Any insight would be appreciated.
# 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(string):
dictionary = {}
words = string.lower().split(" ")
for word in words:
key = word
count = 0
for word in words:
if word == key:
count += 1
dictionary[key] = count
return dictionary
2 Answers
Christopher Shaw
Python Web Development Techdegree Graduate 58,248 PointsThis is a common issue with this challange. Use split(), not split(' '). Your version will only split on single white spaces while split() splits on single or mutiple whitespaces.
eodell
26,386 PointsThank you for that explanation!