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Nicholas Lee
1,098 PointsHi I got the expected return in workspaces but not in the challenge. Please help. ( I used the zip fuction)
def word_count(string):
string = string.lower()
wordlist = string.split()
wordcount = []
worddict = {}
for word in wordlist:
count = 0
for char in word:
count += 1
wordcount.append(count)
worddict = dict(zip(wordlist,wordcount))
'''print (wordlist)
print (wordcount)
print (worddict)'''
return worddict
# 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):
string = string.lower()
wordlist = string.split()
wordcount = []
worddict = {}
for word in wordlist:
count = 0
for char in word:
count += 1
wordcount.append(count)
worddict = dict(zip(wordlist,wordcount))
'''print (wordlist)
print (wordcount)
print (worddict)'''
return worddict
1 Answer
Steven Parker
231,269 PointsYou may have misunderstood the instructions. The challenge wants a dictionary that shows how many times that each unique word appears in the string. But this code seems to be building one that shows how many letters are in each word (with no concern for uniqueness).
Check the example input and output shown in the comments.