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lin huang
4,284 PointsI run this in my IDLE with no problem
def word_count(words): newdic = {} words = words.lower() words = words.split(" ") for word in words: try: newdic[word]+=1 except KeyError: newdic[word]=1
return newdic
# 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(words):
newdic = {}
words = words.lower()
words = words.split(" ")
for word in words:
try:
newdic[word]+=1
except KeyError:
newdic[word]=1
return newdic
2 Answers
Jon Mirow
9,864 PointsDon't worry, everyone gets caught out on this one, the checker is SUPER strict. I even went as far as stripping all not alpha, verifying the split string wasn't empty after filtering lol...
It's the .split(" ") that's causing the problem. As far as I can tell they're scanning the syntax you wrote to see if you used it instead of the default .split(), not just comparing output. My code filtered out everything that using split(" ") would allow through and it still failed lol. So you can provide the correct output to all their tests and still fail if you used split(" ").
The reason I assume is just to remind you that you don't add an argument to split() if you want all whitespace removed, only if you specifically want it split at certain points.
lin huang
4,284 Pointsthank you for encourage.