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Yewon Stephanie Cho
2,220 Pointswhy isnt this going through?
not sure why this code is not accepted .. it gets the results as specified in the comments
# 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):
results = {}
s = s.lower().split(' ')
for item in s:
results[item] = s.count(item)
print(results)
5 Answers
tobiaskrause
9,160 PointsDiffference between split() and split(' ')
split(' ')
def word_count(s):
results = {}
s = s.lower().split(' ')
for item in s:
results[item] = s.count(item)
print(results)
word_count("This is \na \ntest");
OUTPUT: {'\ntest': 1, '\na': 1, 'is': 1, 'this': 1}
split()
def word_count(s):
results = {}
s = s.lower().split()
for item in s:
results[item] = s.count(item)
print(results)
word_count("This is \na \ntest");
OUTPUT: {'this': 1, 'a': 1, 'is': 1, 'test': 1}
And thats why the challanged said you should NOT only split the whitespaces -> Bummer! Hmm, didn't get the expected output. Be sure you're not splitting only on spaces! BTW: if you are not common with \n -> this is a build in String escape character for a new line like
==> https://docs.python.org/2/reference/lexical_analysis.html#string-literals
tobiaskrause
9,160 Points# 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(str1):
str2 = str1.lower()
new_string = str2.split()
result = {}
for s in new_string:
if s in result:
result[s] += 1
else:
result[s] = 1
return result
Alexander Davison
65,469 PointsFirst, you shouldn't print out anything. The challenge didn't ask you to print anything 
Second, your program only splits on spaces.
split function, it will by default split on all whitespace (the challenge expects you to do this). Whitespace includes: spaces, tabs, newlines, etc.
def word_count(s):
results = {}
s = s.lower().split()
for item in s:
results[item] = s.count(item)
return results
I hope this helps. ~Alex
If this answered your question, Please mark is as a "Best Answer". Thank you!
Yewon Stephanie Cho
2,220 Pointsthank you Tobias for providing the solution but im genuinely curious to see why my code is not accepted as the answer !
Yewon Stephanie Cho
2,220 PointsThanks Alexander and Tobias for the helpful feedback !! :)
Alexander Davison
65,469 PointsNo problem :)