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Start your free trialDebajyoti Kar
Front End Web Development Techdegree Student 2,244 Pointscode gave an error in the test but works in the workspace
the code below is what I wrote for the challenge, it gives an error saying that it didn't give the expected output which is supposed to be
{'i': 2, 'do': 1, 'not': 1, 'like': 1, 'it': 1, 'sam': 1, 'am': 1}
if I use the string "I do not like it Sam I Am"
however, using the same code in the treehouse workspace prints exactly what is desired. I literally copy pasted the code with the exception that I use print function in the workspace to print the dictionary to check if the output I get is same as what I wanted.
# 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):
lower_case_string = string.lower()
string_array = lower_case_string.split(" ")
string_dict = {}
for word in string_array:
if word in string_dict.keys():
string_dict[word] += 1
else:
string_dict[word] = 1
return string_dict
1 Answer
Kyle Knapp
21,526 PointsThis one is a bit of a trick question. Your function breaks if there is additional whitespace between any of the words in the string.
Example:
#function
word_count("I\t \n do not like it Sam I Am")
#result
{'i\t': 1, '': 1, '\n': 1, 'do': 1, 'not': 1, 'like': 1, 'it': 1, 'sam': 1, 'i': 1, 'am': 1}
So you'll need to remove ALL whitespace between each word in the string.
Debajyoti Kar
Front End Web Development Techdegree Student 2,244 PointsDebajyoti Kar
Front End Web Development Techdegree Student 2,244 PointsOh!! completely slipped my mind. It makes sense. Thanks, Kyle.