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Start your free trialandrew falcone
8,651 Pointsword_count dictionary question code seems to work but my solutions is not being accepted :(
I think my code is right, but it isn't being accepted with no information about what is wrong! (WTF, why not give error message?)
# 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_array = my_string.split(" ")
word_dictionary = {}
for word in string_array:
lower_word = word.lower()
try:
word_dictionary[lower_word] += 1
except:
word_dictionary[lower_word] = 1
return word_dictionary
2 Answers
KRIS NIKOLAISEN
54,971 PointsThe other has to do with splitting on all whitespace
split(" ") will split on blank spaces only
split() includes tabs, and others
andrew falcone
8,651 PointsThanks!
KRIS NIKOLAISEN
54,971 PointsIf you paste this into a workspace you'll see a better error message.
NameError: name 'my_string' is not defined
You have 'string' as a parameter but split 'my_string'.
andrew falcone
8,651 PointsI just fixed that (that's annoying that python doesn't give you an error message for trying to use a variable defined outside the scope of the function!).
I changed my_string to string and it is not giving an error message but still won't accept my answer :(
andrew falcone
8,651 PointsAlso, is there a place to see best solutions for problems like this? I have no idea if the way I solve a problem is the most efficient way or best practice.
Thanks!
KRIS NIKOLAISEN
54,971 PointsKRIS NIKOLAISEN
54,971 PointsIf you click on the topic path above (in this case Word Count)
Python > Python Collections > Dictionaries > Word Count
you can see how others have solved the same challenge