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Mark Ramos
19,209 Pointsword_count Challenge
My code below seems to work OK in Workspace, but doesn't pass the challenge. Any idea what I've got wrong?
# 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):
dict = {}
count = []
words = string.lower().split(" ")
for word in words:
dict[word] = words.count(word)
return dict
Mark Ramos
19,209 PointsHey Steven - the line you referred to is a way to add a new key and value to a dictionary. For example:
>>> my_dict = {'a': 1, 'b': 2}
>>> print(my_dict)
{'a': 1, 'b': 2}
>>> my_dict['c'] = 3
>>> print(my_dict)
{'a': 1, 'b': 2, 'c': 3}
So 'dict' was just calling my empty dictionary from earlier in the block and [word] was me adding new keys to dict, one for every word in words. I hope that is helpful!
Steven Olick
1,561 PointsSO helpful. Thanks for breaking that down for me!!
1 Answer
Unsubscribed User
9,496 Pointsuse split with no arguments, and it will split at the whitespace. try it and it will pass.
words = string.lower().split()
Steven Olick
1,561 PointsSteven Olick
1,561 PointsHi Mark, your answer helped me wrap my head around this a little better. My answer was not too far off from yours. However, I was wondering if you could explain this line for me. It was where I got stuck in my answer.
dict[word] = words.count(word)
What exactly does the dict[word] do?