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Yuqi Zhou
6,381 PointsCreate a function named word_count() that takes a string. Return a dictionary with each word in the string as the key an
what's the problem, thx
# E.g. word_count("I am that I am") gets back a dictionary like:
# {'i': 2, 'am': 2, 'that': 1}
# Lowercase the string to make it easier.
# Using .split() on the sentence will give you a list of words.
# In a for loop of that list, you'll have a word that you can
# check for inclusion in the dict (with "if word in dict"-style syntax).
# Or add it to the dict with something like word_dict[word] = 1.
def word_count(string):
string_dict = {}
haha=string.lower().split()
for word in haha:
try:
string_dict[word]+=1
except:
string_dict[word] = 1
return string_dict
1 Answer
Keerthi Suria Kumar Arumugam
4,585 PointsThe problem is that the "return string_dict" is not indented. So Python assumes that the return statement is not part of the function and executes it separately.
def word_count(string):
string_dict = {}
haha=string.lower().split()
for word in haha:
try:
string_dict[word]+=1
except:
string_dict[word] = 1
return string_dict