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Phoebe Li
2,278 Pointsgetting 'bummer! If' error and not sure what it means
not sure why code isn't passing
# 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(arg):
string = arg.split()
my_dict = {}
for word in string:
if my_dict[word]:
my_dict[word] = my_dict[word]+1
else:
my_dict[word] = 1
return my_dict
2 Answers
Anish Walawalkar
8,534 Pointsremember to use the "in" keyword, this should solve your problem:
def word_count(sent):
words = sent.split(' ')
count_dict = {}
for word in words:
if word in count_dict.keys():
count_dict[word] += 1
else:
count_dict[word] = 1
return count_dict
Brent Petit
7,860 PointsRemember to use the "in" keyword syntax to determine whether a key is defined in the dictionary as you iterate over the words.