Welcome to the Treehouse Community
Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.
Looking to learn something new?
Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.
Start your free trialSergey Mangov
3,651 Pointswhat's wrong with my word_count function?
def word_count(string):
new_dict = {}
new_string=string.lower()
new_string_list = string.split()
for word in new_sting_list:
if word in new_dict:
new_dict[word] +=1
else:
new_dict[word] = 1
return new_dict
# 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):
new_dict = {}
new_string=string.lower()
new_string_list = string.split()
for word in new_sting_list:
if word in new_dict:
new_dict[word] +=1
else:
new_dict[word] = 1
return new_dict
1 Answer
Russell Sawyer
Front End Web Development Techdegree Student 15,705 Pointsdef word_count(string):
new_dict = {}
new_string=string.lower()
new_string_list = string.split()
for word in new_string_list: # You have a typo. You need a r in string
if word in new_dict:
new_dict[word] +=1
else:
new_dict[word] = 1
return new_dict