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 trialKen Johnson
11,339 PointsBaffled by word count
I've studied some other answers in the community and was sure I had it! Am I close? Thanks!
# 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):
dicto = {}
split_string = string.lower().split()
for word in split_string:
count = 1
if word in dicto:
count += 1
dicto[word] = count
else:
dicto[word] = count
return dicto
1 Answer
Aby Abraham
Courses Plus Student 12,531 PointsTry this
def word_count(string): dicto = {} split_string = string.lower().split(' ') for word in split_string: if word not in dicto: dicto[word] = 1 else: dicto[word] += 1 return dicto