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 trial
Terry Headley-Johnson
30 PointsWhat is wrong with this solution
This is my solution to the problem...and I see it is very different from what other people did.
Other than it being a bit long, is there anything actually wrong with how I did this?
# 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 = {}
final_string = string.lower().split()
counter = 0
for x in range(len(final_string)):
word1 = final_string[x]
for word2 in final_string:
if word1 == word2:
counter +=1
new_dict.update({word1:counter})
counter = 0
return new_dict
1 Answer
Denny Ho
4,472 Pointsi think you should change your first for loop to
for x in final_string
right now, your loop is iterating through a range of numbers, but you want to be able to iterate through your string itself