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
MIkhail Vinogradov
2,669 PointsHello, any ideas where the mistake is? please.
def word_count (string): string = string.lower() string = string.split(" ") dict={} for e in string: dict[e] = 0 for w in string: c=0 for r in string: if r == w: c += 1 dict[w]=c return dict
# E.g. word_count("I do not like it Sam I Am") gets back a dictionary like:
# {'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1}
# Lowercase the string to make it easier.
def word_count (string):
string = string.lower()
string = string.split(" ")
dict={}
for e in string:
dict[e] = 0
for w in string:
c=0
for r in string:
if r == w:
c += 1
dict[w]=c
return dict
1 Answer
Philip Schultz
11,437 PointsHello, If you just delete the quotation in the split function your code will work. Also, the first for loop is not needed since you are assigning both the key and the value in the second for loop.
Here is my solution to the problem
def word_count(args):
args = args.lower().split() # this will take the argument and make it lower case and split it (you don't need quotation)
words_dict = {} # making a dict
for word in args: #looping through the argument
words_dict[word] = args.count(word) # making each word a key and setting equal to count of each word in the arguemnt
return words_dict #returning the dict