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
Pau Diaz Gallifa
1,616 PointsI can't figure out where is the problem with this script. Please help.
Hello, there is a mistake (or some of them) in this script, but I am not able to find them. Could anyone help me? 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.
#Create a function named word_count() that takes a string. Return a dictionary with each word in the string as the key and the number of times it appears as the value.
# 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.
str = "I am that I am"
def word_count(str):
strLC = str.lower()
str_parts = strLC.split()
word_dict = {}
count = 0
for item in str_parts:
if count < len(str_parts):
if item in word_dict:
word = 1
count += 1
word += 1
word_dict.update({str_parts[count - 1] : word})
else:
count += 1
word = 1
word_dict[str_parts[count -1]] = 1
continue
else:
break
return (word_dict)
word_count(str)
2 Answers
Chris Freeman
Treehouse Moderator 68,426 PointsI've marked up your code to a working solution. There was some unnecessary parts that I commented out. For the dictionary assignments I've shown an alternative solution to your assignments. The return statement wasn't inside the word_count() function:
# 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(str):
strLC = str.lower()
str_parts = strLC.split()
word_dict = {}
#count = 0
for item in str_parts:
#if count < len(str_parts):
if item in word_dict:
#word = 1
#count += 1
#word += 1
#word_dict.update({str_parts[count - 1] : word})
word_dict[item] += 1 #<-- look up item and increment the count value
else:
#count += 1
#word = 1
#word_dict[str_parts[count -1]] = 1
word_dict[item] = 1 #<-- Create new dictionary key, set value to 1
continue
#else:
# break
return (word_dict)
#word_count(str)
Pau Diaz Gallifa
1,616 PointsMany thanks again Chis, I really appreciate your comments and help.
Ahmed Elsawey
Courses Plus Student 3,527 PointsAhmed Elsawey
Courses Plus Student 3,527 PointsThank you very much Chris you helped me a lot through this I was only missing a small part of code but you helped me with it still and other questions I really appreciate your help chris freeman