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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 68,454 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