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Python Python Collections (2016, retired 2019) Dictionaries Word Count

Deniz Kahriman
Deniz Kahriman
8,538 Points

What is wrong in this code? Trying to count the words in a string and assign values. Thank you!

Trying to complete the challenge but got stuck. Thank you very much in advance!

wordcount.py
# 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):
    final_list = {}
    for word in string.split():
        if word.lowercase() in string.lowercase():
            final_list[word] += 1
            else:
                final_list[word] =1
    return final_list

2 Answers

Robert Stefanic
Robert Stefanic
35,157 Points

There's an extra indentation in your if/else statement. The else block is an extra level in, so it doesn't have an if statement to match it.

Just bring your else statement out so it lines up with your if statement.

Deniz Kahriman
Deniz Kahriman
8,538 Points

Thanks, Robert! I tried it but still didn't work :(

def word_count(string):
    final_list = {}
    for word in string.split():
        if word.lowercase() in string.lowercase():
            final_list[word] += 1
        else:
            final_list[word] =1
    return final_list
Robert Stefanic
Robert Stefanic
35,157 Points

I see. There are a few things to note here:

First, lowercase() isn't a method strings have. The method you're looking for is lower().

So change and .lowercase() to .lower().

Second, in your condition for your if statement, you're checking the wrong thing:

# Make sure you change the method names here to ".lower()"
if word.lowercase() in string.lowercase():

You're checking to see if the lower case word is in the string that you pulled the word from! You need to check to see if the word is part of your final_list dictionary.

And lastly, when you add the words to your dictionary, you should lowercase them. The example has the word 'sam' in the dictionary as a lower case word, even though it appears capitalized in the string that's passed to the function.

So just make sure that you add .lower() to the keywords that you're adding to your dictionary, like so:

final_list[word.lower()] += 1
Deniz Kahriman
Deniz Kahriman
8,538 Points

Hi Robert,

I tried it again based on your notes below but I'm still getting an error. Would you mind looking at the code below to see what I'm missing here? Thanks!

# 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):
    final_list = {}
    for word in string.split():
        if word.lower() in final_list.lower():
            final_list[word.lower()] += 1
        else:
            final_list[word.lower()] =1
    return final_list
Robert Stefanic
Robert Stefanic
35,157 Points

You're very close. Think about this: you're calling .lower() on final_list. How could you apply .lower() to a dictionary object? You can't.

You've already lower cased the words before adding them to your final_list. Since they're already lowered, there's no need to lowercase the words in final_list because they're already entered into your dictionary as lower cased words!

So just remove .lower() from final list. :)

Deniz Kahriman
Deniz Kahriman
8,538 Points

Finally, completed the challenge! Thank you very much!! :)