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Python Python Collections (Retired) Dictionaries Word Count

Posted by Mrinal Kant
Mrinal Kant
4,373 Points

Challenge task evaluate seem like not working correctly for python . has any one faced similar issue?

I am getting Bummer! try again! for word_count.py .

The code i have written seems correct to me. at-least i was expecting some error reason. It is frustrating when you get this kind of error without any error code.

word_count.py 
# 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_text):
    word_dict=dict()
    word_list_lower = string_text.lower()
    word_list = word_list_lower.split()
    for word in word_list:
        if word in word_dict:
            word_dict[word] += 1
        else:
            word_dict[word] = 1

return word_dict

the issue was return indentation....

2 Answers

Mark Christian Roma
Mark Christian Roma
5,232 Points
Mark Christian Roma
Mark Christian Roma
5,232 Points

I also have another approach for this challenge ...

def word_count(string):
    new_dict = {}
    new_string = string.lower().split()
    for word in new_string:
        if word not in new_dict:
            new_dict.update({word:new_string.count(word)})
    return new_dict
john larson
john larson
16,594 Points

I like it :D

john larson
john larson
16,594 Points
john larson
john larson
16,594 Points

Mrinil, One thing I found, the return is a tab too far to the left. I'll look and see if there's anything else

john larson
john larson
16,594 Points

with that small fix, it passed fine :D