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

Posted by Jirka Michalik
Jirka Michalik
11,573 Points

Alright, this one might be a bit challenging but you've been doing great so far, so I'm sure you can manage it.

What the heck? This code is again working only in my python IDLE, but Treehouse didn't accept this solution :/

Can you help me please?

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):
    dictionary = {}
    list_with_string = string.lower().split(" ") # small chars in list
    for item in list_with_string:
        value = list_with_string.count(item)
        dictionary[item] = value

2 Answers

Krishna Pratap Chouhan
Krishna Pratap Chouhan
15,203 Points

The problem with your code is.... "Incorrect Indentation" Try this... i corrected the indentation...

Plus i also changed "split" function

def word_count(string):
    dictionary = {}
    list_with_string = string.lower().split() # small chars in list
    for item in list_with_string:
        value = list_with_string.count(item)
        dictionary[item] = value
    return dictionary

Hope this helps! if correct then please do accept the answer. Have fun coding.

Swapnil Paralkar
Swapnil Paralkar
3,359 Points
Swapnil Paralkar
Swapnil Paralkar
3,359 Points

Jus a quick question of this . Read somewhere that python Dictionary do not support duplicate keys . If we want to show duplicate keys . What would be the workaround?

Krishna Pratap Chouhan
Krishna Pratap Chouhan
15,203 Points

Swapnil Paralkar , you cannot write duplicate keys in dictionary.

What you can do though, is make dictionary a list of values and then append values to those lists.

If you want this as a functionality, you might want to see into... defaultdict

https://docs.python.org/3/library/collections.html#collections.defaultdict

Jirka Michalik
Jirka Michalik
11,573 Points
Jirka Michalik
Jirka Michalik
11,573 Points

Oh okay - it's the right solution ... thank you