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

Posted by Sébastien VERMANDELE
Sébastien VERMANDELE
4,346 Points

I can't find the good solution to this challenge. My function get the right output but not in the same order.

I try many solutions but my dictonary still output in a different order.

Thanks for help Seb

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(phrase):
    dict = {}
    phrase = phrase.lower()
    tab = phrase.split(" ")
    for item in tab:
        dict.update({ item : tab.count(item)})
    return dict

3 Answers

Alexander Schott
PLUS
Alexander Schott
Courses Plus Student 937 Points
Alexander Schott
Alexander Schott
Courses Plus Student 937 Points
wordcount.py 
def word_count(text):
    word_dict = {}
    word_list = text.lower().split(" ")
    for word in word_list:
        if word in word_dict:
            word_dict[word] += 1
        else:
            word_dict[word] = 1
    return word_dict

that works for me but doesn't get accepted. why? Thank you for your convenient solution!

btw. how did you add the filename to your markdown code example?

Sébastien VERMANDELE
Sébastien VERMANDELE
4,346 Points
Sébastien VERMANDELE
Sébastien VERMANDELE
4,346 Points

Hi Alexender

I had the same return. You just have to modify your split method from split(" ") to slipt().

In split method, the default separator is any white space.

Good luck Seb

Sébastien VERMANDELE
Sébastien VERMANDELE
4,346 Points
Sébastien VERMANDELE
Sébastien VERMANDELE
4,346 Points

I just find the solution myself, it was my split method.