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

Posted by Alex Rogers
Alex Rogers
9,844 Points

Help on this (Python Dictionaries)

I'm very lost.

I tried to use a for loop to iterate how many times a word came up in a string, but I get stuck on making sure that those words that come up more than once are not duplicated. I'm not really sure how to fix.

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):
    comparison = string.lower().split()
    newDict = []
    for word in comparison:
        count = 0
        act = word
        for j in range(len(comparison)):
            if comparison[j] == word:
                count += 1
        if count > 1:

1 Answer

Alexander Davison
Alexander Davison
65,469 Points
Alexander Davison
Alexander Davison
65,469 Points

From reading your code, I believe that you are looking at every word, then checking how many times that word occurs. Because of this, you have duplicates that you do not need. Of course, this solution might work. But wouldn't it be easier if you didn't need to deal with duplicates? What if the duplicates actually can help you? :wink:

You can actually do this. Instead of going through every word and seeing how many times it occurs, than removing duplicates, why don't you simply go through every word, and if the word is not in newDict, set newDict[word] to 0, and for every word add 1 to newDict[word]?

Also note that you made a dictionary using [], you should use {}.

The (easier) solution in code:

def word_count(string):
    words = string.lower().split()
    my_dict = {}
    for word in words:
        if word not in my_dict:
            my_dict[word] = 0
        my_dict[word] += 1
    return my_dict

I hope this helps! :grin:

Happy coding! :zap: ~Alex