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

Mike Le
Mike Le
3,854 Points
Posted by Mike Le
Mike Le
3,854 Points

Word Count

I'm not sure why this code doesn't work. Tried it in the console and the result was similar to the example.

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):
    string_list = string.lower().split(' ')
    count = {x: string_list.count(x) for x in string_list}
    return count

2 Answers

Mohammed Ismail
Mohammed Ismail
7,190 Points
Mohammed Ismail
Mohammed Ismail
7,190 Points

Hi Mike,

Please double check on the count() and append rules applicable to dictionary data type

Please find my code for the word_counts challenge. Hope this gives you some idea.

def word_count(str1): counts = dict() words = str1.lower().split() for word in words: if word in counts: counts[word] += 1 else: counts[word] = 1 return counts

print( word_count("I do not like it Sam I Am"))

Mike Le
Mike Le
3,854 Points
Mike Le
Mike Le
3,854 Points

Thank you Mohammed!

Wade Williams
Wade Williams
24,476 Points
Wade Williams
Wade Williams
24,476 Points

Pretty advanced using a comprehension instead of a loop.

The issue is this line:

string_list = string.lower().split(' ')

This splits the string based on a single space and when they run tests they use various types of white space. To split on all white space just leave split() empty.

This passes

def word_count(string):
    string_list = string.lower().split()
    count = {x: string_list.count(x) for x in string_list}
    return count

Nice job!

Mike Le
Mike Le
3,854 Points
Mike Le
Mike Le
3,854 Points

Thanks Wade!