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

Posted by Tan Yang
Tan Yang
13,163 Points

please help what is the problem of my code

help

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.split(" ")
    length = len(string_list)

    new_string_list = []
    for word in string_list:
        new_string_list.append(word.lower())

    new_dict = {}
    for lowered_word in new_string_list:
        count = 0
        for i in range(length):
            if lowered_word == new_string_list[i]:
                count += 1

        new_dict[lowered_word] = count

    return new_dict

3 Answers

Martynas Matimaitis
Martynas Matimaitis
10,480 Points
Martynas Matimaitis
Martynas Matimaitis
10,480 Points

Hi,

can you specify what do you mean by wrong? I added a function call with a variable and the output seems to be correct in few tests I ran:

def word_count(string):
    string_list = string.split(" ")
    length = len(string_list)

    new_string_list = []
    for word in string_list:
        new_string_list.append(word.lower())

    new_dict = {}
    for lowered_word in new_string_list:
        count = 0
        for i in range(length):
            if lowered_word == new_string_list[i]:
                count += 1

        new_dict[lowered_word] = count

    return new_dict

count = word_count("Ha ha ha ha HA HA HA hi Hi HI")
print(count)

Output: {'ha': 7, 'hi': 3}

Tan Yang
Tan Yang
13,163 Points
Tan Yang
Tan Yang
13,163 Points

great, thanks Grigorij. it works.

Grigorij Schleifer
Grigorij Schleifer
10,365 Points
Grigorij Schleifer
Grigorij Schleifer
10,365 Points

Hi Tan, your code works if you remove the quotation marks from the split method. Just split the string with whitespace which is the default one of the split option.