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Python Python Collections (Retired) Dictionaries Word Count

Posted by dima Dima
dima Dima
901 Points

What can I fix to get in working, any help would help

Create a function named word_count() that takes a string. Return a dictionary with each word in the string as the key and the number of times it appears as the value.

word_count.py 
# E.g. word_count("I am that I am") gets back a dictionary like:
# {'i': 2, 'am': 2, 'that': 1}
# Lowercase the string to make it easier.
# Using .split() on the sentence will give you a list of words.
# In a for loop of that list, you'll have a word that you can
# check for inclusion in the dict (with "if word in dict"-style syntax).
# Or add it to the dict with something like word_dict[word] = 1.
def word_count(str1):
    str2=str1.lower()
    str3=list[str2.split]
    count=0
    for item in str3:
        count=count+1
    return new_dict{item:count}

3 Answers

jacinator
jacinator
11,936 Points
jacinator
jacinator
11,936 Points

Hi dima Dima,

This code should work for you. Comments explain some of it.

def word_count(str1):
    str2 = str1.lower()
    str3 = str2.split()  # list[] doesn't work; list() around str.split() is redundant
    dict1 = {}  # Define dictionary to store information in
    for word in str3:
        try:
            dict1[word] += 1  # Update...
        except KeyError:
            dict1[word] = 1  # ...or create information
    return dict1  # Return stored information

# Alternate function is more concise
def word_count(string):
    return_dict = dict()
    for word in string.lower().split():
        try:
            return_dict[word] += 1
        except KeyError:
            return_dict[word] = 1
    return return_dict
dima Dima
dima Dima
901 Points
dima Dima
dima Dima
901 Points

Thank you very much but can you help me understand, why do you write return_dict[word]=1, I understand the exception when the word realy appears only one time in the string, but what if, for example it appears three times. So for the first time and the second and third time the "try" part will work but then we should again break the loop to stop the counting but we cannot because the except part works only for the number one, and not the number three. So if this except part does not cover all the possible solutions, what does that? or how the except part does that?

Thanks in advance, Dima

jacinator
jacinator
11,936 Points 
try:
    # This block will increase the word count if return_dict[word] already
    # exists. Which means it has been counted once already.
    return_dict[word] += 1
except KeyError:
    # Otherwise this block will create the word as a key in the dictionary and
    # assign it to being counted once.
    return_dict[word] = 1

You shouldn't have to break the for loop because it passes in each word in the string once. If it was a while loop, you would have to break it eventually or it would continue indefinitely.

I hope that this helps explain a bit.

Benedictt Vasquez
Benedictt Vasquez
2,130 Points
Benedictt Vasquez
Benedictt Vasquez
2,130 Points

Aquí otra manera.

def word_count(x):
    y = ((x.lower()).split(' '))
    dic = {}
    for word in y:
        dic.update((dict([(word, y.count(word))])))
    return (dic)