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Python Python Collections (Retired) Dictionaries Word Count

Word count.py

This is not working for me:

# 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(a_string):
  #split the string into a list.
  split_string = a_string.split()

  string_dict = {}
  for word in split_string:
    if string_dict[word]:
        string_dict[word] += 1
        string_dict[word] = 1

  return split_string

If anyone has suggestions that would be great.

6 Answers

Hello Hank,

I'll walk you through my solution:

def word_count(string):
    d = {}
    for char in string.lower().split():
        if char in d:
            d[char] += 1
            d[char] = 1
    return d

So first we define a function that intakes one parameter, assuming a string. We then declare an empty dictionary d to store the letters and their occurrence. Next we loop over the string which we lowercase and split the whitespace and assign each iteration to a variable char. Then we check if the letter, represented in char is present in the dictionary d. If our dictionary d contains our value stored in char then we precede to increment the value in "d" for char. If our value stored in char is not in the dictionary we will assign that key a initial value of one, thus instantiating the entry for char. Finally we return the dictionary d.

I hope this helps you & feel free to ask any follow up questions :)

Thanks man! I was stuck on that for a while and I was trying to figure it out on my own, but sometimes you just need to ask! Thanks a lot man.


I totally know what you mean, but I'm glad I could help :)

sandeep krishnappa
sandeep krishnappa
1,571 Points

Hello Ryan, I have a question about your code as I am still quite new to python and hope you don't mind.

my question is that when the code is running through the for loop first time, d is actually an empty dictionary, how can char be able to match with d? Unless all the char has already been allocated in d as keys? Then I don't understand from which line in the code, char has been allocated as keys in d.

Hope my question makes sense. Thanks.

Lucas Diz
Lucas Diz
4,804 Points

This is great. I was stuck here.

But please, could anyone explain about the (**) double stars? We didnt use it so far.

Mark Christian Roma
Mark Christian Roma
5,232 Points

I also have another approach for this challenge ...

def word_count(string):
    new_dict = {}
    new_string = string.lower().split()
    for word in new_string:
        if word not in new_dict:
    return new_dict

Here's an alternative. It uses the Counter Class.

from collections import Counter

def word_count(a_string):
    counted_dict = Counter(a_string.split())
    return counted_dict

Hopefully that's not considered a cheat, since Classes haven't been covered in this course; I just thought it was a nice, simple solution :)

Can you tell me whats wrong here?


import copy

def word_count(some_string):
    some_string = some_string.lower()
    some_string = some_string.split(" ")
    some_dictionary = {}
    broken = copy.copy(some_string)

    for word in some_string:
        if word in broken:
            some_dictionary[word] = 0

            while True:
                if word in broken:
                    some_dictionary[word] += 1


    return some_dictionary
Adam Boyko
Adam Boyko
1,087 Points

This took me a bit of brain power, new to python but not as new (just really rusty) for other languages.

I kept running into a simple issue where (before converting the string to count against into a list of its own), count() would count each 'word' literal as a substring in "word_to_count".

where "I like this" would return {"i": 3, "like": 1, "this": 1}. Obviously "I" should've only been equal to 1. The behavior of "count()" is interesting when applied to different classes.

def word_count(word_to_count):
    word_dict = {}
    for word in word_to_count.lower().split():
        if not word_dict.__contains__(word):
            word_dict[word] = word_to_count.lower().split().count(word)
    return word_dict
Fernando Alarcon
Fernando Alarcon
16,289 Points

I am stuck on this one and don't know why. It outputs the correct dictionary. Here is my code. Any help will be greatly appreciated:

def word_count(word):
    wl = word.split(" ")
    wc = []
    d = {}
    i = 0
    while i <= len(wl) - 1:
      i += 1
    i = 0
    while i <= len(wl) - 1:
      if i <= len(wl) - 2:
        if wl[i] == wl[i + 1]:
          wc[i] += 1
          del wc[i + 1]
          del wl[i + 1]
      i += 1
    i = 0
    while i <= len(wl) - 1:
      d[wl[i].lower()] = wc[i]
      i += 1