Word count.py

Question

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
    else:
        string_dict[word] = 1

  return split_string

If anyone has suggestions that would be great.

Answer 1 · 2014-09-25T21:46:57Z

September 25, 2014 9:46pm

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
        else:
            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 :)

Answer 2 · 2017-02-24T09:55:42Z

February 24, 2017 9:55am

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:
            new_dict.update({word:new_string.count(word)})
    return new_dict

Answer 3 · 2015-10-31T16:00:59Z

October 31, 2015 4:00pm

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 :)

Answer 4 · 2018-04-22T18:30:26Z

April 22, 2018 6:30pm

Can you tell me whats wrong here?

Code:

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
                    broken.remove(word)
                    continue

                else:
                    break

    return some_dictionary

Answer 5 · 2018-09-14T18:44:11Z

September 14, 2018 6:44pm

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

Answer 6 · 2018-10-06T20:51:01Z

October 6, 2018 8:51pm

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
    wl.sort()
    while i <= len(wl) - 1:
      wc.append(1) 
      i += 1
    i = 0
    while i <= len(wl) - 1:
      print(wl[i])
      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
    return(d)

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Hank de Roover

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Word count.py

6 Answers

Ryan Ackermann

Ryan Ackermann

Hank de Roover

Hank de Roover

Ryan Ackermann

Ryan Ackermann

sandeep krishnappa

sandeep krishnappa

Lucas Diz

Lucas Diz

Mark Christian Roma

Mark Christian Roma

Sylvia Walker-Wavell

Sylvia Walker-Wavell

Adi Strasser

Adi Strasser

Adam Boyko

Adam Boyko

Fernando Alarcon

Fernando Alarcon