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Python Python Collections (Retired) Dictionaries Word Count

Posted by beven nyamande
beven nyamande
9,575 Points

iteration in dictionaries

getting bummer ! some words missing where am i getting it wrong

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(sentence):
  my_dict={}

  for word in sentence.split():
    if word in my_dict.keys():
      my_dict[word]  +=1
    else: my_dict[word] = 1
    return my_dict

1 Answer

Chris Shaw
Chris Shaw
26,676 Points
Chris Shaw
Chris Shaw
26,676 Points

Hi Beven,

You have a couple of issues with your code.

  1. You are using inconsistent spaces and indentation, your else statement should be on it's own line and return should be in the same indent column as for

  2. my_dict.keys() isn't required, all we need to do is simply check if the word is in the dictionary already.

def word_count(sentence):
  words = {}

  for word in sentence.split():
    if word in words:
      words[word] += 1
    else:
      words[word] = 1

  return words

One thing you will notice is I changed my_dict to words, I prefer to use self explanatory variable names which help others understand your code easier.

Happy coding.

beven nyamande
beven nyamande
9,575 Points

still keep getting bummer!"some words missing"

Chris Shaw
Chris Shaw
26,676 Points
Chris Shaw
Chris Shaw
26,676 Points

Works fine from my end, not to sound to forthright but are you sure to copied the code correctly?

beven nyamande
beven nyamande
9,575 Points

oh thanks chris saw its to do with indendation