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Python Python Collections (Retired) Dictionaries Word Count

gibran erlangga
gibran erlangga
5,539 Points

word_count challenge

need a little help here guys.

I got stuck in this challenge for hours, and still cant find any solution. Anyone?

Attached is my latest code. Thanks guys!

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.

some_dict = {'a': 1, 'b': 2, 'c': 3}

key_list = ['a', 'b', 'd']

def word_count(some_dict, key_list):
  counts = 0
  for item in key_list:
    if item in some_dict.keys():
      counts += 1
  return counts

1 Answer

Steven Parker
Steven Parker
230,024 Points

Here's a few hints:

  • you only need to create the function, the challenge will provide the test data
  • your function will take only one argument, a string
  • you will probably want to split the string to help loop through the words
  • your function will build a new dictionary
  • the counts will be kept in the new dictionary
  • you don't need the keys function to check if an item is in the dictionary
  • when the item is not already in the dictionary, a new item with a count of 1 will be added to it
  • remember to return the finished dictionary