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Python Python Collections (Retired) Dictionaries Word Count

Peter Lynch
Peter Lynch
946 Points

Word_count.py

Am I close to the solution or way off base?

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(string):
  string = string.lower()
  string = list(string.split())
  for word in string: 
    count = 0
    if word in string:
      count = count + 1 
      return word_dict[word] = count

2 Answers

Carlos Federico Puebla Larregle
Carlos Federico Puebla Larregle
21,074 Points

You are close, you can do the lower() and split() in one line. Remember that the split() function returns a list so it is a little bit redundant to put the "list()" too. You could use a dictionary variable to check "if" the current value of the list already exists. You could do it like this:

def word_count(string_value):
  list_value = string_value.lower().split()
  dict_value = {}
  for word in list_value:
    if word in dict_value.keys():
      dict_value[word] += 1
    else:
      dict_value[word] = 1
  return dict_value

I hope that helps

Peter Lynch
Peter Lynch
946 Points

Thanks Carlos helped me understand much better.