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Python Python Collections (Retired) Dictionaries Word Count

Posted by MUZ140890 David Duri
MUZ140890 David Duri
2,613 Points

I am stuck on this one, where am i getting it wrong here?

Dictionaries second Challenge

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(a_string):
  string_dict={}
  for word in string.split():
    if word in string_dict:
      string_dict[word]+=1
      else:
        string_dict[word]=1
        return string_dict

1 Answer

Andreas cormack
.a{fill-rule:evenodd;}techdegree seal-36
Andreas cormack
Python Web Development Techdegree Graduate 33,011 Points

Hi David

Most of your code is correct apart from a few typos and indentation.

on line 3 > its should be a_string.split() and not string.split() your else block needs to be indented properly and your return statement need to be outside the for loop.

see amends to your code below.

# 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):
  string_dict={}
  for word in a_string.split():
    if word in string_dict:
      string_dict[word]+=1
    else:
      string_dict[word]=1
  return string_dict

hope this helps

MUZ140890 David Duri
MUZ140890 David Duri
2,613 Points

that worked thanks so much