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Python Python Collections (Retired) Dictionaries Word Count

Dhruv Patel
Dhruv Patel
8,287 Points
Posted by Dhruv Patel
Dhruv Patel
8,287 Points

"IF"?

I have my code written and i'm pretty sure it works but for some reason, it keeps on returning Bummer! "if"

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(arg):
  words = arg.lower().split(" ")
  word_count = {}
  for word in words:
    if word_count[word] >= 1:
      word_count[word] += 1
    else:
      word_count[word] = 0
  return word_count

2 Answers

patrickgardner
patrickgardner
7,395 Points
patrickgardner
patrickgardner
7,395 Points

Hi Dhruv,

I don't know python super well, but this worked okay for me when I changed your if statement to check for the presence of the word in the dictionary, rather than looking for how many times it exists in a dictionary as follows:

def word_count(arg):
  words = arg.lower().split(" ")
  word_count = {}
  for word in words:
    if word in word_count # Check if the word is in the dictionary.
      word_count[word] += 1
    else:
      word_count[word] = 1 # this is also changed because if you're seeing it in the for loop, it has to turn up in results.
  return word_count

Hope this helps!

Dhruv Patel
Dhruv Patel
8,287 Points
Dhruv Patel
Dhruv Patel
8,287 Points

Thanks, this explains much better about why I got some counts for words incorrect.

liang zhou
PLUS
liang zhou
Courses Plus Student 6,567 Points
liang zhou
liang zhou
Courses Plus Student 6,567 Points

Maybe you should notice this line in the description of the challenge

# check for inclusion in the dict (with "if word in dict"-style syntax).

I think the problem is you forget check the existence of the key.

I hope the code below can help you.

def word_count(arg):
  words = arg.lower().split(" ")
  word_count = {}
  for word in words:
    if word in word_count:
      word_count[word] += 1
    else:
      word_count[word] = 1
  return word_count
Dhruv Patel
Dhruv Patel
8,287 Points
Dhruv Patel
Dhruv Patel
8,287 Points

Thanks for the reply