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Python Python Collections (2016, retired 2019) Dictionaries Word Count

Posted by alenwong
alenwong
4,346 Points

Count the number of occurrence of each word in the sentence

...make a function named word_count. It should accept a single argument which will be a string. The function needs to return a dictionary. The keys in the dictionary will be each of the words in the string, lowercased. The values will be how many times that particular word appears in the string.

E.g. word_count("I do not like it Sam I Am") gets back a dictionary like: {'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1} Lowercase the string to make it easier.

print(word_count("I do not like it Sam I Am")) Result --> {'i': 2, 'do': 1, 'not': 1, 'like': 1, 'it': 1, 'sam': 1, 'am': 1}

As the system didn't provide further information - like the test case it use, the expected result... I can't think of a way to debug this...

Thanks in advance :'(

wordcount.py 
# E.g. word_count("I do not like it Sam I Am") gets back a dictionary like:
# {'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1}
# Lowercase the string to make it easier.

def word_count(varStr):
  varList = varStr.split(" ")
  varDict = {}

  idx = 0
  while idx < len(varList):
    varList[idx] = varList[idx].lower()
    varDict.update({varList[idx]:0})
    idx +=1

  for key in varDict.keys():
    idx = 0
    hit = 0
    while idx < len(varList):
      if varList[idx] == key:
        hit +=1
      idx +=1
    varDict.update({key:hit})

  return varDict

2 Answers

Jeff Muday
MOD
Jeff Muday
Treehouse Moderator 28,722 Points
Jeff Muday
Jeff Muday
Treehouse Moderator 28,722 Points

I said before, it looks like you really understand coding-- which is a great start.

The reason your code snippet did not work is very simple... They are asking to split the sentence on "all whitespace characters" and you are splitting on spaces only, not tabs, linefeeds, etc.

If you remove the double-quoted space character-- you get to the default behavior of the split method, which works on all whitespace (the test cases several types of whitespace characters in between words).

Best of luck with Python!

varList = varStr.split(" ") # change this
varList = varStr.split() # to this
alenwong
alenwong
4,346 Points

It works!! Thanks Jeff :)

Jeff Muday
MOD
Jeff Muday
Treehouse Moderator 28,722 Points
Jeff Muday
Jeff Muday
Treehouse Moderator 28,722 Points

That's some nice coding there! It works in my IDE. To me, it looks like you have some background in another language (Java or Javascript). A couple of inefficiencies, but it shows you understand the concept of initialization!

We can simplify it quite a bit approaching it in a more Pythonic way. 

def word_count(varStr):
    varDict = {}
    # iterate over the split string
    for word in varStr.split():
        # the magic line of code below, the "get" method attempts to get the key,
        # if it isn't there, it returns a zero
        # learn this method and it will serve you well!
        varDict[word.lower()] = varDict.get(word.lower(), 0) + 1  
    return varDict
alenwong
alenwong
4,346 Points

Hi Jeff

Thanks for the feedback. but specifically I would like to understand why my code can't pass the "word count" test https://teamtreehouse.com/library/python-collections-2/dictionaries/word-count

I'd like to know if there's something wrong with the logic of my code, or something else.