word_count is causing a few problems!!

Question

This is my code, Any tips!!
wordsList = "I am that I am"
wordsList = wordsList.split()
print(wordsList)
def wordCount(wordsList):
    count = 0
    key =""
    wordCounter ={key: count}
    for word in wordsList:
        for key in wordCount:
            if key != word:
                wordCount[key]=word
                wordCount[count]=1
            else:
                count = wordCount.get(key, count)
                count = count + 1
                wordCount.update(key, count)
    return wordCount
print(wordCount(wordsList))
```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.
```

Steve Horton · Answer

This code below works when I compile it in pycharm, its lists the works and next to a count of how many times the word appears, but in the exercise screen said that some some the words are missing.
wordsList = "I am that I am"
wordsList = wordsList.lower()
wordsList = wordsList.split()
def wordcount(wordsList):
    worddict ={}
    count = 0
for word in wordsList:
    if word in word_dict.keys():
        count = word_dict.get(word, count)
        word_dict[word]=count+1

else:
        word_dict[word]=count+1

return word_dict
print(word_count(wordsList))

Steven Parker · Answer

It's a bit hard to read the unquoted code, but it looks like you have a good bit of work to do:
- you don't need to create a test string, the challenge will pass one to the function
- you don't need to print anything
- you only need to define the function, not call it
- the challenge asks for the function to be named "word_count" (not "wordCount")
- you need to declare  your return dictionary before you assign anything to it
- you might want to split the string to begin with
- you will probably want to use the challenge's hint about "if word in dict"
- you can manage the counts directly in the dictionary, you won't need a local variable
- the "word" is your key
Hopefully, those hints will help get you on the right track.

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Steve Horton

Steve Horton

word_count is causing a few problems!!

2 Answers

Steven Parker

Steven Parker

Steve Horton

Steve Horton