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Python Python Collections (Retired) Dictionaries Word Count

Posted by Sachin Gupta
Sachin Gupta
130 Points

How to solve this? How can I process a local variable outside its scope?

Suppose I want to count the words in a loop and print it outside the loop, I am not able to access it? Please help me to debug this code.

def word_count(string):
      mydict = {}
      mylist = string.split()
      for item in mylist:
        count = 0
        for key in string:
          if(key == item):
            count += 1
        mydict[item] = count
      return mydict  

dict = word_count("Racism is bad is really bad")
print dict

[MOD: added ```python formatting -cf]

1 Answer

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,441 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,441 Points

In your code there are two scopes: the main (top), and inside word_count. Code inside word_count can access variables in both of these namespaces, code outside of word_count can only access the variables in the global namespace.

Global name space: word_count function, dict dictionary Local word_count namespace: mydict dictionary, mylist list, count int, key string.

Within word_count, after the loops have run, but before the return, all of the local variables have their last value.

The last value of key is "d", count is 0. Why is key last value equal to "d"? There are some issues with your code. You are comparing the characters in the string to the words in the string. Will only match single character words.

def word_count(string):
    mydict = {}  #<-- no need to initialize dict
    # split string into word list
    mylist = string.split()
    # loop using each word in wordlist
    for item in mylist:
        # Initialize count
        count = 0
        # loop over each character in string
        for key in string:  #<-- not what you want: try using "string.split()" or "mylist"
            if key == item:  #<-- parens not needed
                # increment count for each matching word
                count += 1
        # update dict with word count results
        mydict[item] = count
    # return dict of all results
    return mydict  

dict = word_count("Racism is bad is really bad")
print dict

Notice this is a bit inefficient as you are scanning the word list with every word in the list. So you would be counting all the "is" words twice. Can you think of a way to loop through the word list one and simply increment the count of each word seen? Remember to test if the word has been see before (hint: try "if item in mydict" to see if the word is there. increment if yes, create new dict item using mydict[item] = 1)

Sachin Gupta
Sachin Gupta
130 Points

Thank you Chris ! I implemented your suggestions and it worked ! But I am still not able to figure out how can I use try and except here. My improved code is :

def word_count(string):
    mydict = {}
    my_list = string.split()
    for item in my_list :
            count = 0
            if item in mydict:
                continue
            for word in my_list:
                if word==item:
                    count += 1

            mydict[item] = count    
    return mydict
dict = word_count("This Answer is Awesome ! Really it is Awesome")
print(dict)

Thank you once again!!

[MOD: wrapped code in ```python formatting -cf]

Chris Freeman
Chris Freeman
Treehouse Moderator 68,441 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,441 Points

Here is how you can use a try statement:

def word_count(string):
    mydict = {}
    my_list = string.split()
    for item in my_list :
        try:
            # increment count if item in dict
            mydict[item] += 1
        except KeyError:
            # item not yet in dictionary, init with count of 1
            mydict[item] = 1
    return mydict

dict = word_count("This Answer is Awesome ! Really it is Awesome")
print(dict)

Returns

{'!': 1, 'This': 1, 'is': 2, 'it': 1, 'Answer': 1, 'Awesome': 2, 'Really': 1}