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

Posted by Aryan Monga
Aryan Monga
4,061 Points

A quiz question!

Can i get to know what is wrong with my code? Thanks!

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(a):
    dictonary={}
    count= 1
    a=a.lower()
    lista=a.split(" ")
    for word in lista:
        if word in dictonary:
            dictonary[word]= count + 1
        else:
            dictonary.append(word)
            dictionary[word]=1
    print dictonary

1 Answer

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

I spot a few issues:

  • to split on any whitespace, leave the argument to split empty
  • you spelled "dictionary" two different ways (also "dictonary") — either will work but it must be consistent
  • the "count" variable is not needed or useful, just increment the directory entry itself
  • you can increment this way "dictonary[word] += 1"
  • dictionaries don't have (or need) an append method
  • the assignment of the value to the dictionary key will create a new entry by itself
  • you need to return the result, you won't need to print anything