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

Posted by Adam Stapleton
Adam Stapleton
1,693 Points

My Code Works In Terminal But Isn't Being Accepted, What Can I Do?

I tested the code as a python script running Python 2.7 from terminal, and it returns a dictionary with the keys as the lower case'd words in the string argument, and the values as the count of the occurrences of that word.

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(stringy):
    dic={}
    listy=stringy.lower().split(" ")
    for el in listy:
        if not el in dic:
            dic[el]=1
        else:
            dic[el]+=1
    return dic
Marcus Sassi
Marcus Sassi
4,394 Points

Your code shouldn't be working because

else:
       dic[el] += 1

Should give you ValueError and that is also where the solution are to be found. Below you can find my code which works, please study it because you are very close.

myword ="I do not like it Sam I Am"

def word_count(sentence):
  my_dict = {}
  sentence = sentence.lower().split()
  for sen in sentence:
    if sen in my_dict:
      count = my_dict[sen] + 1
      my_dict[sen] = count
    else:
      my_dict[sen] = 1
  return my_dict