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

huyen nguyen
huyen nguyen
850 Points
Posted by huyen nguyen
huyen nguyen
850 Points

Please help me debug this code

Thank a lot

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.
a_string=string.split()
def word_count(string):
    my_dict = {}
    string.lower()
    new_str = string.split(' ')
    for words in new_str:
        if words in my_dict:
            my_dict[word] += 1
        else:
            my_dict[word] = 1
    return my_dict
print(word_count(string))
Shaohui Yang
Shaohui Yang
5,009 Points

words => word you have a typo

1 Answer

William Li
PLUS
William Li
Courses Plus Student 26,868 Points
William Li
William Li
Courses Plus Student 26,868 Points

The overall logic of your code is spot-on, however, there're some small bugs here and there.

  • As Shaohui Yang points out, there're inconsistent uses of words and word in your code, this is obviously a typo.
  • string.lower(), calling the lower() method on string merely returns a copy of string, it has no impact on string's value.
  • Lastly, .split(" ") when a single empty space is provided as the argument to the split function, Python will only use one space as the delimiter for splitting the string into word. But if you call the split() without argument, the function will use any number of consecutive spaces (" ", "\t" ... etc) as splitting delimiter. Doing so will give your split function little flexibility to handle the tricky test cases used by the grader.
def word_count(string):
  my_dict = {}
  new_str = string.lower().split() # combine this and previous line into one
  for word in new_str:
      if word in my_dict:
          my_dict[word] += 1
      else:
          my_dict[word] = 1
  return my_dict

def word_count(string):
  my_dict = {}
  for word in string.lower().split():  # or you can combine 3 lines into one, making it one less local variable
      if word in my_dict:
          my_dict[word] += 1
      else:
          my_dict[word] = 1
  return my_dict

hope it helps.