Create a function named word_count() that takes a string. Return a dictionary with each word in the string as the key an

Question

where am l getting this wrong.'i' whats that??

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.
a_string="I am that I am"
a_stringsplit=(a_string.lower()).split()
def word_count(a_string):
    string_dict={}
    for word in a_stringsplit:
      if string_dict[word]:
        string_dict += 1
      else:
        string_dict[word] = 1
    return string_dict

Answer 1 · 2015-06-08T21:03:11Z

June 8, 2015 9:03pm

Hello Ronald.

You are not too far but still I see a few mistakes here.

Please note that you do not need to hard code the string itself, it will automatically be provided to the compiler. So get rid of the 2 lines outside the function.

Then, inside the for loop, you will need to go through the list we get when splitting the string. So something like for word in a_string.split() would do.

Lastly the if statement needs to check if the word is in the dictionary, so "if word in string_dict:" Also, please note that in both cases (if and else) what we need to target is the right key in the dictionary, so in both cases we would go with string_dict[word] like you did in the else statement.

These little changes should do the trick, let me know if you manage to pass the test!

;)

Vittorio

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MUZ141005 Ronald Maravanyika

MUZ141005 Ronald Maravanyika

Create a function named word_count() that takes a string. Return a dictionary with each word in the string as the key an

1 Answer

Vittorio Somaschini

Vittorio Somaschini

MUZ141005 Ronald Maravanyika

MUZ141005 Ronald Maravanyika

Vittorio Somaschini

Vittorio Somaschini