Read list word for word, only reading the last

user_10722339 0
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I have a string that I'm turning into a list. I then count the words in that list and I'm trying to use that number (length) as a counter so that I can add each word to a dictionary with a number as their key. It seems to enumerate correctly but it's only reading the last entry on the list instead of word for word.

Any idea why it's only reading the last entry on the list i.e. 'you?'?

mystring = "Hello mate how are you?"

mylist = mystring.split()
lengh = len(mylist)

class my_dictionary(dict): 

    # __init__ function 
    def __init__(self): 
        self = dict() 

    # Function to add key:value 
    def add(self, key, value): 
        self[key] = value 



# Main Function 
dict_obj = my_dictionary() 
for x in range(lengh):
    for line in mylist:
        for word in line.split():
            dict_obj.add(x, line)

print(dict_obj)

output is:

{0: 'you?', 1: 'you?', 2: 'you?', 3: 'you?', 4: 'you?'}

Output should be:

{0: 'Hello?', 1: 'mate?', 2: 'how', 3: 'are', 4: 'you?'}
python-3.x python
2 Answers

You are making things a little too complex. You are looping over your whole list for every number in the length. You can just use enumerate() to get the index:

mystring = "Hello mate how are you?"

mylist = mystring.split()

class my_dictionary(dict): 

    def add(self, key, value): 
        self[key] = value 

# Main Function 
dict_obj = my_dictionary() 
for x, word in enumerate(mylist):
    dict_obj.add(x, word)

print(dict_obj)
# {0: 'Hello', 1: 'mate', 2: 'how', 3: 'are', 4: 'you?'}

In reality almost none of the above code is needed and could be refactored out. This code does the same thing:

mystring = "Hello mate how are you?"
words = mystring.split()

dict_obj = dict(enumerate(words))

print(dict_obj)
# {0: 'Hello', 1: 'mate', 2: 'how', 3: 'are', 4: 'you?'}
user_38746230

You want to change your code to be something like this

# Main Function
dict_obj = my_dictionary()
x = 0
for line in mylist:
    dict_obj.add(x, line)
    x += 1
user_23722100
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