CODEWITHSUNDEEP
python
Jack
Jack

Reputation: 3057

New to python and confused about get() returning None

So the task i am trying to accomplish is for two string inputs such as '1' + '2' to return '3',

I would like to be able to be able to do something like this,

d = {'1': 1, '2': 2, '3': 3}

So i have a dictionary like this ^, then I can do 

d.get('1')

In hopes it would return 1, except it returns None, How would I work around this?

Thank you for the help

So thanks to your help i got it to work sort of, although for some reason it only accepts digits which add to 4 and lower, Here is the code so you may better understand

def code_char(char, key):
    d = {'1': 1, '2':2 ,'3': 3 ,'4': 4,'5': 5 ,'6': 6 ,'7': 7 ,'8' :8 ,'9':9}
    f = {1: '1', 2: '2' ,3: '3' ,4: '4' ,'5':5 ,'6':6 ,'7':7 ,'8':8 ,'9':9}
    sum = d.get(char)+d.get(key)
    if sum < 9:
        print(f.get(sum))
    else:
        sum = sum % 10
        value = f.get(sum)
        print(value)
code_char('1','5')

For some reason code+char('1','3') will correctly return 3 but any higher and it will just print None.

It's the start to my encrypter, thanks for the help so far!

Upvotes: 0

Views: 103

Answers (3)

dawg
dawg

Reputation: 103844

You can use braces:

>>> d = {'1': 1, '2': 2, '3': 3}
>>> d['1']+d['2']
3

Or, if you want a string result:

>>> str(d['1']+d['2'])
'3'

But really -- no mapping is required to do this:

>>> str(int('22')+int('33'))
'55'

Upvotes: 0

Patricio Molina
Patricio Molina

Reputation: 390

This is how it works: D.get(k[,d=None]) -> D[k] if k in D else d. Be sure to use '1' instead of 1.

Upvotes: 0

Evgeny
Evgeny

Reputation: 3274

No idea why it returns None for you. Is that what you actually tried?

> d = {'1': 1, '2': 2, '3': 3}
> d.get('1')
1

Btw, you can turn string to int by int function.

> int('1') + int('2')
3

Upvotes: 1

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