# -*- coding: utf-8 -*-

def deflt(x):
    print('\nKey', x, 'is not in the dictionary')
    return 'Default'

dicti = {1:'one', 2:'two', 3:'three'}

key = 2
print('\ndicti \t\t', dicti)
print('\t\t key =',key)

print('get \t\t',     dicti.get(key, deflt(key)))
print('setdefault \t',dicti.setdefault(key, deflt(key)))
print('dicti \t\t',   dicti)
print('pop \t\t',     dicti.pop(key, deflt(key)))
print('dicti \t\t',   dicti)
print('setdefault \t',dicti.setdefault(key, deflt(key)))
print('dicti \t\t',   dicti)

'''
Printed outputs:

Python 3.7.1 (v3.7.1:260ec2c36a, Oct 20 2018, 14:57:15) [MSC v.1915 64 bit (AMD64)]
Type "copyright", "credits" or "license" for more information.

IPython 7.1.1 -- An enhanced Interactive Python.

runfile('C:/Temp/Smazat/Dictionary_Defaults.py', wdir='C:/Temp/Smazat')

dicti            {1: 'one', 2: 'two', 3: 'three'}
                 key = 2

Key 2 is not in the dictionary                 #  ???        
get              two

Key 2 is not in the dictionary                 #  ???
setdefault       two
dicti            {1: 'one', 2: 'two', 3: 'three'}

Key 2 is not in the dictionary                 #  ???
pop              two
dicti            {1: 'one', 3: 'three'}

Key 2 is not in the dictionary
setdefault       Default
dicti            {1: 'one', 3: 'three', 2: 'Default'}

'''

'''    
Dictionary Help :
    .get(key[, default])
        Return the value for key
        if key is in the dictionary, else default.
        If default is not given, it defaults to None,
        so that this method never raises a KeyError

    .pop(key[, default])
        If key is in the dictionary, remove it and return its value,
        else return default.
        If default is not given and key is not in the dictionary,
        a KeyError is raised.

    .setdefault(key[, default])
        If key is in the dictionary, return its value.
        If not, insert key with a value of default and return default.
        default defaults to None.

'''