Python : Default Dictionary Initialization

Question

I have created a default dictionary as below :

from collections import defaultdict
dd = defaultdict(lambda : "Key not    found")
dd = {'a':1,'b':2}
print(dd) 
print(dd['a']) # Prints 1
print(dd['c'])  # Throws KeyError

But, the below code snippet works :

from collections import defaultdict
(lambda : "Key not    found")
dd['a']=1
dd['b']=2
print(dd) 
print(dd['a']) # Prints 1
print(dd['c'])  # Prints "Key not found"

Could anyone please explain me why the first code snippet throws error whereas the second one runs fine as expected..

Answer 1

You have overwritten dd = defaultdict(lambda : "Key not found") with dd = {'a':1,'b':2} so that defaultdict() has became a dict().

Answer 2

Please consider to delete the line 3 in the first snippet. That has overwritten your dd whose type is defaultdict.

Answer 3

The problem is in the below two lines of your code snippet -

dd = defaultdict(lambda : "Key not    found")
dd = {'a':1,'b':2}

In the first line, you are creating a defaultdict object and dd is pointing to this object -

In the second line, you are creating a dictionary and now, saving its reference to dd - So, now as you can see in the picture above that dd is pointing to a new dictionary created in second line. And since, this new dictionary which dd is referring to is not having a key named 'c', you are getting a Key not found error.