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 Dictionary in Python
#1
Hi,

I am new to Python so please bear me for this simple question but I really could n't find reason for this.

current = {}
new_dict = current

print("Current Dict:",current)
print("New Dict:",new_dict)


current[1]={}

print("Current Dict:",current)
print("New Dict:",new_dict)
Here is the output of above program.

Current Dict: {}
New Dict: {}

Current Dict: {1: {}}
New Dict: {1: {}}

I understand I am assigning new_dict to current so first time the new_dict is also empty dictionary. But after I assign current[1]={}, why the value of new_dict also got changed?

The reason why I am confused is here is a simple variable assignment program

a=0
b=a
print(a)
print(b)
print('-------------------------------------------')
a=10
print(a)
print(b)

and the output is below

0
0
-------------------------------------------
10
0


The value of 'b' does not get changed even after I assign a=10.

Why there is a different behavior for dictionary
Quote
#2
dict, list, set (and those I've forgotten) are mutable objects.
int, float, str, tuple are immutable objects.

In your second example you assign 0 to a name. The term variable is wrong, because it's a name which holds a reference to the object in memory. First the numeric literal is created as new object in memory and the reference to it is assigned to the name.

The line b = a does following:
  1. Get the reference of the name a
  2. Assign the reference to b

The line a = 10 does following:
  1. Create a new int with the value 10 in memory. (values from -2 - something are preassigned during interpreter startup)
  2. Assign the reference to a
  3. The name a has now the reference to the object which has the value 10


To make a copy from dicts you could use:
my_d1 = {}
my_d2 = my_d1.copy()
With lists:
my_list1 = list(range(10))
my_list2 = my_list1[:]
# get all references from my_list1 and assign them to my_list2
There is also the module copy with the functions copy and deepcopy.
The use case for deepcopy is for example to copy a nested list/dict or something else, which holds elements, which are also containers/sequences.

Replacing the content inside a list with new content:
my_list1 = list(range(10))
print(id(my_list1))
my_list1[:] = list(range(20))
print(id(my_list1))
# no change of ID, this means it is still the same object
With the built-in function id you get the virtual memory address back. Same value means same object.
If you understood this, you should dig deeper into it.

For example beginners are confused with is and ==.
Only use is, if you want to check, if you have the same object.
If you want to check for equality, use ==.
scidam, stullis, snippsat like this post
My code examples are always for Python >=3.6.0
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Quote
#3
I still have n't the answer for my original question. Can someone help answer it?

My question is Why new_dict is changed when I assign current[1]={}, ideally it has to hold an empty dictionary which was assigned originally from 'current'
Quote
#4
(Dec-07-2019, 12:29 PM)kartheek Wrote: I still have n't the answer for my original question. Can someone help answer it?
@DeaD_EyE has aswers it,maybe you did not understand his explanation.
Both point to same object in memory,so all changes will affect both.
>>> current = {}
>>> new_dict = current
>>> id(current)
201281040
>>> id(new_dict)
201281040
To copy the mutable types like dictionaries,use copy/deepcopy.
>>> from copy import deepcopy
>>> 
>>> current = {}
>>> new_dict = deepcopy(current)
>>> id(current)
203313200
>>> id(new_dict)
127766128
from copy import deepcopy

current = {}
new_dict = deepcopy(current)
print("Current Dict:",current)
print("New Dict:",new_dict)
current[1]={}
print("Current Dict:",current)
print("New Dict:",new_dict)
Output:
Current Dict: {} New Dict: {} Current Dict: {1: {}} New Dict: {}
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