[Tut] How to Initialize Multiple Variables to the Same Value in Python?

[Tut] How to Initialize Multiple Variables to the Same Value in Python? - Printable Version

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[Tut] How to Initialize Multiple Variables to the Same Value in Python? - xSicKxBot - 06-30-2022

How to Initialize Multiple Variables to the Same Value in Python?

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Summary: To initialize multiple variables to the same value in Python you can use one of the following approaches:

Use chained equalities as:
Code:
var_1 = var_2 = value
Use
Code:
dict.fromkeys

This article will guide you through the ways of assigning multiple variables with the same value in Python. Without further delay, let us dive into the solutions right away.

[b]Method 1: Using Chained Equalities[/b]

You can use chained equalities to declare the variables and then assign them the required value.

Code:
Syntax: variable_1 = variable_2 = variable_3 = value

Code:

Code:
x = y = z = 100
print(x)
print(y)
print(z)
print("All variables point to the same memory location:")
print(id(x))
print(id(y))
print(id(z))

Output:

Code:
100
100
100
All variables point to the same memory location:
3076786312656
3076786312656
3076786312656

It is evident from the above output that each variable has been assigned the same value and each of them point to the same memory location.

[b]Method 2: Using dict.fromkeys[/b]

Approach: Use the

Code:
dict.fromkeys(variable_list, val)

method to set a specific value (

Code:
val

) to a list of variables (

Code:
variable_list

).

Code:

Code:
variable_list = ["x", "y", "z"]
d = dict.fromkeys(variable_list, 100)
for i in d: print(f'{i} = {d[i]}') print(f'ID of {i} = {id(i)}')

Output:

Code:
x = 100
ID of x = 2577372054896
y = 100
ID of y = 2577372693360
z = 100
ID of z = 2577380842864

Discussion: It is evident from the above output that each variable assigned holds the same value. However, each variable occupies a different memory location. This is on account that each variable acts as a key of the dictionary and every key in a dictionary is unique. Thus, changes to a particular variable will not affect another variable as shown below:

Code:
variable_list = ["x", "y", "z"]
d = dict.fromkeys(variable_list, 100)
print("Changing one of the variables: ")
d['x'] = 200
print(d)

Output:

Code:
{'x': 200, 'y': 100, 'z': 100}

Conceptual Read:

Code:
fromkeys()

is a dictionary method that returns a dictionary based on specified keys and values passed within it as parameters.

Syntax:

Code:
dict.fromkeys(keys, value)

keys is a required parameter that represents an iterable containing the keys of the new dictionary.

value is an optional parameter that represents the values for all the keys in the new dictionary. By default, it is

Code:
None

.

Example:

Code:
k = ('key_1', 'key_2', 'key_3')
my_dictionary = dict.fromkeys(k, 0)
print(my_dictionary) # OUTPUT: {'key_1': 0, 'key_2': 0, 'key_3': 0}

Related Question

Let’s address a frequently asked question that troubles many coders.

Problem: I tried to use multiple assignment as show below to initialize variables, but I got confused by the behavior, I expect to reassign the values list separately, I mean b[0] and c[0] equal 0 as before.

Code:
a=b=c=[0,3,5]
a[0]=1
print(a)
print(b)
print(c)

Output:

Code:
a = b = c = [0, 3, 5]
a[0] = 1
print(a)
print(b)
print(c)

But, why does the following assignment lead to a different behaviour?

Code:
a = b = c = 5
a = 3
print(a)
print(b)
print(c)

Output:

Code:
3
5
5

Question Source: StackOverflow

[b]Solution[/b]

Remember that everything in Python is treated as an object. So, when you chain multiple variables as in the above case all of them refer to the same object. This means,

Code:
a

Code:
b

and

Code:
c

are not different variables with same values rather they are different names given to the same object.

[Image: image-169.png]

Thus, in the first case when you make a change at a certain index of variable a, i.e, a[0] = 1. This means you are making the changes to the same object that also has the names b and c. Thus the changes are reflected for b and c both along with a.

Verification:

Code:
a = b = c = [1, 2, 3]
print(a[0] is b[0]) # True

To create a new object and assign it, you must use the copy module as shown below:

Code:
import copy
a = [1, 2, 3]
b = copy.deepcopy(a)
c = copy.deepcopy(a)
a[0] = 5
print(a)
print(b)
print(c)

Output:

Code:
[5, 2, 3]
[1, 2, 3]
[1, 2, 3]

However, in the second case you are rebinding a different value to the variable

Code:
a

. This means, you are changing it in-place and that leads to a now pointing at a completely different value at a different location. Here, the value being changed is an interger and integers are immutable.

Follow the given illustration to visualize what’s happening in this case:

[Image: img.gif]

Verification:

Code:
a = b = c = 5
a = 3
print(a is b)
print(id(a))
print(id(b))
print(id(c))

Output:

Code:
False
2329408334192
2329408334256
2329408334256

It is evident that after rebinding a new value to the variable

Code:
a

, it points to a different memory location, hence it now refers to a different object. Thus, changing the value of

Code:
a

in this case means we are creating a new object without touching the previously created object that was being referred by

Code:
a

Code:
b

and

Code:
c

.

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