Working with Lists

 Object

In general, anything that can be assigned to a variable in Python is referred to as an object.

Strings, Integers, Floats, Lists etc. are all objects.

Examples

• "A"
• 1.25
• [1,2,3]

Identity of an Object

Whenever an object is created in Python, it will be given a unique identifier (id).

This unique id can be different for each time you run the program.

Every object that you use in a Python Program will be stored in Computer Memory.

The unique id will be related to the location where the object is stored in the Computer Memory.

Finding Id

We can use the

id()

to find the id of a object.

Code

1

print(id("Hello"))

PYTHON

Output

140589419285168

Id of Lists

1

2

3

4

list_a = [1, 2, 3]

list_b = [1, 2, 3]

print(id(list_a))

print(id(list_b))

PYTHON

Output

139637858236800

139637857505984

Modifying Lists

Modifying Lists - 1

When assigned an existing list both the variables

list_a

and

list_b

will be referring to the same object.

Code

1

2

3

4

list_a = [1, 2, 3]

list_b = list_a

print(id(list_a))

print(id(list_b))

PYTHON

Output

140334087715264

140334087715264

Modifying Lists - 2

When assigned an existing list both the variables

list_a

and

list_b

will be referring to the same object.

Code

1

2

3

4

5

list_a = [1, 2, 3, 5]

list_b = list_a

list_b[3] = 4

print("list a : " + str(list_a))

print("list b : " + str(list_b))

PYTHON

Output

list a : [1, 2, 3, 4]

list b : [1, 2, 3, 4]

Modifying Lists - 3

The assignment will update the reference to new object.

Code

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2

3

4

5

list_a = [1, 2]

list_b = list_a

list_a = [6, 7]

print("list a : " + str(list_a))

print("list b : " + str(list_b))

PYTHON

Output

list a : [6, 7]

list b : [1, 2]

Modifying Lists - 4

The assignment will update the reference to a new object.

Code

1

2

3

4

5

list_a = [1, 2]

list_b = list_a

list_a = list_a + [6, 7]

print("list a : " + str(list_a))

print("list b : " + str(list_b))

PYTHON

Output

list a : [1, 2, 6, 7]

list b : [1, 2]

Modifying Lists - 5

Compound assignment will update the existing list instead of creating a new object.

Code

1

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3

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list_a = [1, 2]

list_b = list_a

list_a += [6, 7]

print("list a : " + str(list_a))

print("list b : " + str(list_b))

PYTHON

Output

list a : [1, 2, 6, 7]

list b : [1, 2, 6, 7]

Modifying Lists - 6

Updating mutable objects will also effect the values in the list, as the reference is changed.

Code

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list_a = [1,2]

list_b = [3, list_a]

list_a[1] = 4

print(list_a)

print(list_b)

PYTHON

Output

[1, 4]

[3, [1, 4]]

Modifying Lists - 7

Updating immutable objects will not effect the values in the list, as the reference will be changed.

Code

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5

a = 2

list_a = [1,a]

print(list_a)

a = 3

print(list_a)

PYTHON

Output

[1, 2]

[1, 2]