CODEWITHSUNDEEP
pythonpython-3.x
Sreeram TP
Sreeram TP

Reputation: 11917

Python None list size

I was just checking the size of some datatypes in Python 3 and I observed this.

import sys

val = None
print(sys.getsizeof(val))

The output was 16 as expected.

I tried making a list of 1000 locations of None and I expected the size to be 16*1000 = 16000 or more. But the result I got was different.

import sys

val = [None]*1000

print(sys.getsizeof(val))

The output was 8064. Nearly the half of the size I expected.

What is the reason for this.? Why memory allocated is less.?

Upvotes: 2

Views: 1908

Answers (2)

kartheek
kartheek

Reputation: 696

import sys

val = None

print(sys.getsizeof(val))

Answer:

16

val = []

print(sys.getsizeof(val))

Answer: 

72

val = [None]

print(sys.getsizeof(val))

Answer:

80

so [None]*1000 = 1000* 8 + 72 = 8072

Note: No of bytes may vary depending on the environment

Upvotes: 6

Florian Weimer
Florian Weimer

Reputation: 33719

There is just a single None object referenced a thousand times. So the situation is this:

l[0]   ----> None
         /    ^
l[1]   -/     |
….            |
l[999]  -----/

And not this:

l[0]   ----> None

l[1]   ----> None
….
l[999] ----> None

This is more visible when repeating a mutable object, like this:

>>> l = [set()] * 3
>>> print(l)
[set(), set(), set()]
>>> l[0].add(1)
>>> print(l)
[{1}, {1}, {1}]

There is just a single, shared set object referenced three times, so changes to the set at l[0] also affect l[1] and l[2].

Python data structures such as list, set and dict are reference-based. In your case, most of the 8064 bytes you observed come from the object references (8 bytes per list element).

Upvotes: 5

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