Peephole optimisations for tuple, list, set
- Last edited on Apr 2, 2015, 6:15:18 PM by vaultah
Thanks to Martijn Pieters' Stackoverflow Answer
The CPython interpreter replaces the second form with the first.
That’s because loading the tuple from a constant is one operation, but the list would be 3 operations; load the two integer contents and build a new list object.
Because you are using a list literal that isn’t otherwise reachable, it is substituted for a tuple:
>>> import dis
>>> dis.dis(compile('if number in [1, 2]: pass', '<stdin>', 'exec'))
1 0 LOAD_NAME 0 (number)
3 LOAD_CONST 3 ((1, 2))
6 COMPARE_OP 6 (in)
9 POP_JUMP_IF_FALSE 15
12 JUMP_FORWARD 0 (to 15)
>> 15 LOAD_CONST 2 (None)
18 RETURN_VALUE
Here the second bytecode loads a (1, 2)
tuple as a constant, in one step. Compare this to creating a list object not used in a membership test:
>>> dis.dis(compile('[1, 2]', '<stdin>', 'exec'))
1 0 LOAD_CONST 0 (1)
3 LOAD_CONST 1 (2)
6 BUILD_LIST 2
9 POP_TOP
10 LOAD_CONST 2 (None)
13 RETURN_VALUE
Here N+1 steps are required for a list object of length N.
This substitution is a CPython-specific peephole optimisation; see the Python/peephole.c
source. For other Python implementations then, you want to stick with immutable objects instead.
That said, the best option when using Python 3.2 and up, is to use a set literal:
if number in {1, 2}:
as the peephole optimiser will replace that with a frozenset()
object and membership tests against sets are O(1) constant operations:
>>> dis.dis(compile('if number in {1, 2}: pass', '<stdin>', 'exec'))
1 0 LOAD_NAME 0 (number)
3 LOAD_CONST 3 (frozenset({1, 2}))
6 COMPARE_OP 6 (in)
9 POP_JUMP_IF_FALSE 15
12 JUMP_FORWARD 0 (to 15)
>> 15 LOAD_CONST 2 (None)
18 RETURN_VALUE
This optimization was added in Python 3.2 but wasn’t backported to Python 2.
As such, the Python 2 optimiser doesn’t recognize this option and the cost of building either a set
or frozenset
from the contents is almost guaranteed to be more costly than using a tuple for the test.