I wanted to create a list with some initialized values, because an empty list is not an option in python. So I started thinking which would be faster:
l = [None for i in range(1000)]
or
l = [None] * 1000
And I tried testing it with with timeit:
In [56]: timeit.timeit('l = [None] * 1000', number=10000)
Out[56]: 0.04936316597741097
In [58]: timeit.timeit('l = [None for i in range(1000)]', number=10000)
Out[58]: 0.2318978540133685
I was surprised that [None] * 1000 is faster.
I assume you are using CPython. Let's compare the generated Python bytcode with the dis module. This is the first version:
>>> import dis
>>> def f():
... return [None] * 1000
>>> dis.dis(f)
2 0 LOAD_CONST 0 (None)
2 BUILD_LIST 1
4 LOAD_CONST 1 (1000)
6 BINARY_MULTIPLY
8 RETURN_VALUE
This is pretty clear: a list [None] is built (lines 0-2), and multiplied by 1000 (lines 4-6).
This is the second version:
>>> def g():
... return [None for _ in range(1000)]
>>> dis.dis(g)
2 0 LOAD_CONST 1 (<code object <listcomp> at ..., file "<doctest __main__[3]>", line 2>)
2 LOAD_CONST 2 ('g.<locals>.<listcomp>')
4 MAKE_FUNCTION 0
6 LOAD_GLOBAL 0 (range)
8 LOAD_CONST 3 (1000)
10 CALL_FUNCTION 1
12 GET_ITER
14 CALL_FUNCTION 1
16 RETURN_VALUE
This is more complex: a function named g.<locals>.<listcomp> (line 2) and with a code we'll see below (line 0) is created (line 4). A range(1000) is built (lines 6-8-10) and an iterator created (line 12). This iterator is passed to the g.<locals>.<listcomp> function (line 14) and the result returned (line 16).
Let's look at the g.<locals>.<listcomp> function:
>>> dis.dis(g.__code__.co_consts[1])
2 0 BUILD_LIST 0
2 LOAD_FAST 0 (.0)
>> 4 FOR_ITER 8 (to 14)
6 STORE_FAST 1 (_)
8 LOAD_CONST 0 (None)
10 LIST_APPEND 2
12 JUMP_ABSOLUTE 4
>> 14 RETURN_VALUE
An empty list is created (line 0), the iterator parameter (iter(range(1000))) is pushed on the stack (line 2), and a for loop begins (line 4). The value of the loop index (_) is stored inb the local array (line 6) and None is appended to the list (lines 8-10), until the loop ends (line 12 that loops to line 4).
To summarize:
The second version is indeed slower.
Note Beware the usual pitfall
>>> A = [[0]] * 3
>>> A
[[0], [0], [0]]
>>> A[0].append(1)
>>> A
[[0, 1], [0, 1], [0, 1]]
But:
>>> A = [[0] for _ in range(3)]
>>> A
[[0], [0], [0]]
>>> A[0].append(1)
>>> A
[[0, 1], [0], [0]]
If you wonder why, look at the bytecodes above.
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