Browse Source
Both "bound" (like, length known) and "unbound" (length unknown) are tested. All of list, tuple, bytes, bytesarray offer approximately the same performance, with "unbound" case being 30 times slower.pull/706/merge
Paul Sokolovsky
11 years ago
8 changed files with 64 additions and 0 deletions
@ -0,0 +1,8 @@ |
|||
import bench |
|||
|
|||
def test(num): |
|||
for i in iter(range(num//10000)): |
|||
l = [0] * 1000 |
|||
l2 = list(l) |
|||
|
|||
bench.run(test) |
@ -0,0 +1,8 @@ |
|||
import bench |
|||
|
|||
def test(num): |
|||
for i in iter(range(num//10000)): |
|||
l = [0] * 1000 |
|||
l2 = list(map(lambda x: x, l)) |
|||
|
|||
bench.run(test) |
@ -0,0 +1,8 @@ |
|||
import bench |
|||
|
|||
def test(num): |
|||
for i in iter(range(num//10000)): |
|||
l = [0] * 1000 |
|||
l2 = tuple(l) |
|||
|
|||
bench.run(test) |
@ -0,0 +1,8 @@ |
|||
import bench |
|||
|
|||
def test(num): |
|||
for i in iter(range(num//10000)): |
|||
l = [0] * 1000 |
|||
l2 = tuple(map(lambda x: x, l)) |
|||
|
|||
bench.run(test) |
@ -0,0 +1,8 @@ |
|||
import bench |
|||
|
|||
def test(num): |
|||
for i in iter(range(num//10000)): |
|||
l = [0] * 1000 |
|||
l2 = bytes(l) |
|||
|
|||
bench.run(test) |
@ -0,0 +1,8 @@ |
|||
import bench |
|||
|
|||
def test(num): |
|||
for i in iter(range(num//10000)): |
|||
l = [0] * 1000 |
|||
l2 = bytes(map(lambda x: x, l)) |
|||
|
|||
bench.run(test) |
@ -0,0 +1,8 @@ |
|||
import bench |
|||
|
|||
def test(num): |
|||
for i in iter(range(num//10000)): |
|||
l = [0] * 1000 |
|||
l2 = bytearray(l) |
|||
|
|||
bench.run(test) |
@ -0,0 +1,8 @@ |
|||
import bench |
|||
|
|||
def test(num): |
|||
for i in iter(range(num//10000)): |
|||
l = [0] * 1000 |
|||
l2 = bytearray(map(lambda x: x, l)) |
|||
|
|||
bench.run(test) |
Loading…
Reference in new issue