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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
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import bench |
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def test(num): |
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for i in iter(range(num//10000)): |
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l = [0] * 1000 |
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l2 = list(l) |
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bench.run(test) |
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import bench |
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def test(num): |
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for i in iter(range(num//10000)): |
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l = [0] * 1000 |
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l2 = list(map(lambda x: x, l)) |
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bench.run(test) |
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import bench |
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def test(num): |
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for i in iter(range(num//10000)): |
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l = [0] * 1000 |
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l2 = tuple(l) |
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bench.run(test) |
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import bench |
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def test(num): |
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for i in iter(range(num//10000)): |
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l = [0] * 1000 |
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l2 = tuple(map(lambda x: x, l)) |
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bench.run(test) |
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import bench |
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def test(num): |
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for i in iter(range(num//10000)): |
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l = [0] * 1000 |
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l2 = bytes(l) |
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bench.run(test) |
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import bench |
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def test(num): |
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for i in iter(range(num//10000)): |
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l = [0] * 1000 |
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l2 = bytes(map(lambda x: x, l)) |
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bench.run(test) |
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import bench |
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def test(num): |
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for i in iter(range(num//10000)): |
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l = [0] * 1000 |
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l2 = bytearray(l) |
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bench.run(test) |
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import bench |
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def test(num): |
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for i in iter(range(num//10000)): |
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l = [0] * 1000 |
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l2 = bytearray(map(lambda x: x, l)) |
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bench.run(test) |
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