[issue23290] Faster set copying

[Serhiy Storchaka]

Serhiy Storchaka added the comment:

Here is even faster patch. When there are no dummies in source set we can just dump a table, placing entries at the same indices.

$ ./python -m timeit -s "s = set(range(10**4))" -- "frozenset(s)"
Unpatched: 1000 loops, best of 3: 658 usec per loop
Patched: 1000 loops, best of 3: 631 usec per loop

$ ./python -m timeit -s "s = {i+(j<<64) for i in range(10**3) for j in range(10)}" -- "frozenset(s)"
Unpatched: 100 loops, best of 3: 6.72 msec per loop
Patched: 1000 loops, best of 3: 930 usec per loop

$ ./python -m timeit -s "s = {i+(j<<64) for i in range(10**2) for j in range(10**2)}" -- "frozenset(s)"
Unpatched: 100 loops, best of 3: 14 msec per loop
Patched: 1000 loops, best of 3: 1.12 msec per loop

To test other branch we should add dummy entry: s.add(-1); s.discard(-1).

$ ./python -m timeit -s "s = {i+(j<<64) for i in range(10**3) for j in range(10)}; s.add(-1); s.discard(-1)" -- "frozenset(s)"
Unpatched: 1000 loops, best of 3: 661 usec per loop
Patched: 1000 loops, best of 3: 643 usec per loop

$ ./python -m timeit -s "s = {i+(j<<64) for i in range(10**3) for j in range(10)}; s.add(-1); s.discard(-1)" -- "frozenset(s)"
Unpatched: 100 loops, best of 3: 6.8 msec per loop
Patched: 100 loops, best of 3: 2.1 msec per loop

$ ./python -m timeit -s "s = {i+(j<<64) for i in range(10**2) for j in range(10**2)}; s.add(-1); s.discard(-1)" -- "frozenset(s)"
Unpatched: 100 loops, best of 3: 14 msec per loop
Patched: 100 loops, best of 3: 2.71 msec per loop

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Added file: http://bugs.python.org/file37808/set_faster_copy_2.patch

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<http://bugs.python.org/issue23290>
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