12.03.20 17:02, Eric Wieser пише:
>>> [id(v) for v in itertools.combinations([1, 2, 3], 1)]
[2500926200992, 2500926199072, 2500926200992]
Note that the first id is repeated for third item. And if you use larger
example
>>> [id(v) for v in itertools.combinations(range(10), 1)]
[139701363452336, 139701363451616, 139701363452336, 139701363451616,
139701363452336, 139701363451616, 139701363452336, 139701363451616,
139701363452336, 139701363451616]
you will see that two ids are interleaving.
There are many causes of reusing the id. First, the combinations()
iterator keeps a reference to a tuple and reuse it if it is free.
Second, the tuple allocator keeps a pool of preallocated objects for
small tuples. Third, the Python memory allocator keeps a pool of small
blocks. Fourth, the libc memory allocator also more likely return the
same memory address if repeatedly allocate and free the same amount of
memory.
All this is optimizations and implementation artifacts. If you are
worring that the second tuple produced by combinations() uses more
slower cache, you can make a first-level cache in combinations() keeping
references to two tuples. Or make the second-level cache more efficient.
Of cause if you have evidences that there is a problem and that this
solution fixes it. Did you do any benchmarks?
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