Why is `{*l}` faster than `set(l)` - python sets (not really only for sets, for all sequences)

问题

So here is my timings:

>>> import timeit
>>> timeit.timeit(lambda: set(l))
0.7210583936611334
>>> timeit.timeit(lambda: {*l})
0.5386332845236943

Why is that, my opinion would be equal but it's not.

So unpacking is fast from this example, right?

回答1:

For the same reason [] is faster than list(); the interpreter includes dedicated support for syntax based operations that uses specialized code paths, while constructor calls involve:

1. Loading the constructor from built-in scope (requires a pair of dict lookups, one in global scope, then another in built-in scope when it fails)
2. Requires dispatch through generic callable dispatch mechanisms, and generic argument parsing code, all of which is far more expensive than a single byte code that reads all of its arguments off the stack as a C array

All of these advantages relate to fixed overhead; the big-O of both approaches are the same, so {*range(10000)} won't be noticeably/reliably faster than set(range(10000)), because the actual construction work vastly outweighs the overhead of loading and calling the constructor via generic dispatch.

来源：https://stackoverflow.com/questions/53219640/why-is-l-faster-than-setl-python-sets-not-really-only-for-sets-for