On Sun, 2025-03-23 at 10:17 +0800, Fang Zhang wrote:
> Sebastian,
>
> I think the core issue you pointed out is indeed correct, but your
> detailed
> explanation is backwards, since `maximum(arr[:, 0], arr[:, 1])`
> implements
Ah, right, I explained things thinking of axis=1, thanks! As you said
the arguments should be identical, but reversed.
> `arr.max(axis=1)` instead of `arr.max(axis=0)`. So OP's transpose
> method is
> essentially approach 1, which for this array shape has less overhead
> than
> approach 2 (and since 2 is small enough, it doesn't matter that we
> are
> paying this overhead in Python instead of in C).
>
> > From my tests, when n = 1_000_000, the advantage of the transpose
> > method
> persists up to m = 6. But for smaller n the normal method becomes
> better
> again, so it's not that clear cut (although it would be nice to have
> a way
> to use the transpose method but pay the overhead in C instead of
> Python!)
>
> Test code:
>
> import numpy as np
> from IPython import get_ipython
>
> def calculate_bbox_normal(vertices: np.ndarray):
> return vertices.min(axis=0), vertices.max(axis=0)
>
>
> def calculate_bbox_transpose(vertices: np.ndarray):
> return np.array([col.min() for col in vertices.T]),
> np.array([col.max()
> for col in vertices.T])
>
> n = 1_000_000
> for m in range(4, 9):
> print(f"shape: {n, m}")
> vertices = np.random.random((n, m))
> vmin, vmax = calculate_bbox_normal(vertices)
> vmin_t, vmax_t = calculate_bbox_transpose(vertices)
> assert (vmin == vmin_t).all()
> assert (vmax == vmax_t).all()
> print("Normal: ", end="")
> get_ipython().run_line_magic("timeit",
> "calculate_bbox_normal(vertices)")
> print("Transpose: ", end="")
> get_ipython().run_line_magic("timeit",
> "calculate_bbox_transpose(vertices)")
> print()
>
> - Fang
>
> On Sat, Mar 22, 2025 at 7:10 PM Sebastian Berg
>
> wrote:
>
> > On Fri, 2025-03-21 at 23:22 +0100, Tiziano Zito via NumPy-
> > Discussion
> > wrote:
> > > Hi George,
> > >
> > > what you see is due to the memory layout of numpy arrays. If you
> > > switch your array to F-order you'll see that the two functions
> > > have
> > > the same timings, i.e. both are fast (on my machine 25 times
> > > faster
> > > for the 1_000_000 points case).
> >
> >
> > Memory order is indeed very important, but it should be indirectly
> > so!
> >
> > The thing is that NumPy tries to re-order most operations for
> > memory
> > access order (not particularly advanced).
> >
> > I.e. NumPy can calculate this type of operation in two ways (the
> > inner-
> > loop is always 1-D, so in C first, outer, loop here is separate):
> > 1. for i in arr.shape[0]: res[i] = arr[i].max()
> > 2. for i in arr.shape[1]: res[:] = maximum(res, arr[i])
> > (For i == 0, you would just copy `arr[0]` or initialize `res`
> > first)
> >
> > NumPy chooses the first form here, precisely because it is
> > (normally)
> > better for the memory layout, here.
> > But for 1 `arr[i].max()` is actually a function call inside NumPy
> > (deep
> > in C, not on an actual NumPy array object of course).
> >
> > So for `arr.shape[1] == 2`, that just gets in the way because of
> > overheads!
> > Writing it as 2. is much better (memory order doesn't even matter),
> > calculating it as `maximum(arr[0], arr[1])` would be the best.
> >
> > But even for other small values for `arr.shape[1]` it would
> > probably be
> > better to use the second version, memory order will start to matter
> > more and more (I could imagine even for arr.shape[1] == 3 or 4 it
> > dominates quickly for huge arrays, but didn't try).
> >
> > So what you see should be overheads of using approach 1. Taking
> > the
> > maximum of 2 elements is simply fast compared to even a lightweight
> > outer for loop and just calling the function to take the maximum.
> > (And the core function does a bit of extra work to optimize for
> > many
> > elements additionally [1])
> > You really don't need large inner-loops to hide most of those
> > overheads, but 2 elements just isn't enough.
> >
> > Anyway, of course it could be cool to add a heuristic to pick
> > approach
> > 2 here when we obviously have `arr.shape[1] == 2` or maybe
> > `arr.shape <
> > small_value`.
> > There are also probably some low-hanging fruits to just reduce the
> > overheads here (for this specific case I know the outer iteration
> > can
> > be optimized, although I am not sure it would improve things, also,
> > some inner-loop optimizations could be moved out of the inner-loop
> > since a few year now -- but only casts actually do so).
> >
> > In the end, `maximum(arr[:, 0], arr[:, 1])` will always the fastest
> > way
> > for this specific thing, because it explicitly picks the clearly
> > best
> > approach.
> > It would be good to close that gap at least a bit, though.
> >
> > - Sebastian
> >
> >
> > [1] locally I can actually see a small speed-up if I pick less
> > optimized loops at run-time
