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
`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 <sebast...@sipsolutions.net>
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 via `NPY_DISABLE_CPU_FEATURES=AVX2`.
>
> >
> > Try:
> >
> > vertices = np.array(np.random.random((n, 2)), order='F')
> >
> > When your array doesn't fit in L1-cache anymore, either order 'C' or
> > order 'F' becomes (much) more efficient depending on which dimension
> > you are internally looping through.
> >
> > You can read more about it here:
> >
> https://numpy.org/doc/stable/dev/internals.html#internal-organization-of-numpy-arrays
> > and
> >
> https://numpy.org/doc/stable/reference/arrays.ndarray.html#internal-memory-layout-of-an-ndarray
> >
> > Hope that helps,
> > Tiziano
> >
> >
> > On Fri 21 Mar, 12:10 +0200, George Tsiamasiotis via NumPy-Discussion
> > <numpy-discussion@python.org> wrote:
> > > Hello NumPy community!
> > >
> > > I was writing a function that calculates the bounding box of a
> > > polygon (the smallest rectangle that fully contains the polygon,
> > > and who's sides are parallel to the x and y axes). The input is a
> > > (N,2) array containing the vertices of the polygon, and the output
> > > is a 4-tuple containing the vertices of the 2 corners of the
> > > bounding box.
> > >
> > > I found two ways to do that using the np.min() and np.max()
> > > methods, and I was surprised to see a significant speed difference,
> > > even though they seemingly do the same thing.
> > >
> > > While for small N the speed is essentially the same, the difference
> > > becomes noticeable for larger N. >From my testing, the difference
> > > seems to plateau, with the one way being around 4-5 times faster
> > > than the other.
> > >
> > > Is there an explanation for this?
> > >
> > > Here is a small benchmark I wrote (must be executed with IPython):
> > >
> > > import numpy as np
> > > from IPython import get_ipython
> > >
> > > vertices = np.random.random((1000, 2))
> > >
> > > def calculate_bbox_normal(vertices: np.ndarray) ->
> > > tuple[np.float64]:
> > > xmin, ymin = vertices.min(axis=0)
> > > xmax, ymax = vertices.max(axis=0)
> > > return xmin, ymin, xmax, ymax
> > >
> > > def calculate_bbox_transpose(vertices: np.ndarray) ->
> > > tuple[np.float64]:
> > > xmin = vertices.T[0].min()
> > > xmax = vertices.T[0].max()
> > > ymin = vertices.T[1].min()
> > > ymax = vertices.T[1].max()
> > > return xmin, ymin, xmax, ymax
> > >
> > > bbox_normal = calculate_bbox_normal(vertices)
> > > bbox_transpose = calculate_bbox_transpose(vertices)
> > >
> > > print(f"Equality: {bbox_normal == bbox_transpose}")
> > >
> > > for n in [10, 100, 1000, 10_000, 100_000, 1_000_000]:
> > > print(f"Number of points: {n}")
> > > vertices = np.random.random((n, 2))
> > > 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()
> > >
> >
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>
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