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How do you handle it in numpy if you want something like ((x - y)/2 + z) * w all elementwise over the arrays? Naively that's 3 intermediate unused arrays bein

How do you handle it in numpy if you want something like ((x - y)/2 + z) * w all elementwise over the arrays? Naively that's 3 intermediate unused arrays being created.

zarzavat 4y ago
There's plenty of CPU and GPU numpy accelerators available. * Numba: https://numba.pydata.org/ https://numba.pydata.org/ * JAX: https://jax.readthedocs.io/en/latest/notebooks/quickstart.html https://jax.readthedocs.io/en/latest/notebooks/quickstart.ht...
- kzrdude 4y ago
  numba - instead of writing in Rust, you write it in numba, which is also almost like another language. Not bad, but needs to be taken into account and is not pure numpy.
europeanguy 4y ago
You use Julia and let loop fusion handle it. You get 1 allocation if you want to store the result in a new array, or zero allocations if you already have the array allocated. https://julialang.org/blog/2017/01/moredots/ https://julialang.org/blog/2017/01/moredots/
korijn 4y ago
Inplace operators?
zozbot234 4y ago
Isn't that what Theano/Aesara is for? Still based on a NumPy interface AIUI, but automagically compiling from NumPy-based to efficient code on the CPU or GPU.
CraigJPerry 4y ago
Is there an optimal solution to that for all sizes of the arrays? E.g. I’d expect for small/medium then re-writing as a fold over the input arrays would be fastest because you’d only traverse once & everything would fit in cache. However if the 3 arrays combined size is larger than L1 cache, I’d be strongly tempted to bet on the naïve 3 operation approach to be faster. You’d save so much time on cache line flush / reload activities. But then if 2 arrays are larger than L3 i’d expect going back to the fold / single traversal to win again because the iterating 2 arrays at a time behaviour is no longer any different to iterating 3 at a time. Untested hypothesis.
pmeira 4y ago
You could use NumExpr: https://github.com/pydata/numexpr#what-is-numexpr https://github.com/pydata/numexpr#what-is-numexpr

How do you handle it in numpy if you want something like ((x - y)/2 + z) * w all elementwise over the arrays? Naively that's 3 intermediate unused arrays being created.

zarzavat 4y ago

There's plenty of CPU and GPU numpy accelerators available. * Numba: https://numba.pydata.org/ https://numba.pydata.org/ * JAX: https://jax.readthedocs.io/en/latest/notebooks/quickstart.html https://jax.readthedocs.io/en/latest/notebooks/quickstart.ht...

kzrdude 4y ago
numba - instead of writing in Rust, you write it in numba, which is also almost like another language. Not bad, but needs to be taken into account and is not pure numpy.

europeanguy 4y ago

You use Julia and let loop fusion handle it. You get 1 allocation if you want to store the result in a new array, or zero allocations if you already have the array allocated. https://julialang.org/blog/2017/01/moredots/ https://julialang.org/blog/2017/01/moredots/

korijn 4y ago

Inplace operators?

zozbot234 4y ago

Isn't that what Theano/Aesara is for? Still based on a NumPy interface AIUI, but automagically compiling from NumPy-based to efficient code on the CPU or GPU.

CraigJPerry 4y ago

Is there an optimal solution to that for all sizes of the arrays? E.g. I’d expect for small/medium then re-writing as a fold over the input arrays would be fastest because you’d only traverse once & everything would fit in cache. However if the 3 arrays combined size is larger than L1 cache, I’d be strongly tempted to bet on the naïve 3 operation approach to be faster. You’d save so much time on cache line flush / reload activities. But then if 2 arrays are larger than L3 i’d expect going back to the fold / single traversal to win again because the iterating 2 arrays at a time behaviour is no longer any different to iterating 3 at a time. Untested hypothesis.

pmeira 4y ago

You could use NumExpr: https://github.com/pydata/numexpr#what-is-numexpr https://github.com/pydata/numexpr#what-is-numexpr