Performance: in-place dpctl.tensor.add with strides

[npolina4]

import dpctl.tensor as dpt
a = dpt.ones((8192, 8192), dtype='i4', device='cpu')
b = dpt.ones((8192 + 2, 8192 + 2), dtype='i4', device='cpu')
%timeit b[2:, 2:]+=a
#209 ms ± 36.8 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

import numpy
a_np = numpy.ones((8192, 8192), dtype='i4')
b_np = numpy.ones((8192 + 2, 8192 + 2), dtype='i4')
%timeit b_np[2:, 2:]+=a_np
#75.7 ms ± 1.31 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

[oleksandr-pavlyk]

This was addressed and should be closed.

[ndgrigorian]

Checked timings again, for Xeon CPU:

In [2]: import dpctl.tensor as dpt
   ...: a = dpt.ones((8192, 8192), dtype='i4', device='cpu')
   ...: b = dpt.ones((8192 + 2, 8192 + 2), dtype='i4', device='cpu')

In [3]: q = a.sycl_queue

In [4]: %timeit b[2:, 2:] += a; q.wait()
6.59 ms ± 748 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)

for i7-1185G7:

In [2]: import dpctl.tensor as dpt
   ...: a = dpt.ones((8192, 8192), dtype='i4', device='cpu')
   ...: b = dpt.ones((8192 + 2, 8192 + 2), dtype='i4', device='cpu')

In [3]: q = a.sycl_queue

In [4]: %timeit b[2:, 2:] += a; q.wait()
72.2 ms ± 2.14 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

@oleksandr-pavlyk should this be closed?

[oleksandr-pavlyk]

I agree, the performance had improved.

I'd think systematic way to decide whether there are any improvements to be had is to collect the numpy timing on the same machine, and then with dpctl using taskset -c 0 ipython, taskset -c 0-1 ipython, taskset -c 0-3 ipython, taskset -c 0-7 ipython, and plain ipython to see whether performance of CPU device scales.