xtensor-stack
filter missing broadcasting? view missing selector?
My eye crossed https://stackoverflow.com/q/74100940/2646505 . The goal there is to mimic
import numpy as np
keep = np.array([True, False, True])
A = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
print(A[keep, :])
In xtensor this is not possible it seems.
xt::filter does not broadcast:
#include <xtensor/xarray.hpp>
#include <xtensor/xtensor.hpp>
#include <xtensor/xsort.hpp>
#include <xtensor/xview.hpp>
#include <xtensor/xindex_view.hpp>
#include <xtensor/xio.hpp>
int main()
{
xt::xtensor<bool, 1> keep = {{true, false, true}};
xt::xtensor<int, 2> A = {{1, 2, 3}, {4, 5, 6}, {4, 5, 6}};
auto c = xt::filter(A, keep);
std::cout << c << std::endl;
return 0;
}
Returns only the first column!?
{1, 3}
xt::view does not take logical
A syntax could be
int main()
{
xt::xtensor<bool, 1> keep = {{true, false, true}};
xt::xtensor<int, 2> A = {{1, 2, 3}, {4, 5, 6}, {4, 5, 6}};
auto c = xt::view(A, xt::is_true(keep));
std::cout << c << std::endl;
return 0;
}
or have an overload on keep, but that might be a bit risky. What is the input is meant as logical, but given as ints?
(bonus) View should take xt::missing
I realised that it would be nice to do
import numpy as np
keep = np.array([True, False, True])
A = np.arange(3 * 4 * 5).reshape(3, 4, 5)
print(A[keep, ...])
as follows:
int main()
{
xt::xtensor<bool, 1> keep = {{true, false, true}};
xt::xtensor<int, 2> A = {{1, 2, 3}, {4, 5, 6}, {4, 5, 6}};
auto c = xt::view(A, xt::is_true(keep), xt::missing);
std::cout << c << std::endl;
return 0;
}
Yes, filter doesn't take broadcasting. I 've talked about the implementation of filter in https://github.com/xtensor-stack/xtensor/issues/2584, and IMO, it's natural to have broadcasting in filter. I 'm curious about the intent of this implementation.
Hi,
I think it would be nice to have this over any possible axis, as in NumPy
auto c = xt::view(A, xt::drop(1), xt::keep(keep), xt::all());
Where keep is a dynamic array of bool the size of the relevant dimension, or a subset of indices to keep.
Would broadcasting be able to solve the general (boolean) case?
