Different from numpy?

[msnh2012]

c++ code:

    xt::xarray<int> aad ={{1, 2},{4, 5}};
    xt::xarray<bool> v = {{true,false}};
    xt::filter(aad,v) = 0;
    std::cout<<aad;

result:

{{0, 2},
 {4, 5}}

in numpy:

{{0, 0},
 {4, 5}}

[JohanMabille]

Can you post your numpy code please?

[msnh2012]

@JohanMabille numpy code

import numpy as np
a = np.array(((1,2),(4,5)))
b = np.array((True,False))
a[b]=0

[JohanMabille]

xt::filter is not boolean indexing (like you do in Numpy), it returns a one-dimensional views holding elements of an xexpression that verify a given condition. Internally, the implementation is:

auto indices = argwhere<L>(std::forward<O>(condition));
using view_type = xindex_view<xclosure_t<E>, decltype(indices)>;
return view_type(std::forward<E>(e), std::move(indices));

That is, the selected elements in the view have the same indices as the true elements in the condition. Usually you don't need to bother with the implementation detial because the condition depends on the expression you want to filter:

auto f = filter(a, a >= 5);

So here the condition (a >= 5) and a have the same shape and the result is intuitive.

In your case, the condition and the expression that you want to filter do not have the same shape, and there is a bug in the filter implementation where the condition is not broadcasted (see #963 ).
Therefore, the vector of indices for selecting elements contains only only one index (i.e. (0, 0)). When the broadcasting issue is fixed, your example will be equivalent to:

xt::filter(aad, {{true, false}, {true, false}}) = 0;

resulting in

{{0, 2},
 {0, 5}}

So it will still be different from Numpy.

[msnh2012]

I got it. Thx