Adding `xt::reshape` like `reshape` intrinsic in Fortran about xtensor

[JohanMabille]

@certik That's it. This limitation on xtensor::reshape is because it operates in place, like in NumPy. I think we don't want to mix "copy" and "in-place" behaviors in the same function (at least implicitly), so an additional free function (although the name copy_and_reshape might not be ideal) looks like a good option to me. Also it would work for any kind of expressions (even those which are not evaluated).

EDIT: maybe eval_reshaped could be a good name?

[certik]

@JohanMabille thanks!

Just so that I understand: the code in #2760 (comment) is currently the only way to take a 16x16 xtensor and reshape it into a new 16x8x2 xtensor?

The current facility is the xtensor::reshape, but that must keep the number of dimensions the same (and the number of elements of course), so it can change 16x16 to 8x32, but it can't change it to 16x8x2, correct?

[czgdp1807]

cc: @certik

[JohanMabille]

xtensor provides an API similar to numpy, where reshape changes the shape of the passed array. xtensor is missing a free funcion reshape, but if we add it, it should forward the call to the reshape method of its argument.

For creating a new array of a given shape, you can use the empty free function, which has less restrictions than the reshape funciton in Fortran.

[czgdp1807]

Actually I am trying to port https://github.com/lfortran/lfortran/blob/main/integration_tests/arrays_reshape_14.f90 to C++ using xt::xtensor APIs. For transforming the following line to C++,

https://github.com/lfortran/lfortran/blob/8e4d530121d8ad7d218854fbe37db221b8395500/integration_tests/arrays_reshape_14.f90#L23

I want the following to work.

// xt::xtensor_fixed<double, xt::xshape<256>> b
// xt::xtensor<double, 2>& a
// xt::xtensor_fixed<int, xt::xshape<1>> newshape
b = xt::reshape(a, newshape); // would be great to have xt::reshape which can work for this case.

I am using xtensor and xtensor_fixed because they map well to fortran types. I might be wrong but AFAIK, xarray doesn't store any information related to the rank of the array but Fortran specifies rank of the array in its type itself (analogous to xtensor).

I am open to ideas and different approaches for this problem.

[czgdp1807]

xtensor is missing a free funcion reshape, but if we add it, it should forward the call to the reshape method of its argument.

I see. The return type of xt::reshape would depend on the size of its second argument (a fixed size 1 D array).

[JohanMabille]

I might be wrong but AFAIK, xarray doesn't store any information related to the rank of the array but Fortran specifies rank of the > array in its type itself (analogous to xtensor).

xarray provides the rank information at runtime only, while xtensor and xtensor_fixed can provide it at build time.

I am open to ideas and different approaches for this problem.

I think the implementation could be something like:

template <class E, class S>
auto copy_and_reshape(E&& e, const S& shape)
{
    using value_type = typename std::decay_t<E>::value_type;
    auto res = xt::empty_like<value_type>(shape);
    std::copy(e.cbegin(), e.cend(), res.begin());
    return res;
}

We can probably find a better name, but it should definitely not be reshape to avoid confusion with the existing reshape feature.

xtensor is missing a free function reshape, but if we add it, it should forward the call to the reshape method of its argument.

I see. The return type of xt::reshape would depend on the size of its second argument (a fixed size 1 D array).

Not sure to get you, what I meant is that xt::reshape(tensor, shape) should call tensor.reshape(shape) that reshapes tensor in place (therefore the return type would be void), sorry if my first message was not clear.

[czgdp1807]

Not sure to get you, what I meant is that xt::reshape(tensor, shape) should call tensor.reshape(shape) that reshapes tensor in place (therefore the return type would be void), sorry if my first message was not clear.

For example if we want to reshape, xt::xtensor<double, 2> a (say of shape [16, 16]) to xt::xtensor<double, 1> b (say of shape [256]) then xt::reshape(a, {256}) should return an object of type xt::xtensor<double, 1>. However, if we want to reshape a to xt::xtensor<double, 3> c (say of shape [16, 8, 2]) then xt::reshape(a, {16, 8, 2}) should return xt::xtensor<double, 3>. Basically the return type depends on the size of shape argument. If shape is of size n then returned xt::xtensor would be of rank n. Even if are able to define the signature of xt::reshape, implementing it might be tricky. In Fortran, reshape is an intrinsic and its upto the Fortran compiler on how it implements this feature depending upon the way it handles array internally.

[czgdp1807]

I think the implementation could be something like:

I see. Is this implementation doing the same thing as reshape feature of Fortran as I described in my above comment? Seems like it's creating an object res with shape and then contents of input e are copied to it? If so then it is making sense to me.

[JohanMabille]

Basically the return type depends on the size of shape argument. If shape is of size n then returned xt::xtensor would be of rank n

Ha I see; indeed xtensor does not allow that, since the reshape is done in place, and since the rank of an xtensor object is known at build time, reshape must preserve the rank.

Seems like it's creating an object res with shape and then contents of input e are copied to it? If so then it is making sense to me.

Yes, that's exactly what it is doing. Also notice that the initial tensor values are not initialized before the copy. This avoids iterating over the whole buffer to set 0 everywhere.

[czgdp1807]

Great to hear. It would be great have this functionality in xtensor library.

EDIT: maybe eval_reshaped could be a good name?

Sounds good to me.