xt::xarray<double> a({1, 3, 4, 2});
does not initialize a 4D-array, but a 1D-array containing the values 1
, 3
,
4
, and 2
.
It is strictly equivalent to
xt::xarray<double> a = {1, 3, 4, 2};
To initialize a 4D-array with the given shape, use the :cpp:func:`xt::xarray::from_shape` static method:
auto a = xt::xarray<double>::from_shape({1, 3, 4, 2});
The confusion often comes from the way :cpp:type:`xt::xtensor` can be initialized:
xt::xtensor<double, 4> a = {1, 3, 4, 2};
In this case, a 4D-tensor with shape (1, 3, 4, 2)
is initialized.
Consider the following function:
template <class C> auto func(const C& c) { return (1 - func_tmp(c)) / (1 + func_tmp(c)); }
where func_tmp
is another unary function accepting an xtensor expression. You may
be tempted to simplify it a bit:
template <class C> auto func(const C& c) { auto tmp = func_tmp(c); return (1 - tmp) / (1 + tmp); }
Unfortunately, you introduced a bug; indeed, expressions in xtensor are not evaluated
immediately, they capture their arguments by reference or copy depending on their nature,
for future evaluation. Since tmp
is an lvalue, it is captured by reference in the last
statement; when the function returns, tmp
is destroyed, leading to a dangling reference
in the returned expression.
Replacing auto tmp
with xt::xarray<double> tmp
does not change anything, tmp
is still an lvalue and thus captured by reference.
Using a random number function from xtensor actually returns a lazy generator. That means, accessing the same element of a random number generator does not give the same random number if called twice.
auto gen = xt::random::rand<double>({10, 10}); auto a0 = gen(0, 0); auto a1 = gen(0, 0); // a0 != a1 !!!
You need to explicitly assign or eval a random number generator, like so:
xt::xarray<double> xr = xt::random::rand<double>({10, 10}); auto xr2 = xt::eval(xt::random::rand<double>({10, 10})); // now xr(0, 0) == xr(0, 0) is true.
When :cpp:func:`xt::variance` is passed an expression and an integer parameter, this latter is not the axis along which the variance must be computed, but the degree of freedom:
xt::xtensor<double, 2> a = {{1., 2., 3.}, {4., 5., 6.}}; std::cout << xt::variance(a, 1) << std::endl; // Outputs 3.5
If you want to specify an axis, you need to pass an initializer list:
xt::xtensor<double, 2> a = {{1., 2., 3.}, {4., 5., 6.}}; std::cout << xt::variance(a, {1}) << std::endl; .. Outputs { 0.666667, 0.666667 }
Builder functions such as :cpp:func:`xt::empty` or :cpp:func:`xt::ones` accept an initializer list as argument. If the elements of this list do not have the same type, a curious compilation error may occur on Windows:
size_t N = 10ull; xt::xarray<int> ages = xt::empty<int>({N, 4ul}); // error: cannot convert argument 1 from 'initializer list' // to 'const xt::fixed_shape<> &'
To avoid this compiler bug (for which we don't have a workaround), ensure all the elements in the initializer list have the same type.
Note
If you are using C++ >= 17
you should not have to worry about this.
When building with xsimd (see :ref:`external-dependencies`), if you define a structure
having members of fixed-size xtensor types, you must ensure that the buffers properly
aligned. For this you can use the macro XTENSOR_FIXED_ALIGN
available in
xtensor/xtensor_config.hpp
.
Consider the following example:
template <typename T>
class alignas(XTENSOR_FIXED_ALIGN) Foo
{
public:
using allocator_type = std::conditional_t<XTENSOR_FIXED_ALIGN != 0,
xt_simd::aligned_allocator<T, XTENSOR_FIXED_ALIGN>,
std::allocator<T>>;
Foo(T fac) : m_fac(fac)
{
m_bar.fill(fac);
}
auto get() const
{
return m_bar;
}
private:
xt::xtensor_fixed<T, xt::xshape<10, 10>> m_bar;
T m_fac;
};
Whereby it is important to store the fixed-sized xtensor type (in this case xt::xtensor_fixed<T, xt::xshape<10, 10>>
) as first member.