Tensor.Sqrt 方法

定义

重载

Sqrt<T>(ReadOnlyTensorSpan<T>)

获取 ReadOnlyTensorSpan<T> 的每个元素的平方根,并返回结果的新 Tensor<T>

Sqrt<T>(ReadOnlyTensorSpan<T>, TensorSpan<T>)

获取 x 的每个元素的平方根,并返回结果的新 TensorSpan<T>

Sqrt<T>(ReadOnlyTensorSpan<T>)

Source:
TensorExtensions.cs

获取 ReadOnlyTensorSpan<T> 的每个元素的平方根,并返回结果的新 Tensor<T>

public:
generic <typename T>
 where T : System::Numerics::IRootFunctions<T> static System::Numerics::Tensors::Tensor<T> ^ Sqrt(System::Numerics::Tensors::ReadOnlyTensorSpan<T> % x);
public static System.Numerics.Tensors.Tensor<T> Sqrt<T> (in System.Numerics.Tensors.ReadOnlyTensorSpan<T> x) where T : System.Numerics.IRootFunctions<T>;
static member Sqrt : ReadOnlyTensorSpan -> System.Numerics.Tensors.Tensor<'T (requires 'T :> System.Numerics.IRootFunctions<'T>)> (requires 'T :> System.Numerics.IRootFunctions<'T>)
Public Function Sqrt(Of T As IRootFunctions(Of T)) (ByRef x As ReadOnlyTensorSpan(Of T)) As Tensor(Of T)

类型参数

T

参数

x
ReadOnlyTensorSpan<T>

要获取其平方根的 ReadOnlyTensorSpan<T>

返回

适用于

Sqrt<T>(ReadOnlyTensorSpan<T>, TensorSpan<T>)

Source:
TensorExtensions.cs

获取 x 的每个元素的平方根,并返回结果的新 TensorSpan<T>

public:
generic <typename T>
 where T : System::Numerics::IRootFunctions<T> static System::Numerics::Tensors::TensorSpan<T> ^ Sqrt(System::Numerics::Tensors::ReadOnlyTensorSpan<T> % x, System::Numerics::Tensors::TensorSpan<T> % destination);
public static ref readonly System.Numerics.Tensors.TensorSpan<T> Sqrt<T> (scoped in System.Numerics.Tensors.ReadOnlyTensorSpan<T> x, in System.Numerics.Tensors.TensorSpan<T> destination) where T : System.Numerics.IRootFunctions<T>;
static member Sqrt : ReadOnlyTensorSpan * TensorSpan -> TensorSpan (requires 'T :> System.Numerics.IRootFunctions<'T>)
Public Function Sqrt(Of T As IRootFunctions(Of T)) (ByRef x As ReadOnlyTensorSpan(Of T), ByRef destination As TensorSpan(Of T)) As TensorSpan(Of T)

类型参数

T

参数

x
ReadOnlyTensorSpan<T>

要获取其平方根的 ReadOnlyTensorSpan<T>

destination
TensorSpan<T>

返回

适用于