ILNumerics - Technical Application Development
Assembly: ILNumerics.Toolboxes.Interpolation (in ILNumerics.Toolboxes.Interpolation.dll) Version: 5.5.0.0 (5.5.7503.3146)
Interpolated data, see remarks for size details.
Computes interpolated values using this cubic interpolator for new query points Xn.
[ILNumerics Interpolation Toolbox]
Namespace: ILNumerics.Toolboxes
Assembly: ILNumerics.Toolboxes.Interpolation (in ILNumerics.Toolboxes.Interpolation.dll) Version: 5.5.0.0 (5.5.7503.3146)
Syntax
Parameters
- Xn
- Type: ILNumericsInArraySingle
Query points, specified as a scalar, vector or n-dim array of real numbers. - arguments (Optional)
- Type: ILNumericsInCell
(Ignored)
Return Value
Type: RetArraySingleInterpolated data, see remarks for size details.
Exceptions
Exception | Condition |
---|---|
ArgumentNullException | If an input parameters are NULL |
InvalidOperationException | If Interpolation object has been disposed before Apply() calls |
Remarks
Xn defines new positions for the interpolation based on the trained object state. The shape of Xn is not taken into account. Values of Xn are read in sequential, column-major order.
Interpolation values are computed from the original values V used to train this interpolator object. The array returned will have the same shape as V, except the working dimension of V is replaced with n interpolated values, n = Xn.S.NumberOfElements. The 'working dimension' index of V is 1 if V is a row vector, 0 otherwise.
[ILNumerics Interpolation Toolbox]
Examples
// Define some sample points as a function: f(x) = sin(x); Array<float> X = linspace<float>(-pi, pi, 10); Array<float> V = sin(X); // Create Interpolator Object order 5 using (var interp = new CubicInterpolatorSingle(V, X)) { // Get some random scalar values float val1 = (float)interp.Apply(0.024); float val2 = (float)interp.Apply(-0.21); // Get 10 linear spaced values defined on range (-0.5, 1.25) Array<float> valRange = interp.Apply(linspace<float>(-0.5, 1.25, 10)); // Get 30 interpolated values using Xn query points defined as array[3 x 2 x 5] Array<float> Xn = counter(0.0, 0.1, 3, 2, 5); Array<float> Vn = interp.Apply(Xn); }
See Also