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.
method returns linear interpolated values at specific query points X.
[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
- X
- 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
X defines new positions for the interpolation based on the trained object state. The shape of X is not taken into account. If X is not a vector, its values are read in sequential order.
Interpolation values are computed from the original values V used to train this interpolator objects. The array returned will have the same shape as V, except the working dimension of V is replaced with n interpolated values, n = X.S.NumberOfElements. The 'working dimension' 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 LinearInterpolatorSingle(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