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lspoly_n

Calculates a set of coefficients for a weighted least squares polynomial fit to the given data on the given dimension.

Available in version 6.2.0 and later.

Prototype

	function lspoly_n (
		x       : numeric,  
		y       : numeric,  
		wgt     : numeric,  
		n   [1] : integer,  
		dim [1] : integer   
	)

	return_val  :  float or double

Arguments

x

Abscissa values of the data.

This can be one-dimensional or multi-dimensional. If one-dimensional, it must be the same length as the rightmost dimension of y. If multi-dimensional, it must be the same dimensionality as y.

y

Ordinate values of the data.

wgt

Weights for a weighted least squares model. If all data values are to be assigned equal weights, then setting the argument equal to a scalar 1.0 will result in all the weights being set to 1.0. Note: if x or y is equal to _FillValue (if present), the weight will be set to 0.0 for that coordinate pair.

This can be one-dimensional or multi-dimensional. If one-dimensional, it must be the same length as the rightmost dimension of y. If multi-dimensional, it must be the same dimensionality as y.

n

The number of coefficients desired (i.e., n-1 will be the degree of the polynomial). It is suggested that n be less than or equal to five.

dim

A scalar integer indicating which dimension of y to do the calculation on. Dimension numbering starts at 0.

Return value

The return array will have the same dimensionality as y, except the dim-th dimension will be of length n.

If either x or y are of type double, then the return array is returned as double. Otherwise, the returned coefficients are returned as type float.

Description

Given a set of data (x(i),y(i)), i = 1,...,m, lspoly_n calculates a set of coefficients for a weighted least squares polynomial fit to the given data on the given dimension. It is necessary that the number of data points) be greater than n (the number of coefficients).

See Also

lspoly, regCoef, regline, regcoef, reg_multlin

Examples

Example 1

Note: we don't need to use lspoly_n in this example, since we are only using 1D arrays, and hence there's no reordering necessary. 

  x = (/-4.5, -3.2, -1.4, 0.8, 2.5, 4.1/)
  y = (/ 0.7,  2.3,  3.8, 5.0, 5.5, 5.6/)

  n = 4
  c = lspoly_n(x, y, 1, n, 0)    ; all weights are set to one 
  print(c)
The 3rd degree polynomial is 
         Y = c(0) + c(1)*x + c(2)*x^2 + c(3)*x^3
The coefficients (which agree with those returned from Mathematica) are: 
(0)      4.66863          
(1)      0.489392        
(2)     -0.0742387      
(3)      0.00267663

Example 2

Similar as the previous example, but this time assume x and y are 3D arrays and that we want to solve the equation for the middle (dim=1) dimension: 

  n = 4
  c = lspoly_n(x, y, 1, n, 1)    ; all weights are set to one 
  print(c)