Scilab 5.5.1

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# nanreglin

Linear regression

### Calling Sequence

[a,b]=nanreglin(x,y)

### Arguments

- x, y, a, b
numerical vectors or matrices.

### Description

Solve the regression problem y=a*x+b in the least square sense.
`x`

and `y`

are two matrices of size
`x(p,n)`

and `y(q,n)`

, where `n`

is the number of samples.

The estimator `a`

is a matrix of size `(q,p)`

and `b`

is a
vector of size `(q,1)`

.

Each line of `y`

is treated as an independent problem, if `x`

or `y`

contain a NaN (`x(i,j) = %nan`

or `y(i,j) = %nan`

),
then `x(:,j)`

and `y(i,j)`

are ignored,
as if the point [x(:,j); y(i,j)] did not exist.

### Examples

Graphical example #1:

// In the following example, both problems represent two straight lines: // one goes from (0,0) to (10,10) and the other one goes from (0,20) to (10,30). // reglin and nanreglin find the same values because all the points are aligned and the NaNs have been ignored. subplot(211) x = 0:10; y = 20:30; [a1, b1] = reglin(x, [x ; y]); plot(x', (a1*x+repmat(b1,1,11))', "red") subplot(212) y2 = y; y2(2:10) = %nan; // Leaving y2(1) and y2(11) unchanged. [a2, b2] = nanreglin(x, [x ; y2]) plot(x', (a2*x+repmat(b2,1,11))', "blue")

Graphical example #2:

// Now both problems represent one straight line (reglin(x, x)) from (0,0) to (2,2), // but while the second argument of the first problem (reglin(x, y)) represents // a flat line (with equation y = 1), the second argument of the second problem // (reglin(x, y2)) ignores the central point of y (set to %nan) so the flat line // now has equation y = 0, because the two remaining points are (0,0) and (2,0). subplot(211) x = 0:2; y = [0 3 0]; [a1, b1] = reglin(x, [x ; y]); plot(x', (a1*x+repmat(b1,1,3))', "red") subplot(212) y2 = y; y2(2) = %nan; // y2 = [0 %nan 0]; [a2, b2] = nanreglin(x, [x ; y2]); plot(x', (a2*x+repmat(b2,1,3))', "blue")

### See Also

### History

Versão | Descrição |

5.5.0 | Introduction |

## Comments

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