MATLAB Central Newsreader - Interpolation

Interpolation

Jianbin — Mon, 16 Jul 2007 14:45:56 -0400

Does anyone have the code to do interpolation so that it always takes the lower value.

For example, here is a known distribution
X Y
0.01 200
0.02 500
0.03 700
0.04 900
0.05 1000
... ...
0.99 20000
1 20300

if x1=0.015, then y1=200; x1=0.019, y1=200; x1=0.039, y1=700;...

Thanks

Re: Interpolation

Per Sundqvist — Mon, 16 Jul 2007 20:39:02 -0400

help sort. Try this

xd=0.15;%your data point
[dummy,ix]=sort(abs(xd-x));
val_closest=[y(ix(1)) y(ix(2))]
[yy,ix0]=sort(val_clostest);
%yy is the closest smallest value y
yy
%and correspond to closest datapoint xx
xx=x(ix0)

/Per

Re: Interpolation

John D'Errico — Mon, 16 Jul 2007 22:38:26 -0400

> Does anyone have the code to do interpolation so that it always takes the lower value.
>
> For example, here is a known distribution
> X Y
> 0.01 200
> 0.02 500
> 0.03 700
> 0.04 900
> 0.05 1000
> ... ...
> 0.99 20000
> 1 20300
>
> if x1=0.015, then y1=200; x1=0.019, y1=200; x1=0.039, y1=700;...

In your example, its simple.

ind = floor(x1/0.1);
y1 = Y(ind);

If X is not so simply spaced, then you can use
interp1.

ind = floor(interp1(X,1:length(X),x1,'linear'));
y1 = Y(ind);

HTH,
John

Re: Interpolation

us — Tue, 17 Jul 2007 00:36:36 -0400

<SNIP apparent interpolation...

one of the many solutions

% your dist and input
     d=[
          .01 100
          .02 200
          .03 300
          .09 900
     ];
     in=[
          .01
          .011
          .015
          .025
          .07
          .09
          .091
          .1
     ];
% the engine
     [ix,ix]=histc(in,sort([d(:,1);in(end)+1]));
% the result
     r=[in,d(ix,:)]

us

Re: Interpolation

Stephen — Tue, 17 Jul 2007 18:20:08 -0400

Thanks for your help.

I try to write this as a function:

===========================================================
function y1=NoInterp(x,Y,x1)

if x1 < min(x)
x1=min(x);
elseif x1>max(x)
x1=max(x);
end

ind = floor(interp1(x,1:length(x),x1,'linear'));
y1 = Y(ind);

===========================================================

When x1 is in the range, the function runs OK. When x1 is out of range, the if statement I added to your code doesn't seem to work properly and the function fails. Could you help me figure out what's wrong here?

Thanks a lot.

Stephen

Re: Interpolation

John D'Errico — Tue, 17 Jul 2007 18:30:33 -0400

> I try to write this as a function:

===========================================================
function y1=NoInterp(x,Y,x1)

if x1 < min(x)
      x1=min(x);
elseif x1>max(x)
     x1=max(x);
end

ind = floor(interp1(x,1:length(x),x1,'linear'));
y1 = Y(ind);

===========================================================

> When x1 is in the range, the function runs OK. When x1 is out of range, the > if statement I added to your code doesn't seem to work properly and the
> function fails. Could you help me figure out what's wrong here?

The problem is that if does not work as you are using
it. An if statement is not "vectorized".

However, why bother? Min and max ARE vectorized.

What does this do for you?

  x1 = min(x1,max(x));

How about this?

  x1 = max(x1,min(x));

John

Re: Interpolation

Stephen — Tue, 17 Jul 2007 18:51:34 -0400

Thanks for the prompt response.

I realized the error in my code as x1 is supposed to be a vector. So I changed the code to be:
===============================================
function y1=NoInterp(x,Y,xd)
xd(xd<min(x))=min(x);
xd(xd>max(x))=max(x);
ind = floor(interp1(x,1:length(x),xd,'linear'));
y1 = Y(ind);
================================================

And it works fine now.

Thanks,
Stephen