MATLAB Central Newsreader - interp1 issue

interp1 issue

Ning — Thu, 13 Nov 2008 06:28:02 -0500

For the example in matlab help 'interp1',
t = 1900:10:1990;
p = [75.995 91.972 105.711 123.203 131.669...
150.697 179.323 203.212 226.505 249.633];
year=1975;
interp1(t,p,year)
'ans =214.8585'

What if the t is a 3-d matrix (2x3x10), p remains constant, and year is a 2-d matrix (2x3), the answer would be a 2-d matrix (2x3) with various p values. How to do this in matlab?

Re: interp1 issue

Titus — Thu, 13 Nov 2008 10:39:34 -0500

"Ning" <ning.robin@gmail.com> schrieb im Newsbeitrag
news:gfghdh$405$1@fred.mathworks.com...
> For the example in matlab help 'interp1',
> t = 1900:10:1990;
> p = [75.995 91.972 105.711 123.203 131.669...
> 150.697 179.323 203.212 226.505 249.633];
> year=1975;
> interp1(t,p,year)
> 'ans =214.8585'
>
> What if the t is a 3-d matrix (2x3x10), p remains constant, and year is a
> 2-d matrix (2x3), the answer would be a 2-d matrix (2x3) with various p
> values. How to do this in matlab?

Hi,
as long as your data are not too large, I guess the simplest approach would
be a loop here:

res = zeros(size(year));
for i=1:size(res,1)
  for j=1:size(res,2)
    res(i,j) = interp1(t(i,j,:), p, year(i,j));
  end
end

Although: interesting, the t changes but the values are the same? How come?

Titus

Re: interp1 issue

Ning — Thu, 13 Nov 2008 11:44:02 -0500

Thank you, Titus. Here's the for loop code.
t = randn(2,3,5);
p = [10:10:50];
year=randn(2,3);
for row=1:2
    for col=1:3
        out(row,col)=interp1(squeeze(t(row,col,:)), p,year(row,col),'linear','extrap')
    end
end
But I can't use the for loop, since the 't' may be lare, like randn(600,500,5).
Any idea about this without loop?

"Titus" <titus.edelhofer@mathworks.de> wrote in message <gfh057$fbt$1@fred.mathworks.com>...
>
> "Ning" <ning.robin@gmail.com> schrieb im Newsbeitrag
> news:gfghdh$405$1@fred.mathworks.com...
> > For the example in matlab help 'interp1',
> > t = 1900:10:1990;
> > p = [75.995 91.972 105.711 123.203 131.669...
> > 150.697 179.323 203.212 226.505 249.633];
> > year=1975;
> > interp1(t,p,year)
> > 'ans =214.8585'
> >
> > What if the t is a 3-d matrix (2x3x10), p remains constant, and year is a
> > 2-d matrix (2x3), the answer would be a 2-d matrix (2x3) with various p
> > values. How to do this in matlab?
>
>
> Hi,
> as long as your data are not too large, I guess the simplest approach would
> be a loop here:
>
> res = zeros(size(year));
> for i=1:size(res,1)
> for j=1:size(res,2)
> res(i,j) = interp1(t(i,j,:), p, year(i,j));
> end
> end
>
> Although: interesting, the t changes but the values are the same? How come?
>
> Titus
>

Re: interp1 issue

John D'Errico — Thu, 13 Nov 2008 12:32:01 -0500

"Ning" <ning.robin@gmail.com> wrote in message <gfh3u2$ol1$1@fred.mathworks.com>...
> Thank you, Titus. Here's the for loop code.
> t = randn(2,3,5);
> p = [10:10:50];
> year=randn(2,3);
> for row=1:2
> for col=1:3
> out(row,col)=interp1(squeeze(t(row,col,:)), p,year(row,col),'linear','extrap')
> end
> end
> But I can't use the for loop, since the 't' may be lare, like randn(600,500,5).
> Any idea about this without loop?

Sorry, but you need a loop if the t variable varies.

Some things are not easily vectorizable. This is one
of them, unless you choose to rewrite interp1. The
real problem lies in the use of histc to efficiently
assign your points into bins for the interpolation.
If the third dimension of t is small for your problem,
it would be possible to write that part to be efficient
without the use of histc. For that to work, t must be
monotonic in the third dimension, which it is not in
your random data.

John