Thread Subject: How to find a list of floats in another list of floats

Hi:

I was wondering if there are any clever tricks to search for sets of real
numbers in a larger list, especially without using a for loop. I only know
how to do the search for one element at a time.

A = [1.0, 2.0];
B = [1.0, 2.0, 3.0];

for i = 1:length(A)
  [t, ind(i)] = min(abs(A(i)-B));
end

Just curious if there is a better way to write this code.

Thanks in advance,

Pete

"Peter de B. Harrington" <peter.harrington@ohio.edu> wrote in message
<47af565c$0$1103$4c368faf@roadrunner.com>...
> Hi:
>
> I was wondering if there are any clever tricks to search for sets of real
> numbers in a larger list, especially without using a for loop. I only know
> how to do the search for one element at a time.
>
> A = [1.0, 2.0];
> B = [1.0, 2.0, 3.0];
>
> for i = 1:length(A)
> [t, ind(i)] = min(abs(A(i)-B));
> end
>
> Just curious if there is a better way to write this code.

I'd compute an inter-point distance matrix,
limited by those points with a distance
less than some tolerance.

Simplest is to do this, assuming that both A
and B are row vectors...

d = abs(bsxfun(@minus,A',B));

% pick some appropriate tolerance
tol = 1e-12;

[i,j] = find(d < tol);

If your vectors are really long, say at least
a few thousand elements in each, then it gets
harder to do, since the intermediate distance
matrix will be a full matrix. There are still
solutions, but it takes a bit more work.

John

"Peter de B. Harrington" <peter.harrington@ohio.edu> wrote in message
<47af565c$0$1103$4c368faf@roadrunner.com>...
> Hi:
>
> I was wondering if there are any clever tricks to search for sets of real
> numbers in a larger list, especially without using a for loop. I only know
> how to do the search for one element at a time.
>
> A = [1.0, 2.0];
> B = [1.0, 2.0, 3.0];
>
> for i = 1:length(A)
> [t, ind(i)] = min(abs(A(i)-B));
> end
>
> Just curious if there is a better way to write this code.
>
> Thanks in advance,
>
> Pete
---------
  Here is how I would have performed just the operation you described in your
for-loop construct using my antiquated version of matlab. I'm not at all sure
that it would be faster on a newer version, however.

 [a,b] = meshgrid(A,B);
 [ignore,ind] = min(abs(a-b),[],1);

Roger Stafford

"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in
message <fonsi6$bp4$1@fred.mathworks.com>...
> "Peter de B. Harrington" <peter.harrington@ohio.edu> wrote in message
> <47af565c$0$1103$4c368faf@roadrunner.com>...
> > Hi:
> >
> > I was wondering if there are any clever tricks to search for sets of real
> > numbers in a larger list, especially without using a for loop. I only know
> > how to do the search for one element at a time.
> >
> > A = [1.0, 2.0];
> > B = [1.0, 2.0, 3.0];
> >
> > for i = 1:length(A)
> > [t, ind(i)] = min(abs(A(i)-B));
> > end
> >
> > Just curious if there is a better way to write this code.
> >
> > Thanks in advance,
> >
> > Pete
> ---------
> Here is how I would have performed just the operation you described in
your
> for-loop construct using my antiquated version of matlab. I'm not at all
sure
> that it would be faster on a newer version, however.
>
> [a,b] = meshgrid(A,B);
> [ignore,ind] = min(abs(a-b),[],1);
>
> Roger Stafford

As it turns out, using bsxfun can help quite
a bit here.

n = 1000;
A = rand(n,1);B = rand(n,1);

tic,[x,y] = meshgrid(A,B);d=abs(x-y);toc
Elapsed time is 0.221196 seconds.

d = abs(bsxfun(@minus,A,B'));toc
Elapsed time is 0.072752 seconds.

Unfortunately, bsxfun appeared only in the
last release or so. It never has to create
those intermediate arrays.

John

"John D'Errico" <woodchips@rochester.rr.com> wrote in message <font4t$ifr
$1@fred.mathworks.com>...
> As it turns out, using bsxfun can help quite
> a bit here.
>
> n = 1000;
> A = rand(n,1);B = rand(n,1);
>
> tic,[x,y] = meshgrid(A,B);d=abs(x-y);toc
> Elapsed time is 0.221196 seconds.
>
> d = abs(bsxfun(@minus,A,B'));toc
> Elapsed time is 0.072752 seconds.
>
> Unfortunately, bsxfun appeared only in the
> last release or so. It never has to create
> those intermediate arrays.
>
> John
--------
  My matlab 4a (1994) is weeping, John!

Roger Stafford

"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid>
wrote in message <fonupf$7fu$1@fred.mathworks.com>...
> "John D'Errico" <woodchips@rochester.rr.com> wrote in
message <font4t$ifr
> $1@fred.mathworks.com>...
> > As it turns out, using bsxfun can help quite
> > a bit here.
> >
> > n = 1000;
> > A = rand(n,1);B = rand(n,1);
> >
> > tic,[x,y] = meshgrid(A,B);d=abs(x-y);toc
> > Elapsed time is 0.221196 seconds.
> >
> > d = abs(bsxfun(@minus,A,B'));toc
> > Elapsed time is 0.072752 seconds.
> >
> > Unfortunately, bsxfun appeared only in the
> > last release or so. It never has to create
> > those intermediate arrays.
> >
> > John
> --------
> My matlab 4a (1994) is weeping, John!
>
> Roger Stafford
>

You might consider my NEARESTPOINT on the FEX:

A = rand(N,1) ; B = rand(N,1) ; tol = 1e-12 ;
tic ; [i,d] = nearestpoint(A,B) ; i = i(d<tol) ; toc ;

Some timing on my PC, Matlab 7.0
N = 1000 ;
Elapsed time is 0.004309 seconds.

N = 1e6 ;
Elapsed time is 4.651996 seconds.

Nearestpoint can be found here:
http://www.mathworks.com/matlabcentral/fileexchange/loadFile.do?objectId=8939&objectType=FILE

hth
Jos

"Peter de B. Harrington":
<SNIP lost his/her numbers...

one of the other solutions

% the data
     a=pi*(1:10);
     b=pi*(11:20);
     b([2,4,5,6])=a(1:4);
     b(1)=3;
% the engine
     tol=.2;
     as=sort([a-tol,a,a+tol]);
     [ix,ix]=histc(b,as);
     r=b(ix>0);
% the results
     find(ix>0)
% 1 2 4 5 6
     r
% 3 3.14159 6.28319 9.42478 12.5664

us

One among many solutions:

A=rand(1,3);
B=rand(1,50);

[Bsorted I J]=unique(B);
inearest=I(interp1(Bsorted,(1:length(Bsorted)),A,...
           'nearest','extrap'));

A - B(inearest)

Bruno