interp2 with huge matrix

4 views (last 30 days)
Michael
Michael on 6 Jul 2011
Hello
Problem: I have a matrix (M) with 6000 by 11000 elements of complex or double type. The task is to preform a second order polynomial transformation (shift and warp) with interp2().
[X Y] = meshgrid(1:11000,1:6000); % very fast
% A and B are the coefficients
transY = A(1) + A(2)*Y + A(3)*X + A(4)*X.*Y + A(5)*X.^2 + A(6)*Y.^2; % very fast
transX = B(1) + B(2)*Y + B(3)*X + B(4)*X.*Y + B(5)*X.^2 + B(6)*Y.^2; % very fast
ZI = interp2(X,Y,M,transX,transY); % out of memory
The interp2() is very slow and as you can see there are four very large matrices and ZI would be the 5th, so running out of memory is always the case.
How to solve this problem?
So far I have tried to cut the matrix into tiles, but this does not work because of the NaN values for each tile.
Thanks in advance!

Answers (1)

the cyclist
the cyclist on 6 Jul 2011
By "cut the matrix into tiles" do you mean doing the interpolation by sections (such as X=1:1000, then X=1001:1100, etc)? I would expect that to work, if handled properly and being careful about edge cases. Maybe you should post the code you used to try to do that part.
  2 Comments
Michael
Michael on 6 Jul 2011
Thanks!
My code is:
function [OUT] = RESAMPLE(DATA)
% this function cuts a 2D Matrix into tiles and applies a +200 shift in Y
% and +300 in X
[imagewidth, imagelength] = size(DATA)
tilewidth = 800;
tilelength = 800;
OUT = zeros([imagewidth, imagelength]);
num_tiles_width = int32(fix(imagewidth/tilewidth));
num_tiles_length = int32(fix(imagelength/tilelength));
restwidth = mod(imagewidth, tilewidth);
restlength = mod(imagelength, tilelength);
if restwidth ~= 0
num_tiles_width + 1
end
if restlength ~= 0
num_tiles_length + 1
end
for i = 0:num_tiles_width
%fprintf('ROW: %i\n',i);
if i == num_tiles_width
%fprintf('STOP\n')
left = (i * tilewidth) + 1;
right= (i * tilewidth) + restwidth;
else
left = (i * tilewidth) + 1;
right= (i+1) * tilewidth;
end
for j = 0:num_tiles_length
if j == num_tiles_length
%fprintf('STOP\n')
top = (j * tilelength) + 1;
bottom = (j * tilelength) + restlength;
else
top = (j * tilelength) + 1;
bottom = (j + 1) * tilelength;
end
fprintf('left: %i \t right: %i \t top: %i \t bottom: %i \n',left,right,top,bottom);
OUT(left:right,top:bottom) = interpolatetile(left,right,top,bottom);
end
end
function [ZI] = interpolatetile(l,r,t,b);
M = (DATA(l:r,t:b));
[meshX meshY] = meshgrid(t:b,l:r);
meshX = double(meshX);
meshY = double(meshY);
M = DATA(l:r,t:b);
%transY = A(1) + A(2)*meshY + A(3)*meshX + A(4)*meshX.*meshY + A(5)*meshX.^2 + A(6)*meshY.^2;
%transX = B(1) + B(2)*meshY + B(3)*meshX + B(4)*meshX.*meshY + B(5)*meshX.^2 + B(6)*meshY.^2;
transY = meshY + 200;
transX = meshX + 300;
M = M + double(r)*double(b);
ZI = interp2(meshX,meshY,M,transX,transY);
end
end
The NaN values of each tile (because of the shift every tile has NaNs) are the problem, and i do not know how to handle that.
Michael
Ashish Uthama
Ashish Uthama on 6 Jul 2011
Michael, consider editing your question to add this code with proper formatting. Aside: blkproc or blockproc might be useful(?).

Sign in to comment.

Categories

Find more on Interpolation in Help Center and File Exchange

Products

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!