Saving matrix data from nested for loop

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DC
DC 2016-11-30
评论： DC 2016-12-1
I have a 18624*3 matrix called 'vertices2' that I wish to break up into separate matrices of 24*3 e.g. rows 1-24, 25-48 and so on. I have the following code which loops through my matrix to (almost) do what I want but it has two problems:
[x,y] = size(vertices2);
for i = 1:24:x
for j = 1:3:y
vertices3(i,j) = vertices2(i:(i+23), j:(j+2));
end
end
As it stands I get the following error... 'Assignment has more non-singleton rhs dimensions than non-singleton subscripts'. This isn't an error message I'm used to getting from using for loops.
If I remove the (i,j) then the code works but as you'd expect, the loop only saves that last iteration of the loop e.g. data from rows 18601-18624, which is of no use to me!
I've tried a number of variations of the code to get it to run but I'm hitting a brick wall and was hoping someone could help! Maybe the step I'm missing is to save each iteration as a new matrix before the loop moves onto the next??
Thanks in advance

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Jan
Jan 2016-11-30
The error message is as expected. With the debugger you can find out what happens in the first iteration:
for i = 1:24:x
for j = 1:3:y
vertices3(i,j) = vertices2(i:(i+23), j:(j+2));
==>
vertices3(1,1) = vertices2(1:24, 1:3);
Now you have a scalar on the left side and a matrix on the right side. This must fail. In the next iteration you would try to write "vertices(1,3)", but what should be in "vertices(1,2)"?
It is more efficient to use a 3D array to store the values:
vertices3 = reshape(vertices2, 24, [], 3);
Now the 2nd index means the specific block of data. Perhaps a permute is useful for resorting.
  1 个评论
DC
DC 2016-12-1
Hi Jan,
Thanks for both the help and explanation - makes perfect sense now. And vert = permute(vertices3,[1 3 2]); worked a treat! I was unaware of the permute command but looks like it's a good one to know. Now to crack on with the rest of my work...

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