How to create a multidimensional array of fixed dimensions?

I would like to create an array in 4 dimensions. Each dimension has a fixed size.

x =0 to 200 -->step 1

y =-25 to 25 --> step 0.5

z = 0 to 180 --> step 1

h = -10 to 10 --> step 0.1

The idea behind is to save one value on a specifc position.

For example if I have an input array [30,-0.5,100,2 ] = 21

At that location I want to save the value.

At a specific amount I would like to read the values inside this array.

Hope you can help me! :)

Answer 1

[Walter Roberson]

xmin = 0; xmax = 200; xincr = 1;
ymin = -25; ymax = 25; yincr = 0.5;
zmin = 0; zmax = 180; zincr = 1;
hmin = -10; hmax = 10; hincr = 0.1;

xvec = xmin : xincr : xmax;
yvec = ymin : yincr : ymax;
zvec = zmin : zincr : zmax;
hvec = hmin : hincr : hmax;

x2xidx = @(xval) round((xval - xmin)/xincr) + 1;
y2yidx = @(yval) round((yval - ymin)/yincr) + 1;
z2zidx = @(zval) round((zval - zmin)/zincr) + 1;
h2hidx = @(hval) round((hval - hmin)/hincr) + 1;

nx = length(xvec); 
ny = length(yvec);
nz = length(zvec);
nh = length(hvec);

M = zeros(nx, ny, nz, nh);

%example of use

M(x2xidx(17), y2yidx(-6.5), z2zidx(93), h2hidx(-4) ) = 1;

[XIDX, YIDX, ZIDX, HIDX] = ind2sub(size(M), find(M));
disp([XIDX, YIDX, ZIDX, HIDX])
disp([xvec(XIDX), yvec(YIDX), zvec(ZIDX), hvec(HIDX)])

You can store by index or you can use the helper functions to convert numeric value to index.

[Walter Roberson]

You can generalize this by creating arrays of bounds and increments, and generate anonymous functions to do the conversions.

However, keep in mind that the above array is already close to 6 gigabytes. If you extend to 8 dimensions, you are likely to run out of memory.

[Mariana]

Yeah its quite a lot. Any ideas on how to reduce the memory usage? I thought it was using 3GB not 6

[Mariana]

x2xidx = @(xval) round((xval - xmin)/xincr) + 1;
y2yidx = @(yval) round((yval - ymin)/yincr) + 1;
z2zidx = @(zval) round((zval - zmin)/zincr) + 1;
h2hidx = @(hval) round((hval - hmin)/hincr) + 1;

What exactly are these lines doing?

[Walter Roberson]

you could use single precision by adding 'single' as an option to the zeros() call.

Those are helper functions that accept numeric values of the coordinates and calculate the corresponding index.

base + (index-1)*increment == value implies that (index-1)*increment = value - base which implies that index-1 == (value-base)/increment which implies that index == (value-base)/increment + 1

[Mariana]

ok. Thanks.

How to find all the possible combinations between those ranges and display them?

xmin = 0; xmax = 10; xincr = 1;
ymin = -5; ymax = 5; yincr = 0.5;
zmin = 0; zmax = 30; zincr = 1;
hmin = -10; hmax = 10; hincr = 0.5;
xvec = xmin : xincr : xmax;
yvec = ymin : yincr : ymax;
zvec = zmin : zincr : zmax;
hvec = hmin : hincr : hmax;
for xmin=xincr:xmax
    for ymin=yincr:ymax
        for zmin=zincr:zmax
            for hmin=hincr:hmax
            end
        end
    end
end

[Walter Roberson]

xmin = 0; xmax = 10; xincr = 1;
ymin = -5; ymax = 5; yincr = 0.5;
zmin = 0; zmax = 30; zincr = 1;
hmin = -10; hmax = 10; hincr = 0.5;
for x = xmin:xincr:xmax
    for y = ymin:yincr:ymax
        for z = zmin:zincr:zmax
            for h = hmin:hincr:hmax
                disp([x y z h])
            end
        end
    end
end

or

xmin = 0; xmax = 10; xincr = 1;
ymin = -5; ymax = 5; yincr = 0.5;
zmin = 0; zmax = 30; zincr = 1;
hmin = -10; hmax = 10; hincr = 0.5;
xvec = xmin : xincr : xmax;
yvec = ymin : yincr : ymax;
zvec = zmin : zincr : zmax;
hvec = hmin : hincr : hmax;
nx = length(xvec);
ny = length(yvec);
nz = length(zvec);
nh = length(hvec);

for xidx = 1 : nx
    x = xvec(xidx);
    for yidx = 1 : ny
        y = yvec(yidx);
        for zidx = 1 : nz
            z = zvec(zidx);
            for hidx = 1 : nh
                h = hvec(hidx);
                disp([x y z h]);
            end
        end
    end
end

If you are going to loop over values instead of doing vectorized calculations, then it is typically much better to loop over the indices, in order to permit you to store at appropriate output locations, such as

output(xidx, yidx, zidx, hidx) = x.^2 - y.^3/3 - (x.*z).^h;

[Mariana]

What if I have a variable step per vector. How will you suggest to find the index?

rmin = 0; rmax = 90; rincr = 0.5;
r2min = 91; r2max = 180; r2incr = 1;
r3min =181.5; r3max = 250 ; r3incr = 1.5;
rvec = rmin : rincr : rmax;
r2vec = r2min : r2incr : r2max;
r3vec = r3min : r3incr : r3max;
relx = single([rvec,r2vec,r3vec]);

Answer 2

[Steven Lord]

There's no such thing as the -25th column in an array in MATLAB, nor is there such a thing as the 1.5th column.

You can make an array with separate coordinates that you can use like indices (though as usual, be careful when performing exact equality comparisons with == on floating-point numbers. What I wrote below is safe because all the numbers in x and y can be represented exactly in double precision.)

x = (-5:5).';
y = (-3:0.5:3);
z = x.^2 + y.^3;
z(x == -4, y == 1.5) % (-4)^2+(1.5)^3

If you tell us a little more detail about what you're trying to do with this array we may be able to offer alternate suggestions.

[Mariana]

This is the kind of the idea that I would like to do. Just to save a value at a specific space of this array. The array has 8 dimensions or 8 indices. How would you suggest to build and add values to this storage array?