How to create a multidimensional array of fixed dimensions?

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Mariana
Mariana 2020-5-8
评论： Mariana 2020-5-14
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! :)

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采纳的回答

Walter Roberson
Walter Roberson 2020-5-9
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.
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Walter Roberson
Walter Roberson 2020-5-9
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
Mariana 2020-5-14
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]);

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更多回答（1 个）

Steven Lord
Steven Lord 2020-5-8
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.
  1 个评论
Mariana
Mariana 2020-5-9
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?

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