Finding index location in volume?

Asked by Syed Abdul Salam

Syed Abdul Salam (view profile)

on 16 Sep 2019
Latest activity Commented on by Syed Abdul Salam

Syed Abdul Salam (view profile)

on 16 Sep 2019
Accepted Answer by Andrei Bobrov

Andrei Bobrov (view profile)

For instance, I have
x = 2000:0.5:2004; y = 45000:0.3:45009; z = 10:0.2:12;
which are axes of a volume, x has length of 9, y has 31 and z has 11.
Now a volume filled of nan values and given by;
C = rand(9,31,11)*nan;
I want to fill this volume with values when new values come up, it will have x, y, z and the corresponding value. So, I need to find the closest index location and fill the value there. E.g.
x =2002.13, y = 45006.811, z = 11.36 have value of 10000 so i want to fill the corresponding C(x,y,z) = 10000;
here I need to find the index for x = 2002, y = 45006.9, z = 11.4 and fill the value 10000 there in C.
how can I use the values converted to closed index in the volume. Thanks

Release

R2019a

Answer by Andrei Bobrov

Andrei Bobrov (view profile)

on 16 Sep 2019

x = 2000:0.5:2004;
y = 45000:0.3:45009;
z = 10:0.2:12;
x0 = 2002.13;
y0 = 45006.811;
z0 = 11.36;
C = nan(9,31,11);
C0 = 1000;
[~,ii] = min(abs(x - x0));
[~,jj] = min(abs(y - y0));
[~,k] = min(abs(z - z0));
C(ii,jj,k) = C0;

Syed Abdul Salam

Syed Abdul Salam (view profile)

on 16 Sep 2019
work perfectly. Thanks

Answer by Nicolas Broch

Nicolas Broch (view profile)

on 16 Sep 2019

Hi,
for your case, I would write your code like that:
x = 2000:0.5:2004; y = 45000:0.3:45009; z = 10:0.2:12;
% no need to use rand() to generate a table of NaN and it's more flexible with automatic size
C = NaN(length(x), length(y), length(z));
pts = [2002.13, 45006.811, 11.36];
[~, xc] = min(abs(x - pts(1)));
[~, yc] = min(abs(y - pts(2)));
[~, zc] = min(abs(z - pts(3)));