How to implement this function in a single line in MATLAB - matlab

Question

The resulting indices of the transformed light field data are ub, vb, sb, tb. Each of them is depending on variables u,v,s,t.
Sorry for being unclear, let me mention that I am trying to transform a 4D dataset through some sort of matrix. In the code below M is simply a 3D transformation matrix.
f=0.1;
n = 11;
[u,v,s,t] = ndgrid([1:Size(3)],[1:Size(4)],[1:Size(1)],[1:Size(2)]);
alpha = M(3,1)*s+M(3,2)*t+M(3,3)*nf;
beta1 = M(1,1)*u+M(1,2)*v+M(1,4);
beta2 = M(2,1)*u+M(2,2)*v+M(2,4);
C = M(3,1)*u+M(3,2)*v+M(3,4);
D1 = M(1,1)*s+M(1,2)*t+M(1,3)*nf;
D2 = M(2,1)*s+M(2,2)*t+M(2,3)*nf;
ub = -D1.*C./alpha+beta1;
vb = -D2.*C./alpha+beta2;
sb = nf*D1./alpha;
tb = nf*D2./alpha;
for s = 1:Size(1)
for t = 1:Size(2)
for u = 1:Size(3)
for v = 1:Size(4)
newLF(sb(u,v,s,t),tb(u,v,s,t),ub(u,v,s,t),vb(u,v,s,t)) = LF2(s,t,u,v);
end;
end;
end;
end;
Now since ub,vb,sb and tb are depending on u,v,s,t therefore, it is not possible to assign it like newLF = LF2;
Now the question is how to minimize these for loops to a single line.

Answer 1

The answer is
newLF = LF2;
Meaning, that code does nothing but copy LF2 to newLF.
To check that I'm right, just let the code run with some random matrix LF2 and then evaluate
all(newLF(:) == LF2(:))
and you'll find it always evaluates to "true".
First of all, your use of sb, tb, ub, vb is redundant. You are indexing into a grid, but this just reproduces the indices. The line
newLF(sb(u,v,s,t),tb(u,v,s,t),ub(u,v,s,t),vb(u,v,s,t)) = LF2(s,t,u,v);
is equivalent to the line
newLF(s,t,u,v) = LF2(s,t,u,v);
This of course is just element-wise copying.
The impression of permutation noted by Shai is given by the line
[ub,vb,sb,tb] = ndgrid([1:Size(3)],[1:Size(4)],[1:Size(1)],[1:Size(2)]);
which looks like you are preparing to permute dimensions (1, 2) with dimensions (3,4). However, you use this index grid in the form sb, tb, ub, vb, assigning the value from s, t, u, v, so the permutation is not actually performed.
Assuming you actually do want to do that permutation of dimensions, the correct code would be
for s = 1:Size(1)
for t = 1:Size(2)
for u = 1:Size(3)
for v = 1:Size(4)
newLF(u,v,s,t) = LF2(s,t,u,v);
end;
end;
end;
end;
In this case Shai would be right, the corresponding one-liner is
newLF = permute(LF2, [3 4 1 2]);