Interpn - changing output - matlab

Question

I have 4 grids:
kgrid which is [77x1]
x which is [15x1]
z which is [9x1]
s which is [2x1]
Then I have a function:
kprime which is [77x15x9x2]
I want to interpolate kprime at some points ksim (750 x 1) and zsim (750 x 1) (xsim is a scalar). I am doing:
[ks, xs, zs, ss] = ndgrid(kgrid, x, z, [1;2]);
Output = interpn(ks, xs, zs, ss, kprime, ksim, xsim, zsim, 1,'linear');
The problem with this interpolation is that the output given is for all combinations of ksim and zsim, meaning that the output is 750x750. I actually need an output of 750x1, meaning that instead of interpolation at all combinations of ksim and zsim I only need to interpolate at ksim(1,1) and zsim(1,1), then ksim(2,1) and zsim(2,1), then ksim(3,1) and zsim(3,1), etc.
In other words, after getting Output I am doing:
Output = diag(squeeze(Output));
I know I can use this output and then just pick the numbers I want, but this is extremely inefficient as it actually interpolates on all other points which I actually do not need. Any help appreciated.

Answer 1

tl;dr: Change xsim and (ssim) from scalars to vectors of the same size as ksim and zsim
Output = interpn (ks, xs, zs, ss, ...
kprime, ...
ksim, ...
repmat(xsim, size(ksim)), ... % <-- here
zsim, ...
repmat(1, size(ksim)), ... % <-- and here
'linear');
Explanation:
The ksim, xsim, zsim, and ssim inputs all need to have the same shape, so that at each common position in that shape, each input acts as an "interpolated subscript" component to the interpolated object. Note that while they all need to have the same shape, this shape can be arbitrary in terms of size and dimensions.
Conversely, if you pass vectors of different sizes (after all, a scalar is a vector of length 1), these get interpreted as the components of an ndgrid construction instead. So you were actually telling interpn to evaluate all interpolations on a grid defined by the vectors ksim, and zsim (and your singletons xsim and ssim). Which is why you got a 2D-grid-looking output.
Note that the same scheme applies with the constructing vectors as well (i.e. ks, xs, zs and ss) i.e. you could have used "vector syntax" instead of "common shape" syntax to define the grid instead, i.e.
Output = interpn(kgrid, x, z, s, kprime, % ...etc etc
and you would have gotten the same result.

Answer 2

From the documents:
Query points, specified as a real scalars, vectors, or arrays.
If Xq1,Xq2,...,Xqn are scalars, then they are the coordinates of a single query point in Rn.
If Xq1,Xq2,...,Xqn are vectors of different orientations, then Xq1,Xq2,...,Xqn are treated as grid vectors in Rn.
If Xq1,Xq2,...,Xqn are vectors of the same size and orientation, then Xq1,Xq2,...,Xqn are treated as scattered points in Rn.
If Xq1,Xq2,...,Xqn are arrays of the same size, then they represent either a full grid of query points (in ndgrid format) or scattered points in Rn.
Answer
You want the usage highlighted in bold. As such, you have to make sure that xsim and ssim ('1' in your code sample) are of size 750x1 also. Then all the query vectors are same length and orientation, such that it can be recognized as a vector of scattered points in Rn. The output will then be a 750x1 vector as needed.

Answer 3

This is to elaborate on #tvo/#Tasos answers, to test the fastest way to create a vector from a scalar:
function create_vector(n)
x = 5;
repm_time = timeit(#()repm(x,n))
repe_time = timeit(#()repe(x,n))
vrep_time = timeit(#()vrep(x,n))
onesv_time = timeit(#()onesv(x,n))
end
function A = repm(x,n)
for k = 1:10000
A = repmat(x,[n 1]);
end
end
function A = repe(x,n)
for k = 1:10000
A = repelem(x,n).';
end
end
function A = vrep(x,n)
v = ones(n,1);
for k = 1:10000
A = x*v;
end
end
function A = onesv(x,n)
for k = 1:10000
A = x*ones(n,1);
end
end
And the results are (for n = 750):
repm_time =
0.049847
repe_time =
0.044188
vrep_time =
0.0041342
onesv_time =
0.0024869
which means that the fastest way to create a vector from a scalar is simply writing x*ones(n,1).