# [Solved] Upsampling multidimensional array with interpn

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Johnny Scalera on 9 Jul 2020
Edited: Johnny Scalera on 10 Jul 2020
Hi all,
I'm trying to upsample a multidimensional array. Here is the code that I used for a monodimensional array:
sf1 = 100; %original sampling frequency
time = (0:(1/sf1):10-(1/sf1))'; %time array
x = randn(length(time), 1);
sf2 = 200; %final sampling frequency
upTime = (0:(1/sf2):10-(1/sf2))';
y = interp1(time, x, upTime);
Now I would like to upsample a 5D array, such as:
z = [randn(length(time), 1) randn(length(time), 1) randn(length(time), 1) randn(length(time), 1) randn(length(time), 1)];
How Can I use the interpn function properly?
Thank You all!

John D'Errico on 9 Jul 2020
Edited: John D'Errico on 9 Jul 2020
That is NOT a 5 dimensional array.
While you may THINK of it as a set of points that live in 5 dimensions, it is NOT a 5-d array. You CANNOT use interpn to interpolate that data. NOT. Period. NOT.
THIS is a 5-dimensional array:
A = rand(2,3,4,5,6);
size(A)
ans =
2 3 4 5 6
You can then use tools like griddatan or scatteredInterpolant to interpolate such data.
If you want to view the data as 5 independent streams of ONE dimensional data, then you can still use interp1 to interpolate. For example:
A = rand(100,5);
t = (1:100)';
interp1(t,A,7.25)
ans =
0.35284 0.051081 0.32818 0.34913 0.75639
Only you know what the array represents. By themselves, numbers are just numbers. Only you know what they represent.

Johnny Scalera on 10 Jul 2020
You are right! This is the correct solution
y1 = interp1(time, z, upTime);
Thank you!
John D'Errico on 10 Jul 2020
Excellent. It almost had to be one of those two cases.