How can you save a 1 by 3 array in each position of a 4D array?

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I am working on an interpolation project where I have 4 variables, each representing a dimension of a 4D array. There are three seperate values that coincide with each coordinate of the 4 dimensional array. The only way I have been able to make this work so far is that I currently have 3 seperate 4D arrays, each with the same set of variables but with one of the 3 values stored at each coordinate. The issue is that this creates a lot of reduntanat data and significantly increases the file size, while also increasing the computational resource requirement by running the interpolation 3 seperate times. I would like to only have one 4D array, with each coordinate housing a 1x3 with all 3 data points needed. Is this possible? If so, what is the best approach? I have approached this using Cell arrays, but I still run into errors of trying to assign 1x3 in a 1x1 space or with trying to assign doubles in the cell array.
  3 Comments
Catalytic
Catalytic 7 minutes ago
Edited: Catalytic 7 minutes ago
@Mitchell We can't understand your question clearly. Too many words. Not enough code.
Mitchell
Mitchell 5 minutes ago
@Matt J thank you, this was more of a conceptual issue of higher dimensions arrays for me. Between your comment here and your answer I have my head wrapped around it now.

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Accepted Answer

Matt J
Matt J about 2 hours ago
Edited: Matt J about 1 hour ago
I would like to only have one 4D array, with each coordinate housing a 1x3 with all 3 data points needed. Is this possible?
You could have a 5D array, where the 5th dimension are the 3 values. Then you can interpolate the 5D array in one go, e.g.,
V=rand(4,4,4,4,3);
xq=linspace(1,4,8);
F=griddedInterpolant(V);
Vq = F({xq,xq,xq,xq,1:3}) ; %upsample by interpolation
whos V Vq
Name Size Kilobytes Class Attributes V 4x4x4x4x3 6.000 double Vq 8x8x8x8x3 96 double
I'm not optimistic this will do much for you beyond keeping everything compactly in a single array variable, but on the other hand, I don't see any signs you had any inefficiency to begin with (see my comment above).
  1 Comment
Mitchell
Mitchell less than a minute ago
Perfect. Yeah I am seeing that now, with your answer here and @the cyclist below, I compared my 3 seperate 4D arrays to a 5D array and to a 4D cell array and they all come out to essentially the same storage requirements, there are just differences in how to interpolate/access the data. For my application going to the 5D array will work best now that I have my head wrapped around it. Simple enough, sometime you just need someone else to point it out. Thanks!

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More Answers (1)

the cyclist
the cyclist about 2 hours ago
This code creates a 4-dimensional cell array, and stores a 1x3 numeric vector in one of the elements
c = cell(2,3,5,7);
c{1,2,3,4} = rand(1,3);

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