Deleting/selecting and subsetting specific slices in a multi-dimensional array

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Sai Prasanth 2020-5-29

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评论： Sai Prasanth 2020-5-31

Hello,

I have three variables that are three-dimensional arrays of the size 699 (say, x or rows) x 699 (say, y or columns) x 60 (say, z or slices). I have first computed certain indices using certain conditions. E.g.,

idxs = Z > 5000 & W > 1 & M_tot > 0.05; % Idx is now a logical array of size 699 x 699 x 60.

Now, for further analysis, I am interested in keeping only those slices that satisfy the above conditions. In other words, I want to delete those slices where the idxs at a given x,y are zeros for all z. I used the any function below to create a subset:

subset_idx = any(idxs,3);

This creates a 699 x 699 logical array where I would like to keep all the points corresponding to 1 and delete all the rows and columns corresponding to a 0. This should give me a subset array of size nx, ny, 60 where nx and ny are much less than 699. How do I achieve this?

I tried something like this:

todelete_idx = ~(any(idxs,3)); Size : 699 x 699 logical

idxs(todelete_idx,:) = [];

However, this produces an error since the todelete_idx is a two-dimensional array and I am unable to subset it like this. What is the way around this?

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Sai Prasanth 2020-5-29

The deletion of the rows/columns is not as important as subsetting. That is, after this, I would like to perform operations on the nx, ny, 60 matrix where the x and y satisfy the condition. The intent is that I increase the computational efficiency by working on only those xs and ys that satisfy the condition.

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Answer 1

Sindar 2020-5-31

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编辑：Sindar 2020-5-31

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This may not be the most efficient way, but you can take that 2D logical array and rebuild your 3D logical array:

subset_3D = and(subset_idx,true(1,1,size(idxs,3)));

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Sindar 2020-5-31

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Note that this should definitely work for subsetting, but deletion is more complicated. Consider a simple 2D case:

>> x = [1 2;3 4]
x = 
     1     2
     3     4

Now, if you want to remove the '3', what do you get?

On paper, we could naively leave a blank there:

     1     2
           4

but that doesn't make sense either mathematically or to Matlab:

>> x(2,1)=[]
A null assignment can have only one non-colon index.

Matlab would have us use NaN:

>> x(2,1)=NaN
x =
     1     2
   NaN     4

and that may be what you want.

But, there's another option: we work on the subset directly:

>> idx=[true true; false true]
idx =
  2×2 logical array
   1   1
   0   1
>> x(idx)
ans =
     1
     2
     4
>> x(idx) = x(idx).^2
x =
     1     4
     3    16

since the subset is treated as a vector, it doesn't matter how many dimensions the original matrix had.

But, you have to be careful that you consistently work with corresponding subsets:

>> x = [1 2;3 4]
x = 
     1     2
     3     4
>> y = [1 1;2 3]
y = 
     1     1
     2     3
>> x(idx) = x(idx) + y
Matrix dimensions must agree.
>> x(idx) = x(idx) + y(idx)
x =
     2     3
     3     7

Sindar 2020-5-31

编辑：Sindar 2020-5-31

在 MATLAB Online 中打开

facepalm yeah, just repeating the matrix is definitely better than my convoluted and process. Note though that while conform_dims is handy, it might lose some of the efficiency and coding directly:

subset_3D = repmat(subset_idx,[1 1 size(idxs,3)])

could be faster. It's also possible that Matlab does some magic in the background that will save you time/memory throughout, since it knows that subset_3D(:,:,15) is the same as subset_3D(:,:,1)

I'm also not sure whether conform_dims will return a logical matrix

Sai Prasanth 2020-5-31

Makes sense, thank you so much!

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