# Accessing Elements in a 3D matrix using Linear Indexing ?

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Venkata Khambhammettu 2020 年 8 月 11 日

Hello,
I would like to know if it is possible to access the individual element of a 3D matrix of size M*N*P using linear indexing?

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### 採用された回答

Bruno Luong 2020 年 8 月 11 日

Given sz = [m,n,p] ans linidx is the linear index of an element, here is one way of computing the subindexes (row, col, page). This works also for generic nd-array.
sz = [m,n,p];
tmp = linidx-1;
nd = max(length(sz),2);
subidx = zeros(1,nd);
for k=1:nd
subk = mod(tmp, sz(k));
subidx(:,k) = subk;
tmp = (tmp-subk) / sz(k);
end
subidx = subidx+1;
You can also see TMW implementation by
>> edit ind2sub

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Venkata Khambhammettu 2020 年 8 月 11 日
Thanks. This definetly helped.

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### その他の回答 (2 件)

Sudheer Bhimireddy 2020 年 8 月 11 日
Yes, it is possible. Read this.
A = rand(10,10,10);
B = A(1:2,1:2,1:2); %<- indexing the first two values in all dimensions so it creates a 2x2x2 matrix
C = A(1,1,1); %<- indexing the first value in all dimensions so it points to a single value

#### 4 件のコメント

Venkata Khambhammettu 2020 年 8 月 11 日
Thanks Stephen. So the concept of Linear Indexing doesn't work for 3D matrices ?
Image Analyst 2020 年 8 月 11 日
Not sure how you deduced that. Of course it does work. You just want to know "how MatLab able to use linear indexing to calculate the location of an element in a 3D matrix." and it's in column major order, basically by iterating the left most index first, then the second index, then finally the third index the slowest.
Stephen Cobeldick 2020 年 8 月 11 日
"So the concept of Linear Indexing doesn't work for 3D matrices ?"
Of course linear indexing works with 3D arrays, just as the documentation that I linked to clearly states: "Another method for accessing elements of an array is to use only a single index, regardless of the size or dimensions of the array. This method is known as linear indexing" (bold added).
This answer does not show any linear indexing though, so it is unrelated to your question.

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Image Analyst 2020 年 8 月 11 日
Yes, it's possible but you'd need 3 dimensions for the linear array, not 2. So not M-by-N but rows-by-columns-by-slice.
[rows, columns, slices] = size(your3DArray);
mask = whatever; % Needs to be a "rows by columns by slices" 3-D array.

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Image Analyst 2020 年 8 月 11 日
I told you the algorithm. As you go from linear index 1 to 2 to 3 to 4, etc. it starts with slice 1 and goes down rows first (see the left column of A(:,:,1)) and then moves over to the second column - see where it says 4,5,6 in column 2 of A(:,:,1)? Then it moves over to the 3rd column, and finally the 4th/last column. Then it repeats that order for the second slice so that's how it goes from 13 to 24. Basically it iterates the left most index first/quickest in an array, then the right-most index last/slowest. You can use sub2ind() and ind2sub() to convert between a linear index and actual separated indexes.
Steven Lord 2020 年 8 月 11 日
You can work backwards.
First, which page contains 20? It can't be the first, because that only contains linear indices 1 through 12. We know this because the size of the array in the first two dimensions is [3 4]. So we know that element 20 is at subscripts (?, ?, 2).
Now we subtract off 12 to find the linear index of the corresponding element in the first page of a matrix. Now we're trying to find the row and column indices in a 3-by-4 matrix corresponding to linear index 8. It can't be in the first column (linear indices 1-3) or the second (4-6) so we know it's in the third column, at (?, 3, 2).
We subtract off 6 to find the linear index of the corresponding element in the first column. This tells us that a linear index of 20 in a 3-by-4-by-2 array corresponds to the element at subscripts (2, 3, 2).
I've done this handwavingly, but you can formalize it using remainders (rem) as ind2sub does.
Venkata Khambhammettu 2020 年 8 月 11 日
Thanks Steven. I am pretty much looking at it like you explained to comeup with a mathematical equation to calculate all the sub-indexes.

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