How does mat2cell work?

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Diego Hens
Diego Hens am 15 Jan. 2021
Kommentiert: Diego Hens am 15 Jan. 2021
Hello,
I need your help understanding how cell2mat works. I have tried so hard, but I just don't get it.
I have:
  • a Matrix called Matrix_A with a size of 18x46092.
I would like to:
  • separate it so, that I get 15364 cells (this is the amount of points, 46092/3, for the x, y and z coordinates), each of these cells containing a 18x3 matrix (18 points each).
I have already done it with a line someone gave me. It is for the mean of Matrix_A. It works.
Matrix_A = transpose(mat2cell(mean(cell2mat(Matrix_A)), 1, 3*ones(size(Matrix_A,2),1)));
Can someone please explain how cell2mat works, so that I can use it myself? A solution (a line of code) solves the problem right now, but this is the second or third time I want to do this and can't. That's why I would prefere an easy to understand explanaition.
Thank you!
Edit: I cannot post the Matrix_A data as it is too big.

Akzeptierte Antwort

Steven Lord
Steven Lord am 15 Jan. 2021
Let's try this with a slightly smaller matrix, so you can see what's going on.
A = reshape(1:24, [3 8])
A = 3×8
1 4 7 10 13 16 19 22 2 5 8 11 14 17 20 23 3 6 9 12 15 18 21 24
Let's slice this into 3-by-2 chunks. We don't actually need to slice it row-wise (since we want each chunk to have 3 rows just like A has three rows) but we do need to slice it column-wise.
rowSliceSizes = 3; % Each cell has three rows like A does
columnSliceSizes = [2 2 2 2]; % Each cell contains 2 columns of A
Now slice.
C = mat2cell(A, rowSliceSizes, columnSliceSizes)
C = 1x4 cell array
{3×2 double} {3×2 double} {3×2 double} {3×2 double}
celldisp(C)
C{1} = 1 4 2 5 3 6 C{2} = 7 10 8 11 9 12 C{3} = 13 16 14 17 15 18 C{4} = 19 22 20 23 21 24
Now let's slice A into a 2-by-3 cell array. Each cell in the first row of the result should contain 2 rows of A and each cell in the second row should contain the third row of A. Each cell in a column of the result contains 3, 2, and 3 columns of A respectively:
D = mat2cell(A, [2 1], [3 2 3])
D = 2x3 cell array
{2×3 double} {2×2 double} {2×3 double} {1×3 double} {1×2 double} {1×3 double}
celldisp(D)
D{1,1} = 1 4 7 2 5 8 D{2,1} = 3 6 9 D{1,2} = 10 13 11 14 D{2,2} = 12 15 D{1,3} = 16 19 22 17 20 23 D{2,3} = 18 21 24
  1 Kommentar
Diego Hens
Diego Hens am 15 Jan. 2021
Thank you so much for your long and detailed answer. I have understood it and have applied it to my work.
Thank you for helping me to understand it and for not telling me the answer beforhand.
I am confident I will be able to correctly use this command next time.
Thanks!

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