How to make a m*2 matrix into n number of 2x2 matrices

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Hey,
I have, A = 208x2 matrix. I wish to to spit this matrix into 104 2x2 matrices. I have tried using num2cell and mat2cell but have had no luck. Any help would be appreicated.
Thanks.

采纳的回答

Stephen23
Stephen23 2022-9-7
A = rand(208,2);
C = mat2cell(A,2*ones(104,1),2)
C = 104×1 cell array
{2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double}
  6 个评论
Dyl
Dyl 2022-9-8
Hey I have another question I reshaped the 2x1x104 matrix into a 13x8 matrix where I only took one value for each val(:,:,1) , valu(:,:,2) etc..
13x8 = [-28,-64 ... ;-33,-74...]
I wanted the values to insert into the 13x8 matric column wise so 1-8 for row 1 then 1-8 for row 2 so on, but they inserted row wise. How can I make the matrix I am after? Thanks in advance.

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更多回答(2 个)

KSSV
KSSV 2022-9-7
A = rand(208,2) ;
[r,c] = size(A);
nlay = 104 ;
out = permute(reshape(A',[c,r/nlay,nlay]),[2,1,3]);

Abderrahim. B
Abderrahim. B 2022-9-7
Split A
A = randi(10, 208, 2) ; % a mtarix of size 208x2
size(A)
ans = 1×2
208 2
B = reshape(A, 2, 2, []) ;
Access 2x2 matrices
B1 = B(:,:,1)
B1 = 2×2
5 7 9 6
B2 = B(:,:,2)
B2 = 2×2
8 8 4 1
Hope this helps
  2 个评论
Stephen23
Stephen23 2022-9-7
编辑:Stephen23 2022-9-7
Note that this method does not keep the 2x2 matrices of the original matrix:
A = randi(10, 208, 2)
A = 208×2
3 1 9 7 9 3 4 9 8 7 5 5 4 4 9 1 8 10 10 8
B = reshape(A, 2,2,[]) % not the same matrices
B =
B(:,:,1) = 3 9 9 4 B(:,:,2) = 8 4 5 9 B(:,:,3) = 8 5 10 4 B(:,:,4) = 2 1 10 10 B(:,:,5) = 2 5 7 9 B(:,:,6) = 4 7 4 9 B(:,:,7) = 6 9 4 6 B(:,:,8) = 10 9 2 3 B(:,:,9) = 10 4 6 4 B(:,:,10) = 8 5 2 5 B(:,:,11) = 4 2 6 4 B(:,:,12) = 5 1 6 8 B(:,:,13) = 7 4 4 6 B(:,:,14) = 3 7 6 6 B(:,:,15) = 4 4 7 9 B(:,:,16) = 9 6 2 10 B(:,:,17) = 8 9 7 6 B(:,:,18) = 3 3 2 4 B(:,:,19) = 6 1 6 3 B(:,:,20) = 4 3 10 10 B(:,:,21) = 5 9 6 5 B(:,:,22) = 2 2 8 10 B(:,:,23) = 4 8 3 7 B(:,:,24) = 9 7 2 3 B(:,:,25) = 8 10 6 9 B(:,:,26) = 6 7 8 4 B(:,:,27) = 7 10 10 1 B(:,:,28) = 3 9 7 4 B(:,:,29) = 4 3 9 10 B(:,:,30) = 4 5 3 2 B(:,:,31) = 2 5 9 5 B(:,:,32) = 2 9 9 4 B(:,:,33) = 2 1 6 9 B(:,:,34) = 8 1 3 4 B(:,:,35) = 2 4 4 3 B(:,:,36) = 7 1 5 4 B(:,:,37) = 10 7 1 10 B(:,:,38) = 8 1 1 10 B(:,:,39) = 10 4 1 8 B(:,:,40) = 2 7 9 7 B(:,:,41) = 3 10 2 7 B(:,:,42) = 8 5 9 5 B(:,:,43) = 3 10 10 1 B(:,:,44) = 7 9 2 3 B(:,:,45) = 2 10 9 10 B(:,:,46) = 8 1 9 6 B(:,:,47) = 5 8 7 9 B(:,:,48) = 2 2 3 2 B(:,:,49) = 3 2 3 5 B(:,:,50) = 4 2 9 3 B(:,:,51) = 7 10 10 3 B(:,:,52) = 9 5 3 3 B(:,:,53) = 1 3 7 9 B(:,:,54) = 7 4 5 1 B(:,:,55) = 10 2 8 10 B(:,:,56) = 7 8 8 4 B(:,:,57) = 3 3 6 1 B(:,:,58) = 9 2 5 6 B(:,:,59) = 10 9 6 3 B(:,:,60) = 1 9 5 4 B(:,:,61) = 5 9 4 3 B(:,:,62) = 7 9 8 5 B(:,:,63) = 3 7 3 10 B(:,:,64) = 8 10 10 8 B(:,:,65) = 6 10 9 4 B(:,:,66) = 8 2 7 3 B(:,:,67) = 3 6 5 3 B(:,:,68) = 7 5 5 6 B(:,:,69) = 3 6 6 10 B(:,:,70) = 4 4 6 1 B(:,:,71) = 2 7 6 8 B(:,:,72) = 1 6 1 5 B(:,:,73) = 7 1 10 1 B(:,:,74) = 1 3 4 6 B(:,:,75) = 1 4 5 9 B(:,:,76) = 10 10 9 1 B(:,:,77) = 10 2 4 4 B(:,:,78) = 4 8 8 3 B(:,:,79) = 10 1 10 8 B(:,:,80) = 6 5 7 9 B(:,:,81) = 10 9 8 5 B(:,:,82) = 5 5 5 8 B(:,:,83) = 1 8 5 1 B(:,:,84) = 10 4 4 8 B(:,:,85) = 8 7 6 4 B(:,:,86) = 4 10 4 2 B(:,:,87) = 9 6 3 2 B(:,:,88) = 9 7 9 7 B(:,:,89) = 10 8 10 7 B(:,:,90) = 1 3 9 4 B(:,:,91) = 7 4 1 8 B(:,:,92) = 10 9 5 9 B(:,:,93) = 2 4 6 8 B(:,:,94) = 6 3 7 10 B(:,:,95) = 10 9 10 2 B(:,:,96) = 10 5 3 2 B(:,:,97) = 10 4 9 4 B(:,:,98) = 10 5 9 8 B(:,:,99) = 5 7 6 7 B(:,:,100) = 7 10 1 9 B(:,:,101) = 1 2 7 9 B(:,:,102) = 8 3 5 6 B(:,:,103) = 10 10 4 8 B(:,:,104) = 6 10 3 4
To keep the original matrices requires taing into account the order of elements stored in memory:
B = permute(reshape(A.',2,2,[]),[2,1,3])
B =
B(:,:,1) = 3 1 9 7 B(:,:,2) = 9 3 4 9 B(:,:,3) = 8 7 5 5 B(:,:,4) = 4 4 9 1 B(:,:,5) = 8 10 10 8 B(:,:,6) = 5 2 4 10 B(:,:,7) = 2 7 10 8 B(:,:,8) = 1 8 10 4 B(:,:,9) = 2 3 7 6 B(:,:,10) = 5 3 9 1 B(:,:,11) = 4 9 4 5 B(:,:,12) = 7 2 9 6 B(:,:,13) = 6 10 4 6 B(:,:,14) = 9 9 6 3 B(:,:,15) = 10 1 2 5 B(:,:,16) = 9 9 3 4 B(:,:,17) = 10 5 6 4 B(:,:,18) = 4 9 4 3 B(:,:,19) = 8 7 2 8 B(:,:,20) = 5 9 5 5 B(:,:,21) = 4 3 6 3 B(:,:,22) = 2 7 4 10 B(:,:,23) = 5 8 6 10 B(:,:,24) = 1 10 8 8 B(:,:,25) = 7 6 4 9 B(:,:,26) = 4 10 6 4 B(:,:,27) = 3 8 6 7 B(:,:,28) = 7 2 6 3 B(:,:,29) = 4 3 7 5 B(:,:,30) = 4 6 9 3 B(:,:,31) = 9 7 2 5 B(:,:,32) = 6 5 10 6 B(:,:,33) = 8 3 7 6 B(:,:,34) = 9 6 6 10 B(:,:,35) = 3 4 2 6 B(:,:,36) = 3 4 4 1 B(:,:,37) = 6 2 6 6 B(:,:,38) = 1 7 3 8 B(:,:,39) = 4 1 10 1 B(:,:,40) = 3 6 10 5 B(:,:,41) = 5 7 6 10 B(:,:,42) = 9 1 5 1 B(:,:,43) = 2 1 8 4 B(:,:,44) = 2 3 10 6 B(:,:,45) = 4 1 3 5 B(:,:,46) = 8 4 7 9 B(:,:,47) = 9 10 2 9 B(:,:,48) = 7 10 3 1 B(:,:,49) = 8 10 6 4 B(:,:,50) = 10 2 9 4 B(:,:,51) = 6 4 8 8 B(:,:,52) = 7 8 4 3 B(:,:,53) = 7 10 10 10 B(:,:,54) = 10 1 1 8 B(:,:,55) = 3 6 7 7 B(:,:,56) = 9 5 4 9 B(:,:,57) = 4 10 9 8 B(:,:,58) = 3 9 10 5 B(:,:,59) = 4 5 3 5 B(:,:,60) = 5 5 2 8 B(:,:,61) = 2 1 9 5 B(:,:,62) = 5 8 5 1 B(:,:,63) = 2 10 9 4 B(:,:,64) = 9 4 4 8 B(:,:,65) = 2 8 6 6 B(:,:,66) = 1 7 9 4 B(:,:,67) = 8 4 3 4 B(:,:,68) = 1 10 4 2 B(:,:,69) = 2 9 4 3 B(:,:,70) = 4 6 3 2 B(:,:,71) = 7 9 5 9 B(:,:,72) = 1 7 4 7 B(:,:,73) = 10 10 1 10 B(:,:,74) = 7 8 10 7 B(:,:,75) = 8 1 1 9 B(:,:,76) = 1 3 10 4 B(:,:,77) = 10 7 1 1 B(:,:,78) = 4 4 8 8 B(:,:,79) = 2 10 9 5 B(:,:,80) = 7 9 7 9 B(:,:,81) = 3 2 2 6 B(:,:,82) = 10 4 7 8 B(:,:,83) = 8 6 9 7 B(:,:,84) = 5 3 5 10 B(:,:,85) = 3 10 10 10 B(:,:,86) = 10 9 1 2 B(:,:,87) = 7 10 2 3 B(:,:,88) = 9 5 3 2 B(:,:,89) = 2 10 9 9 B(:,:,90) = 10 4 10 4 B(:,:,91) = 8 10 9 9 B(:,:,92) = 1 5 6 8 B(:,:,93) = 5 5 7 6 B(:,:,94) = 8 7 9 7 B(:,:,95) = 2 7 3 1 B(:,:,96) = 2 10 2 9 B(:,:,97) = 3 1 3 7 B(:,:,98) = 2 2 5 9 B(:,:,99) = 4 8 9 5 B(:,:,100) = 2 3 3 6 B(:,:,101) = 7 10 10 4 B(:,:,102) = 10 10 3 8 B(:,:,103) = 9 6 3 3 B(:,:,104) = 5 10 3 4
Abderrahim. B
Abderrahim. B 2022-9-7
Thanks @Stephen23. But he does not mention that the order must be te same as in the original matrix!

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