How to transform a matrix in Matlab?

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Moe
Moe 2016 年 2 月 29 日
コメント済み: Priti Gujar 2020 年 6 月 15 日
Matrix A is as follows:
A = [0 240 245 250
25 1 2 1
63 3 2 1];
I want matrix A to be transformed to B (like follows):
B = [0 240 245 250 240 245 250 240 245 250
25 1 0 1 0 1 0 0 0 0
63 0 0 1 0 1 0 1 0 0];
there are three different variables in matrix A, so, 204 to 250 (first row) in matrix B is repeated 3 times (e.g. if there were 5 variables, then 240 to 250 should be repeated 5 times). Then, value of ID = 25 has 1, so 1 is added to B(2,2). Again, A(2,3) = 2, then B(2,6) should by =1 and A(2,4)=1, then A(2,10) should be equal by =1. And same for ID#63
  4 件のコメント
Priti Gujar
Priti Gujar 2020 年 6 月 15 日
Use the readall function to import all the data. Check that the preprocessing function was applied to each file by plotting the Y variable as a function of Time.

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回答 (2 件)

John BG
John BG 2016 年 2 月 29 日
Hi Mohammad
MATLAB has the command linsolve to solve linear equation systems of the type A*x=b
Your question has one A:
A = [0 240 245 250; 25 1 2 1; 63 3 2 1]
and ten b:
B= [0 240 245 250 240 245 250 240 245 250;
25 1 0 0 0 1 0 0 0 1;
63 0 0 1 0 1 0 1 0 0]
One way to solve them is with a for loop:
C=zeros(4,10)
for k=1:1:10
C(:,k)=linsolve(A,B(:,k))
end
answer:
C =
Columns 1 through 6
1.00 0.00 -0.04 -0.04 -0.04 0.00
-0.00 -0.55 0.77 1.30 0.75 -0.02
0 0 0 0 0 0
0.00 1.49 0.24 -0.25 0.24 1.00
Columns 7 through 10
-0.04 -0.04 -0.04 0.00
0.78 1.27 0.77 -0.52
0 0 0 0
0.25 -0.26 0.24 1.50
test it's the correct answer
A*C
ans =
Columns 1 through 6
0.00 240.00 245.00 250.00 240.00 245.00
25.00 1.00 -0.00 -0.00 -0.00 1.00
63.00 -0.00 0 1.00 0 1.00
Columns 7 through 10
250.00 240.00 245.00 250.00
0 -0.00 -0.00 1.00
0 1.00 0 -0.00
note the type (class) has changed to double. To bring it back to, for instance, range [0 255] use uint8.
does this answer help? if so click on the thumbs-up icon link on the top of this page, thanks in advance
John
  2 件のコメント
John BG
John BG 2016 年 2 月 29 日
understood, had to read the previous question to realize that A is a table and the kind of the variable tagging sought.
B has to be built by the answer, not used by it to find a transform matrix.
Thanks for mentioning Mohammad's comment.

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Andrei Bobrov
Andrei Bobrov 2016 年 2 月 29 日
編集済み: Andrei Bobrov 2016 年 2 月 29 日
[m,n] = size(A);
[ii,k] = ndgrid(1:n-1,1:m-1);
jj = A(2:end,2:end)';
b0 = accumarray([ii(:),jj(:),k(:)],1,[n-1,n-1,m-1]);
B = [nan,repmat(A(1,2:end),1,n-1);[A(2:end,1),reshape(b0,[],m-1)']];
or with bsxfun
[m,n] = size(A);
n1 = n - 1;
A0 = A(2:end,2:end)';
b0 = reshape( bsxfun(@eq,1:n1,reshape(A0,n1,[],m-1)),[],m-1)';
B = [nan,repmat(A(1,2:end),1,n-1);[A(2:end,1),b0]];
  2 件のコメント
Andrei Bobrov
Andrei Bobrov 2016 年 2 月 29 日
corrected

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