How to save Matrices which i created in a "for" loop?
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This is my current code:
clc M = dec2bin(0:2^15-1, 15); A=zeros(5);
for i=1:2^15
A(1,1:5)=[str2num(M(i,1)),str2num(M(i,2)),str2num(M(i,3)),str2num(M(i,4)),str2num(M(i,5))];
A(2,2:5)=[str2num(M(i,6)),str2num(M(i,7)),str2num(M(i,8)),str2num(M(i,9))];
A(3,3:5)=[str2num(M(i,10)),str2num(M(i,11)),str2num(M(i,12))];
A(4,4:5)=[str2num(M(i,13)),str2num(M(i,14))];
A(5,5)=[str2num(M(i,15))];
end;
At the moment i am not able to use the matrices i created during the loop. Is there a way to save them, so that i can later use them again?
Could you please include the answer into this code, because i am quite new to matlab.
Thank you very much for your help.
Accepted Answer
Evan
on 2 Jul 2013
Edited: Evan
on 2 Jul 2013
You could create a 3D matrix in order to not lose each value on the next iteration. Just add a third dimension in your indexing:
for i=1:2^15
A(1,1:5,i) =[str2num(M(i,1)),str2num(M(i,2)),str2num(M(i,3)),str2num(M(i,4)),str2num(M(i,5))];
A(2,2:5,i)=[str2num(M(i,6)),str2num(M(i,7)),str2num(M(i,8)),str2num(M(i,9))];
A(3,3:5,i)=[str2num(M(i,10)),str2num(M(i,11)),str2num(M(i,12))];
A(4,4:5,i)=[str2num(M(i,13)),str2num(M(i,14))];
A(5,5,i)=[str2num(M(i,15))];
end
You'll now have a 5x5x2^15 size matrix, where each "layer" is the result from each iteration of your loop.
1 Comment
James Tursa
on 3 Jul 2013
Just be sure to preallocate for this case, e.g.,
A=zeros(5,5,2^15);
More Answers (2)
Jan
on 3 Jul 2013
Note: You can omit the large number of STR2NUM calls, if you convert M initially:
M = dec2bin(0:2^15-1, 15) - '0'; % Implicite conversion to DOUBLE
A = zeros(5, 5, 2^15);
for i = 1:2^15
A(1,1:5,i) = M(i,1:5); % Do we need a RESHAPE here?
...
end
Of course you could look into the code of DEC2BIN and avoid the temporary conversion to a CHAR array also. And finally the FOR loop is not required also:
A(1, :, :) = reshape(M(:, 1:5)', 1, 5, 2^15);
etc.
Sorry, I cannot test this currently.
3 Comments
Jan
on 3 Jul 2013
@andreas: In the posted code the index is moved from the FOR loop directly into the assignement. So in "A(1, 1:5, i)" the "i" is replaced by "1:2^15". And because A has the required size already, "1:2^15" can be replaced by ":".
I try to test this in the evening and post a complete code then.
You are welcome in the forum and it is the nature of beginning that details have to be learned.
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