How to save Matrices which i created in a "for" loop?

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.

[Jan]

Please do not cross-post. Thanks.

Answer 1

[Evan]

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.

[James Tursa]

Just be sure to preallocate for this case, e.g.,

A=zeros(5,5,2^15);

Answer 2

[Jan]

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.

[andreas]

Thank you Jan. If i get you right, i don't have to use a the for loop and just use that reshape statement. I didn#t use it at this point, because i don#t undertsand exactly how i have to use it. Perhaps you can be a bit more precise.

PS: Sorry that i don#t know how to behave on the forum yet. I will try to improve that.

[Jan]

@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.

[andreas]

Thank you very much this a huge help for me.

Answer 3

[andreas]

Thank you very much

[Jan]

Please post comments to answers in the corresponding comment field, not as new answer. If your problem is solved, accept the corresponding answer, such that it is clear, that you do not need further help. Thanks.