Transpose matrix and your transposition

How can I transpose Thais Matrix: Matrix = zeros(3500000,3);

Answer 1

[David Goodmanson]

HI Airto,

Rather than actually transpose it with the ' (apostrophe) command, you can use

zeros(3,3500000);

The transpose

a = zeros(3.5e6,3);
b = a';

is reasonably fast, (about 17 millisec on my PC). You get up around a = zeros(3.5e8,3), though, then b = a'; takes about 1.5 sec, whereas

b = zeros(3,3.5e8)

is sub-millisec because no actual transpose is being done.

[Walter Roberson]

My tests show that

b = a.';

seems to be a hair faster.

a' and a.' produce identical results for real-valued data, but different results for complex-valued data.

Answer 2

[AIRTON]

Hi David and Hi Walter

I thank by yours attention. But when I carry out this:

a = zeros(3.5e6,3);
b = a';

or this

b = a.';

I get this:

Cannot display summaries of variables with more than 524288 elements.

How can I resolve this, please

Thank!

[Chun]

Hi Airto

I believe that both methods have transposed the matrix correctly, and you can use 'b' to continue your work since it has been saved.

This matrix can't be displayed because it has too many elements so that it reach the limit of variable editor.

If you want to check whether the transposition is correct, you might use

size(b)

If you want to see the element in first 10 rows or first 10 columns, you might use

b(:, 1:10)
b(1:10, :)

If you have to check all the elements within it, you might try

writematrix(b, 'filename.csv')%shoule be opened with excel or python
% or
save('filename.mat', 'b', '-v7.3')

However, since the number of elements within the matrix is 3 times 3500000, both documents will be very huge (maybe more than 6GB, I guess). In addition, they might also be hard to open...

[AIRTON]

Hy Chun

great, very good!

I have got to access the data.

Thank you very much!

[David Goodmanson]

Hi Chun, what you did aided the OP, but there is something going on with the 6GB estimate. A number in double precision has 64 bits, 8 bytes. So the a or b matrix runs to 3.5e6*3*8 = 84MB and should not be leagues larger with different software or platform.

Answer 3

[AIRTON]

Hi David, Walter and Chun

I am here again!

With yours helps I have done this scritp to filter two columns.

clear all 
clc
tic
b = 1;
r = zeros(3268760,1); 
load ('transp1.mat');
for i = 1:2
    z1 = a(a(:,b) == 0, :); 
    s1 = sum(z1, 1);
    s2 = max(s1);
    r(b,1) = max(s1);
    b = b + 1;
end
toc

Elapsed time is 209.340817 seconds.

My object is to gain to the results in "r" for all columns.

What I want of you is to decrease the time elapsed in the execution of script.

Can you help me please?

thank!