Transpose matrix and your transposition

2025 9 20
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更新时间：2025 9 21
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David Goodmanson
David Goodmanson 2025-9-20
编辑：David Goodmanson 2025-9-21

1 个投票

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.

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Walter Roberson
Walter Roberson 2025-9-20
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.

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更多回答（2 个）

AIRTON
AIRTON 2025-9-20
编辑：Walter Roberson 2025-9-20

0 个投票

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!

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Chun
Chun 2025-9-20
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
AIRTON 2025-9-20
Hy Chun
great, very good!
I have got to access the data.
Thank you very much!
David Goodmanson
David Goodmanson 2025-9-21
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.

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AIRTON
AIRTON 2025-9-20
编辑：Walter Roberson 2025-9-20

0 个投票

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!

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