How can I write such a large matrix?
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esraa
el 11 de En. de 2015
Comentada: Walter Roberson
el 29 de Mzo. de 2020
thank you very much, how can I construct the matrix b, b: no. of column=20 and
no. of rows=2^20=1048576.
the first column=[0 1 0 1 0 1 ........]
the second column=[0 0 1 1 0 0 1 1 0 0 1 1 ........]
the third column=[0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 ..........]
the fourth column=[0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1..................]
and so on. instead of the following command
for j=1:20
for k=0:(2^(ns-j))-1
a((2^(j-1))+1+k*(2^j):(2^j)+k*(2^j))=1;
end
b(j,:)=a;
clear a
end
b=b';
to take less time compared with the commands I use
2 comentarios
Stephen23
el 11 de En. de 2015
This is unreadable. Please help the other people here by formatting your text correctly. You should edit your question and use the {} Code button above the text box.
You should also read these:
Oleg Komarov
el 11 de En. de 2015
@esraa: how can you fill a 1.5L bottle with 10L? The answer is you can't. If you keep asking the same question we won't be able to help. You need to tell us what do you need those 10L to begin with, so maybe we can start filling the bottle with the first 1.5L, use it, empty the bottle, re-fill and repeat until we used all 10L.
This is block processing. You need to split your problem in subproblems, in blocks whose results can then be combined at the end.
Please reformulate you question keeping specifically my suggestion in mind.
Respuesta aceptada
Roger Stafford
el 11 de En. de 2015
b = zeros(2^20,20);
for k = 1:20
b(:,k) = repmat([zeros(2^(k-1),1);ones(2^(k-1),1)],2^(20-k),1);
end
4 comentarios
John D'Errico
el 11 de En. de 2015
He is asking the same question OVER AND OVER AGAIN. The answer won't change. If the matrix is too big for his computer then get a faster computer. Get more memory. Use 64 bit MATLAB.
Más respuestas (1)
Raquel Taijito
el 29 de Mzo. de 2020
How do I create a large matrix of 96 by 96 with each element being 0.6?
1 comentario
Walter Roberson
el 29 de Mzo. de 2020
ones(96)*0.6
If I recall correctly, there are approaches that are slightly faster but take more code. The difference in timing is small unless you have to do a lot of this.
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