How can I write such a large matrix?

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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 Comments
Stephen23
Stephen23 on 11 Jan 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
Oleg Komarov on 11 Jan 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.

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Accepted Answer

Roger Stafford
Roger Stafford on 11 Jan 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 Comments
John D'Errico
John D'Errico on 11 Jan 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.
Stephen23
Stephen23 on 11 Jan 2015
Edited: Stephen23 on 11 Jan 2015
The code could also do with some improvement...

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More Answers (1)

Raquel Taijito
Raquel Taijito on 29 Mar 2020
How do I create a large matrix of 96 by 96 with each element being 0.6?
  1 Comment
Walter Roberson
Walter Roberson on 29 Mar 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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