Question about assigning value in a matrix.

Hi, all, again, I want to do the work quickly: I have a large matrix A, and I want to assign value in some entry in each row.
For example:
A=zeros(4);
a=[1,2;
2,3;
3,2;
4,1];
I want (1,2), (2,3),(3,2), (4,1) of matrix A be 1. How shall I make it work in no more than 3 commands?
What I am doing now is using loops:
for i=1:4
A(a(i,1),a(i,2)=1;
end
However, as i is very large like 3^14, the program is very slow. So I wonder is there a way to speed the programming?
Thanks.

 Accepted Answer

idx = sub2ind(size(A), [1 2 3 4], [2 3 2 1]);
A(idx) = 1;

3 Comments

Thanks, Walter! So sub2ind can find the index of A quickly by loops, right? I will definitely try this code. I like it!
I've always found for-loops to be significantly faster than sub2ind and ind2sub. Especially as the number of indices gets large.
It's been a few years since I was timing it for my own work though...
Oh, really? I always found that MATLAB is not efficient for large loops. That is why I tried to avoid large loops.

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More Answers (2)

A=zeros(4);
a=[1,2;
2,3;
3,2;
4,1];
idx= sub2ind(size(A),a(:,1),a(:,2));
A(idx)=1
I've always found for-loops to be significantly faster than sub2ind and ind2sub. Especially as the number of indices gets large.
It's been a few years since I was timing it for my own work though...

12 Comments

Sean, let me do a simple experiment and let u know.
The for loop, in this case, is almost twice faster
function testFORIND2SUB
n = 1000;
idx = ceil(rand(n,2)*n); %two dimensions
[t1,t2] = deal(0);
for ii = 1:100
A = magic(n);
tic
for jj = 1:n
A(idx(jj,1),idx(jj,2)) = 0;
end
t1=t1+toc;
A = magic(n);
%With SUB2IND
tic
idxLinear = sub2ind([n n],idx(:,1),idx(:,2));
A(idxLinear) = 0;
t2 = t2+toc;
end
t2./t1
So for 1000x2 I'm seeing a 60% speed up.
Now this might be somewhere that MEXing this with MATLAB Coder could buy you even more speedup. I'll try that tomorrow.
Oh, I see. Yes, you may be right. Interested in more speedup!
For longer idx I am seeing the sub2ind approach be faster.
Sean, yes, my test result also shows it. For 3^10 rows (I assume it is large): Using for loops takes ~565 seconds while sub2ind procedure takes ~446 seconds.
Do you think if there is room for improvement? Ideally I hope it can compute for 3^14.
How many times do you have to do this? I ran it yesterday with 3^14 and it took circa 25 seconds.
2 times each, Sean. My problem is complex, M is a matrix with 3^10 rows, and around 400 columns. Here is the sample code:
for i=1:M1
pr=(rR(i:i+M-1)-1)/10; %pr is the vector of probabilities
prMatrix=zeros(M2);
for j=1:M2
prMatrix(j,j)=pr(j);
end
x=Y*prMatrix; % x is sample yield for each polnicy
x=round(x); % only consider yield as integer
% matrix computation(faster!)
%
TotalCollect=sum(x,2); % total collection for each policy
TotalCollect=TotalCollect+1;
TempMatrix=zeros(3^M2, n);
%{
for j=1:3^M2 %---slow!
TempMatrix(j,TotalCollect(j))=1;
end
%}
%use sub2ind procedure
row=(1:1:3^M2)';
idx=sub2ind(size(TempMatrix),row,TotalCollect);
TempMatrix(idx)=1;
%}
Pr=Pr+TempMatrix;
end
Well a few things real quick:
prMatrix = diag(pr);
Skips the first for-loop and call to zeros()
Second, you migh twant to consider taking the outer for-loop and converting it to a parfor with the Parallel Computing Toolbox. This would allow this to work in Parallel, over a few workers and speed things up that way.
Alternatively, just run it over the weekend or overnight, at two times total and 556 seconds, that's still only 20 minutes...
Thanks, Sean!
Let me try it again. I have to run large data on server, because 3^12 is memory out here.
Sean, thanks for your advice. My program got speedup. sub2ind will be faster if the size is large.

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