Can anyone show me how I can avoid following for loops

2012 10 8
2 个回答
回答已采纳
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0 个投票

*Hello everyone,
Can anyone show me how I can avoid following for loops.
Thanks!*
mth=0;
nth=0;
for i=1:1000
ML = [1,2;3,4];
for ix=1:size(ML,1)
for iy=1:size(ML,2)
M(mth+ix,nth+iy)=ML(ix,iy);
end
end
mth = mth+size(ML,1);
nth = nth+size(ML,2);
end

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 采纳的回答

Matt Fig
Matt Fig 2012-10-8
编辑：Matt Fig 2012-10-8

0 个投票

Here is a one approach:
M = diag(repmat([1 4],1,1000)) +...
diag([repmat([2 0],1,999) 2],1) +...
diag([repmat([3 0],1,999) 3],-1);
And another:
M = diag(repmat([1 4],1,1000));
M(2001:2001:end) = [repmat([2 0],1,999) 2];
M(2:2001:end) = [repmat([3 0],1,999) 3];

更多回答（1 个）

Walter Roberson
Walter Roberson 2012-10-8

0 个投票

for iy=1:size(ML,2)
M(mth+ix,nth+iy)=ML(ix,iy);
end
can be written as
MLc = size(ML,2);
M(mth+ix, nth:nth+MLc-1) = ML(ix, 1:MLc);

3 个评论
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Daniel Shub
Daniel Shub 2012-10-8
编辑：Daniel Shub 2012-10-8
Are you sure? If nth is 0 and MLc is 2, nth:nth+MLc-1 gives [0, 1] which causes indexing problems. I think it is nth+(1:MLc) on the RHS.
Walter Roberson
Walter Roberson 2012-10-8
编辑：Walter Roberson 2012-10-8
Yes, that would make sense on the LHS.

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