0 个投票

Hey, i have a problem with writing a loop, which will be able to do this:
phi(1)=a_locs(1)/((a_odd(1)));
phi(2)=a_locs(2)/((a_odd(1)));
phi(3)=a_locs(3)/((a_odd(1)));
phi(4)=a_locs(4)/((a_odd(1)));
phi(5)=(a_locs(5)-a_odd(1))/((a_odd(2)));
phi(6)=(a_locs(6)-a_odd(1))/((a_odd(2)));
phi(7)=(a_locs(7)-a_odd(1))/((a_odd(2)));
phi(8)=(a_locs(8)-a_odd(1))/((a_odd(2)));
phi(9)=(a_locs(10)-(a_odd(1)+a_odd(2)))/((a_odd(3)));
phi(10)=(a_locs(11)-(a_odd(1)+a_odd(2)))/((a_odd(3)));
Someone can help me?

7 个评论
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James Tursa
James Tursa 2015-4-3
编辑：James Tursa 2015-4-3
Why do you want a loop? Is your real problem actually bigger than this? Is everything shown variables, or are some of them functions?
Michael Monka
Michael Monka 2015-4-3
All are variables, and i have to evaluate a large number (450) of equations, which looks as i write before, and i thought it is possible to write (one or two?) loops which will be do that thing...?
James Tursa
James Tursa 2015-4-3
So what are the sizes of the variables involved, and can you be more explicit in the pattern of the calculations for these larger sizes?
Image Analyst
Image Analyst 2015-4-3
The thing I can't figure out is why a_locs(9) is totally missing from any of the right hand side expressions. And there are not enough equations to figure out which elements need to be "skipped" in the other equations, and what equations the "skipping" would happen at.
Stephen23
Stephen23 2015-4-3
It seems like this might be able to be written using vectorized code, thus avoiding any loops... but it would be nice to know the general algorithm here.
Michael Monka
Michael Monka 2015-4-3
Okay, so the size of variable which i have to obtain (which is phi) is 450. The pattern of the calculations goes like:
phi(1)=a_locs(1)/((a_odd(1)));
phi(2)=a_locs(2)/((a_odd(1)));
phi(3)=a_locs(3)/((a_odd(1)));
phi(4)=a_locs(4)/((a_odd(1)));
phi(5)=(a_locs(5)-a_odd(1))/((a_odd(2)));
phi(6)=(a_locs(6)-a_odd(1))/((a_odd(2)));
phi(7)=(a_locs(7)-a_odd(1))/((a_odd(2)));
phi(8)=(a_locs(8)-a_odd(1))/((a_odd(2)));
phi(9)=(a_locs(9)-(a_odd(1)+a_odd(2)))/((a_odd(3)));
phi(10)=(a_locs(10)-(a_odd(1)+a_odd(2)))/((a_odd(3)));
phi(11)=(a_locs(11)-(a_odd(1)+a_odd(2)))/((a_odd(3)));
phi(12)=(a_locs(12)-(a_odd(1)+a_odd(2)))/((a_odd(3)));
phi(13)=(a_locs(13)-(a_odd(1)+a_odd(2)+a_odd(3)))/((a_odd(4)));
phi(14)=(a_locs(14)-(a_odd(1)+a_odd(2)+a_odd(3)))/((a_odd(4)));
phi(15)=(a_locs(15)-(a_odd(1)+a_odd(2)+a_odd(3)))/((a_odd(4)));
phi(16)=(a_locs(16)-(a_odd(1)+a_odd(2)+a_odd(3)))/((a_odd(4)));
phi(17)=(a_locs(17)-(a_odd(1)+a_odd(2)+a_odd(3)+a_odd(4)))/((a_odd(5)));
phi(18)=(a_locs(18)-(a_odd(1)+a_odd(2)+a_odd(3)+a_odd(4)))/((a_odd(5)));
phi(19)=(a_locs(19)-(a_odd(1)+a_odd(2)+a_odd(3)+a_odd(4)))/((a_odd(5)));
phi(20)=(a_locs(20)-(a_odd(1)+a_odd(2)+a_odd(3)+a_odd(4)))/((a_odd(5)));
phi(21)=...
Jan
Jan 2015-4-3
@Michael: It looks like phi consists of blocks of the length 4. But 450 is not evenly divisable by 4?

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 采纳的回答

Jan
Jan 2015-4-3
编辑：Jan 2015-4-3

1 个投票

With a loop:
phi = zeros(1, 452); % Instead of 450?
s = 0;
m = 1;
for k = 1:4:452
phi(k:k+4) = (a_locs(k:k+4) - s) / a_odd(m);
s = s + a_odd(m);
m = m + 1;
end
And vectorized - assuming that a_odd is a row vector:
phi = reshape(a_locs, 4, 113);
phi = bsxfun(@minus, phi, [0, cumsum(a_odd(1:112)]);
phi = bsxfun(@rdivide, phi, a_odd(1:113));
phi = reshape(phi, 1, 452);

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