How to do this efficiently?
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Hello all,
I have this variable a[k,m]=max(k*Ts+taup,m*Ts+tauq) for k,m=0,1,...,N-1. I want to find the matrix A where [A]_{k,m}=a[k,m] efficiently. To do so, I define two matrices RowInc and ColInc as
RowInc =(0:N-1)'*ones(1,N);
ColInc =transpose(RowInc);
Then I write
for pp=1:Np
for qq=1:Np
A=max(RowInc*Ts+tau(pp),ColInc*Ts+tau(qq));
end
end
Does this give me what I want?
Thanks
4 个评论
Image Analyst
2014-8-9
Nothing is in gray. But basically you mean that A=a everywhere. The A and a matrices are identical.
回答(2 个)
dpb
2014-8-9
...I want to find the matrix A where [A]{k,m}=a[k,m]..._
A=max(RowInc*Ts+tau(pp),ColInc*Ts+tau(qq));
will end up w/ just a single value for A at the last loop of pp and qq since it overwrites the previous A each iteration. But, I don't believe it does what you want, anyway.
Should be simply
A(A==a);
if I understand the query correctly.
5 个评论
dpb
2014-8-9
IA, I couldn't figure out what his last comment said (and little of the rest) if after "I have this variable a[k,m]..." there isn't an array a.
Guess I'll leave the field bloodied on this one...
dpb
2014-8-10
OK, from the loop solution one can write
[x,y]=ndgrid(0:N-1,0:N-1);
A=max(Ts.*x+taup,Ts.*y+tauq);
trading memory for the loop. The loop solution could be simplified since there's only a dependence upon kk for the one term and mm for the other, they could be precomputed outside the loops. Preallocating also would help, of course. Not sure how the timings would come out in the end.
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