question about vectorization using indexes

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Hello, I am trying to do the following operations in matlab but I have a problem with how to properly write my code using vectorization. This is just an example, m, n and the values of the vectors and matrices are just to illustrate my problem. In reality m and n can go up to 1000.
n=5; m=8;
a=4*ones(m,1); a(2)=2;a(n)=3;
b=2*ones(n,2);b(1,1)=5;b(3,1)=1;
ind=3*ones(n,2);
ind(1,2)=0;ind(3,2)=0; b(1,2)=0;b(3,2)=0;
non=zeros(1,n);c=zeros(1,n);
for i=1:n
non(i)=nnz(ind(i,:));
c(i)=prod(a(ind(i,1:non(i)))'.^b(i,1:non(i)),2);
end
I tried the following but it does not give correct results.
i=1:n;c=prod(a(ind(i,1:non(i))).^b(i,1:non(i)),2);
Thank you in advance
  4 Comments
AM
AM on 30 Jul 2019
Edited: AM on 30 Jul 2019
Thank you both for taking time to answer.
The values taken are just examples (poorly chosen I see but the main focus was that line where c is calculated) , what I am trying to do is avoid using a loop seeing as in reality n is equal to 1300. I thought using vectorization would make my code run faster since I am solving odes (I used profile viewer realized that part of the code is run over 8000 times).
What I want is to write a code that allows me to detemine c (i've changed ind so it doesn't always pick element 3) :
n=5; m=8;
a=4*ones(m,1); a([2,3,4,5,m])=[2,1,3,1,0];
b=2*ones(n,2);b([1,3],1)=[5,1];
ind=3*ones(n,2); ind([1,2,3,5],1)=[1 2 4 5];
ind([1,3],2)=0; b([1,3],2)=0;
c=zeros(1,n);
non = sum(ind ~= 0,2)';
for i=1:n
c(i)=prod(a(ind(i,1:non(i)))'.^b(i,1:non(i)),2);
end
I do not know if there is a simpler way to calculate c, I see it's not clear to either of you what that line does so maybe I chose a convoluted way to calculate it.
Basically, with this new example
>> a
a =
4
2
1
3
1
4
4
0
>> ind
ind =
1 0
2 3
4 0
3 3
5 3
>> b
b =
5 0
2 2
1 0
2 2
2 2
c(1)=a(1)^b(1,1)
c(2)=a(2)^b(2,1)*a(3)^b(2,2)
c(3)=a(4)^b(3,1)
and so on
c(i)=prod(a(ind(i,1:non(i)))'.^b(i,1:non(i)),2);
I use a(ind(i,1:non(i))) as the element of a because otherwise if i coded it as a(ind(i,:)) it would try to read a(0) for i=1 for example and give the following error
Array indices must be positive integers or logical values.
Is there a simpler way to do it and one that does not require a loop?

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Accepted Answer

Stephen23
Stephen23 on 30 Jul 2019
Edited: Stephen23 on 30 Jul 2019
Note that ind and b must be transposed for this to work:
>> a = [4;2;1;3;1;4;4;0]; % must be column!
>> ind = [1,0;2,3;4,0;3,3;5,3].'; % transposed!
>> b = [5,0;2,2;1,0;2,2;2,2].'; % transposed!
>> idx = b~=0;
>> XC = ind(idx);
>> bC = b(idx);
>> [~,idc] = find(idx);
>> out = accumarray(idc,a(XC).^bC,[],@prod)
out =
1024
4
3
1
1

More Answers (2)

Guillaume
Guillaume on 30 Jul 2019
Edited: Guillaume on 30 Jul 2019
Another option is to append a 0 (or any finite value) to the start of a and increase ind by 1, so a(ind+1) is always valid. Assuming that b is 0 when ind is 0 as in your example (if not, it's trivially fixed), then anything.^0 is 1 and multiplying by 1 doesn't affect the result, so:
apadded = [0; a];
c = prod(apadded(ind + 1) .^ b, 2)
As a bonus, c is a column vector matching the rows of b.
If b can be non-zero when ind is 0:
c = prod(apadded(ind + 1) . ^ (b .* (ind ~= 0)), 2)
to compensate.
edit: actually, if b can be non-zero when ind is 0, the easiest is to pad a with a 1 instead of a zero. Since 1.^anything is 1, it doesn't affect anything:
apadded = [1; a];
c = prod(apadded(ind + 1) .^b, 2) %b can be zero or non-zero where ind is 0. It'll result in 1.^something
  1 Comment
Guillaume
Guillaume on 31 Jul 2019
Note: although both options are very fast even for very large inputs, this option is about twice as fast as the accepted answer. And much simpler and using less memory.

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Andrei Bobrov
Andrei Bobrov on 31 Jul 2019
ind(ind == 0) = 1;
c = prod(a(ind).^b,2);
  2 Comments
Andrei Bobrov
Andrei Bobrov on 31 Jul 2019
lo = ind == 0;
ind(lo) = 1;
c = prod(a(ind).^(b.*~lo),2);

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