# 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

### Accepted Answer

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

##### 0 Comments

### More Answers (2)

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
on 31 Jul 2019

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