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
댓글 수: 4
Guillaume
2019년 7월 30일
편집: Guillaume
2019년 7월 30일
Note that :
non = nnz(ind(i,:));
x = a(ind(i,1:non);
can be written more simply as:
x = a(nonzeros(ind(i, :)));
which is a lot easier to understand (particularly given the poorly named variables).
--edit:--
I reiterate Adam's question, what is the intent of the line
c(i)=prod(a(ind(i,1:non(i)))'.^b(i,1:non(i)),2)
We now that it's what you want to vectorise. It'd be a lot easier to do if we knew what you're trying to do with it.
In particular, I'll point out that with the example given, the above will always pick element a(3), regardless of i.
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Stephen23
2019년 7월 30일
편집: Stephen23
2019년 7월 30일
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
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추가 답변 (2개)
Guillaume
2019년 7월 30일
편집: Guillaume
2019년 7월 30일
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
댓글 수: 1
Guillaume
2019년 7월 31일
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.
Andrei Bobrov
2019년 7월 31일
ind(ind == 0) = 1;
c = prod(a(ind).^b,2);
댓글 수: 2
Guillaume
2019년 7월 31일
Yes, that will also works as long as b is 0 when ind is 0, since a(1).^0 is 1.
Andrei Bobrov
2019년 7월 31일
lo = ind == 0;
ind(lo) = 1;
c = prod(a(ind).^(b.*~lo),2);
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