Indexing an array with a vector

2015 8 7
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更新时间：2015 8 7
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I would like to index an N-dimensional array with a vector of length N. In particular, for the 2-dimensional case I am currently doing the following.
A = rand(10);
v = randi(10,1,2);
v = num2cell(v);
A(v{:})
However, this approach seems horribly inefficient. Is there not a smarter way to convert the vector v to a proper index (comma-separated) for the array A?
Edit Let's say v = randi(10,1,2) = [ 3 5 ]. In that case, I want to obtain A(3,5).

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Stephen23
Stephen23 2015-8-7
编辑：Stephen23 2015-8-7
This could be the best solution already, for an array of arbitrary number of dimensions. Why do you think that this method "seems horribly inefficient"? Did you run the profiler on your code and find that this operation took the most time?
Jori
Jori 2015-8-7
编辑：Jori 2015-8-7
I did indeed run the profiler on the code and I found that the operation took quite some time. If I compare it against Walter Roberson his method, I obtain a nice improvement, see the figure. (p1 and p2 are identical)
Walter Roberson
Walter Roberson 2015-8-7
You can improve performance slightly by assigning the transpose to p1_off ahead of time, equivalent to if I had written
Aoff = cumprod([1 As(1:end-1)]) .';
and then just use * Aoff rather than * Aoff.'
Also the "-1" was important to add to the indices. It probably can be rolled into the calculation of the offset at the expense of clarity.
Jori
Jori 2015-8-7
Thanks, that is a nice remark.
In these computations, data(n,2:end) has values starting at 0. I need to add 1 to adjust for the minimal index 1 in Matlab. This is followed by the "-1" that you had and so the "-1" disappears in line 25 of the profiler figure.
I verified the results, and this is indeed the correct indexing.

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Walter Roberson
Walter Roberson 2015-8-7

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%setup
As = size(A);
Aoff = cumprod([1 As(1:end-1)]);
%calculation at run-time
vidx = (v-1) * Aoff.' + 1;
A(vidx)
This will work when v is an M x N array of indices, producing a column array of M values. If your array has trailing singular dimensions being indexed then the code will not work as-is (the last entry of Aoff needs to be duplicated for all trailing singular dimensions.)
The calculation here converts the index components into linear indices using some trivial matrix multiplication.

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Azzi Abdelmalek
Azzi Abdelmalek 2015-8-7
编辑：Azzi Abdelmalek 2015-8-7

0 个投票

A = rand(10);
v = randi(10,1,2)
Why converting to cell array? just type
A(v)
If v is a cell array, convert it to a double array
A = rand(10);
v = randi(10,1,2);
v = num2cell(v);
A([v{:}])

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Jori
Jori 2015-8-7
Because that is not what I want. Let's say v = [ 3 5 ]. A(v) gives me [ A(3) A(5) ]. However, I want A(3,5). Hope that clears it up. I'll edit my post.
Azzi Abdelmalek
Azzi Abdelmalek 2015-8-7
You can use sub2ind function
Azzi Abdelmalek
Azzi Abdelmalek 2015-8-7
编辑：Azzi Abdelmalek 2015-8-7
A = rand(10)
v = randi(10,1,2)
idx=sub2ind(size(A),v(1),v(2))
A(idx)
If v is nx2 array
idx=sub2ind(size(A),v(:,1),v(:,2))
A(idx)
Jori
Jori 2015-8-7
编辑：Jori 2015-8-7
The size of v is 1 x N. I do not know N beforehand.
Using idx = sub2ind(size(A),v) does not work. So I would need to use idx = sub2ind(size(A),v(1),v(2),...,v(N)), but I do not know the value of N beforehand. Thus, I would need to use
v = num2cell(v);
idx = sub2ind(size(A),v{:});
A(idx)
Which brings me back to the original problem.

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