# how to extract elements along specified dimension of array

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Peter on 15 Mar 2012
when e.g. selecting all elements in the 2nd dimension and the first element of all remaining dimensions of an array, for a 3-dimensional array one would write: A(1,:,1)
how to program this elegantly when the dimension over which I want the elements (2 in the example) is a variable itself, whose value is unknown until runtime ?
thanks!!

Jonathan Sullivan on 16 Mar 2012
A = rand(100,100,10,10);
dim = 4;
sz = size(A);
inds = repmat({1},1,ndims(A));
inds{dim} = 1:sz(dim);
A(inds{:})
Peter on 16 Mar 2012
thanks a lot. exactly what i needed.
just for my understanding: inds is a cell array. what exactly does inds{:} give? looks like seperate doubles.
thanks again!

per isakson on 15 Mar 2012
A(ixi,ixj,ixk)
where ixi,ixj,and ixk are numerical vectors (/scalar) of integers. ":" would correspond to (1:number_of_element_in_dimension). Or with logical indexing
A(isi,isj,isk)
where isi,isj,isk are logical vectors all with the length "number_of_element_in_dimension". ":" would be true(1,number_of_element_in_dimension).
The elegance will be in the calculation of ixi,ixj,and ixk or isi,isj,isk.

Walter Roberson on 15 Mar 2012
Something like this,
DIM = 2; %changed at runtime
idxexpr = { repmat({1}, 1, DIM - 1), {':'}, repmat({1}, 1, ndim(A)-DIM) };
A(idxexpr{:})
Peter on 16 Mar 2012
num2mat doesn't seem to exist. can you maybe comment on what class idxexpr should be to make it work as a subscript expression? that would give some direction to my own attempts. it seems like a cell array doesn't work. ('ndim' should be 'ndims' btw.)
thanks again!

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