How to access only the first element of the variables in a structure array as a vector?

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Darcy Cordell
Darcy Cordell 2023-5-17
评论： Paul 2023-5-23
Not sure I worded this question correctly, but I will try to explain.
I have a (1 x N) structure array (S) which contains the field "Location" like this:
%Pretend N = 3:
S(1).Location = [1 4];
S(2).Location = [4 15];
S(3).Location = [3 7];
I want to get a vector of the first entry of each "Location" variable, like this:
Location_1_Vector = [1 4 3]
Is there an efficient way to do this in one line without producing an intermediate variable?
The following do not work:
Location_1_Vector = [S(:).Location]; %Concatenates the Location variables together into [1 4 4 15 3 7]
Location_1_Vector = [S(:).Location(1)]; %Error: intermediate dot indexing produced comma separated ...
Of course, I can do it like this:
Location = reshape([S(:).Location],2,length(S));
Location_1_Vector = Location(1,:);
But I would prefer to avoid creating an intermediate "Location" variable with the additional clunkiness of "reshape".
Any help is appreciated.

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采纳的回答

Paul
Paul 2023-5-17
Ran in:
One option
S(1).Location = [1 4];
S(2).Location = [4 15];
S(3).Location = [3 7];
L = arrayfun(@(S) S.Location(1),S)
L = 1×3
1 4 3
  1 个评论
Paul
Paul 2023-5-23
Ran in:
Just realized the Question also asked for an efficient solution. arrayfun is one line and does not produce an intermediate variable, but not necessarily the most effiicient, at least in terms of run time.
S(1).Location = [1 4];
S(2).Location = [4 15];
S(3).Location = [3 7];
S = repmat(S,1,1000);
isequal(test1(S),test2(S),test3(S))
ans = logical
1
timeit(@() test1(S))
ans = 0.0058
timeit(@() test2(S))
ans = 1.2132e-04
timeit(@() test3(S))
ans = 7.1075e-04
function L = test1(S)
L = arrayfun(@(S) S.Location(1),S);
end
function L = test2(S)
Location = reshape([S(:).Location],2,length(S));
L = Location(1,:);
end
function L = test3(S)
L = nan(1,numel(S));
for ii = 1:numel(S)
L(ii) = S(ii).Location(1);
end
end

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