Using the fields of a structure to index through a vector and generate a resulting structure

dataArray = [21,22,23,24,25,26,27,28,29,30];

structureOfIndexes.a1 = [2,5,8];
structureOfIndexes.a2 = [3,4];
structureOfIndexes.a3 = [1,2,3,5,9];
% How do I use the structure of indexes to generate the 'resultStructure'(seen below)

resultStructure.a1 = [22,25,28];
resultStructure.a2 = [23,24];
resultStructure.a3 = [21,22,23,25,29];
% without using a for-loop to index through a1,a2,a3

Answer 1

[Eric Delgado]

Hey @Scorp, structfun is the answer for your issue! :)

dataArray = [21,22,23,24,25,26,27,28,29,30];
structureOfIndexes.a1 = [2,5,8];
structureOfIndexes.a2 = [3,4];
structureOfIndexes.a3 = [1,2,3,5,9];

resultStructure = structfun(@(x) dataArray(x), structureOfIndexes, "UniformOutput", false)

resultStructure = struct with fields:
    a1: [22 25 28]
    a2: [23 24]
    a3: [21 22 23 25 29]

[Scorp]

%Thank you for your reply, a further question:
%if I want to grab the next 3 values in dataArray from each index such that:
dataArray = [21,22,23,24,25,26,27,28,29,30];
structureOfIndexes.a1 = [2,8];
% Trying the code below yields a1: [22 23 24]
resultStructure = structfun(@(x) dataArray(x:x+2), structureOfIndexes, "UniformOutput", false)

resultStructure = struct with fields:
    a1: [22 23 24]

% but I want the output to be: resultStructure.a1 = [22,23,24,28,29,30];