Indexing through a structure to get subsets of data with no looping

5 vues (au cours des 30 derniers jours)
Scorp
Scorp le 27 Sep 2022
Commenté : Scorp le 27 Sep 2022
dataArray = [21,22,23,24,25,26,27,28,29,30];
structureOfIndexes.a1 = [2,8];
structureOfIndexes.a2 = [1,5,7];
%if I want to grab the next 3 values in dataArray starting from each index such that:
%resultStructure.a1 = [22 23 24 28 29 30]
%resultStructure.a2 = [21 22 23 25 26 27 27 28 29]
% Trying the code below yields:
resultStructure = structfun(@(x) dataArray(x:x+2), structureOfIndexes, "UniformOutput", false)
resultStructure = struct with fields:
a1: [22 23 24] a2: [21 22 23]

Réponse acceptée

Eric Delgado
Eric Delgado le 27 Sep 2022
Look... why you don't want to use a loop?! It's something that you can't avoid sometimes. The result you are looking for is weird, maybe you can work with tables and dict (new Matlab data type) instead of struct.
In "my solution" you have two loops! :)
% input
dataArray = [21,22,23,24,25,26,27,28,29,30];
structureOfIndexes.a1 = [2,8];
structureOfIndexes.a2 = [1,5,7];
% Loops... thanks God!
fn = fieldnames(structureOfIndexes);
for ii = 1:numel(fn)
tempValues = [];
for jj = 1:numel(structureOfIndexes.(fn{ii}))
idx = structureOfIndexes.(fn{ii})(jj);
tempValues = [tempValues, dataArray(idx:idx+2)];
end
resultStructure.(fn{ii}) = tempValues;
end
resultStructure
resultStructure = struct with fields:
a1: [22 23 24 28 29 30] a2: [21 22 23 25 26 27 27 28 29]
  1 commentaire
Scorp
Scorp le 27 Sep 2022
My data sets are very large. The dataArray is in the order of 20GB and I need to pull out 20 or so subsets of about 200MB each all with different indexing. I will look at the dictionary option or I will just have to accept the time hit. Thank you for your help.

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