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

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Scorp
Scorp 2022 年 9 月 27 日
コメント済み: Scorp 2022 年 9 月 27 日
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]

採用された回答

Eric Delgado
Eric Delgado 2022 年 9 月 27 日
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 件のコメント
Scorp
Scorp 2022 年 9 月 27 日
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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