sprintf into a number

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Julien Maxime
Julien Maxime 2022 年 10 月 12 日
回答済み: Jan 2022 年 10 月 15 日
Hello,
I have the simple following code and it does not work of course as sprintf makes a string and cannot be entered as a value in the matrix.
Mat=zeros(64,64,round(Zlen));
for i = 1:64
for j = 1:64
for k = 1:Zlen
Mat(i,j,k)=sprintf('Ch0%d_0%d(1,%d)', i, j, k)
end
end
end
I know how Matlab work and that I cannot just hope to supress the ' by magic but I have these Data Matrix that are name as the following Ch01_01 / Ch01_02 / ... / Ch01_64 / Ch02_01 / Ch02_02 / Ch02_64 / Ch03_01 / etc..
However I still cannot find a turnaround that issue. Could you help me ?
Thank you
  5 件のコメント
Stephen23
Stephen23 2022 年 10 月 15 日
"What is stopping me is that when i open the .mat matrix, i have more than 4000 files opening in the workspace and to open them all with load command I would need a loop and call them one by one, coming back to my initial issue."
So far everything you describe sounds like it could be avoided by LOADing into an output variable, as already shown.

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採用された回答

Jan
Jan 2022 年 10 月 15 日
Stephen an Walter hit the point.
% Store imported data in an array instead of polluting the workspace:
FileData = load('YourInputFile.mat');
% No loop over k required:
Mat = zeros(64, 64, round(Zlen));
for i = 1:64
for j = 1:64
Mat(i, j, :) = FileData.(sprintf('Ch0%d_0%d', i, j));
end
end

その他の回答 (1 件)

Walter Roberson
Walter Roberson 2022 年 10 月 12 日
Please read http://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval for information about why we strongly recommend against creating variable names dynamically.

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