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How to iterate a structure name and save directory in a for loop?

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I am trying to use a for loop that saves parameters for 10 different files into a structure, but I am not sure how to get the name of each structure to iterate. My final result should be 10 different structures saved into a directory with the name for the structures being rate100,rate150,rate200,..... So far my for loop has two issues. The first is I am unable to use ['rate' num2str(cnt)] for the structure name, and the second is I am unsure how to get the save path to be my pathname + 'rate' + #. How can I change my for loop to get the structure name to iterate and why does my format for the save command not work?
saveDir = 'my/path/name
cnt = 50;
for j=1:10
cnt = cnt+50;
['rate' num2str(cnt)].minTime = tmin;
['rate' num2str(cnt)].maxTime = tmax;
% Save structure
save([saveDir '/rate' num2str(cnt) '.mat'],['rate' num2str(cnt)]);
end
  1 Kommentar
Stephen23
Stephen23 am 22 Jan. 2023
Bearbeitet: Stephen23 am 22 Jan. 2023
"...with the name for the structures being rate100,rate150,rate200,..... So far my for loop has two issues."
The main issue is that you are forcing meta-data into variable names.
Forcing meta-data into variable names is usually a sign that you are doing something wrong:
If you simply stored that meta-data inside the structure and used exactly the same variable names in every MAT file, then your code would be simpler, more efficient, and much more robust:
saveDir = 'my/path/name';
for k = ..
S.minTime = tmin;
S.maxTime = tmax;
S.rate = whatever_meta_data_you_want;
% Save structure
F = [saveDir,'/rate',num2str(cnt),'.mat'];
save(F, 'S'); % simpler and much more robust
end
In contrast, your approach of forcing meta-data into the variable names suffers from a number of problems:
(that explanation is for fieldnames, but it applies just as much to forcing meta-data into variable names)

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Akzeptierte Antwort

Walter Roberson
Walter Roberson am 22 Jan. 2023
saveDir = 'my/path/name
for j=1:10
cnt = 50 * (j+1);
clear savestruct
ratefield = "rate" + cnt;
savestruct.(ratefield).minTime = tmin;
savestruct.(ratefield).maxTime = tmax;
filename = fullfile(saveDir, ratefield + ".mat");
save(filename, '-struct', 'savestruct');
end
Notice this contains no dynamic variable names -- but it does contain dynamic field names within a structure.
When you save a structure with the -struct option, then instead of saving the struct as a variable, it saves each field of the struct as a variable.

Weitere Antworten (1)

Voss
Voss am 22 Jan. 2023
saveDir = 'my/path/name';
cnt = 50;
for j=1:10
cnt = cnt+50;
S = struct('minTime',tmin,'maxTime',tmax);
% Save structure
filename = fullfile(saveDir,sprintf('rate%d.mat',cnt));
save(filename,'-struct','S');
end
  5 Kommentare
Voss
Voss am 22 Jan. 2023
Yes, but the file name is the same as the structure name, and it'll be a struct when she load()s it.
Stephen23
Stephen23 am 22 Jan. 2023
Alternative avoiding the intermediate structure:
for k = ..
..
save(filename,'tmin','tmax');
end

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