Get the filename of a loaded file

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Fabian Grodek
Fabian Grodek 2022 年 8 月 3 日
コメント済み: Fabian Grodek 2022 年 8 月 3 日
I have a script that loads a file and do some data plotting. I am loading different files each time I run the script. I manually edit the file name since every time the name is different and comming from an unrelated system.
My script reads the file as follows:
How can I register the file name in a variable? I want to later use the file name in a string (after trimming it) so that I can automatically include the file name on the plot title.
Thank you.


Star Strider
Star Strider 2022 年 8 月 3 日
Save the fille name as a string or character array first, then refer to it as any other variable —
filename = '\Data_record_file_20220714T121615.mat';
One option might be to store all the various file names in a cell array or string array, and then just subscript into it —
load_data = load(filecell{k});
where ‘k’ is the desired element of the cell array ‘filecell’ storing the file names.
It is always best to load into a structure output. This avoids problems such as overwriting existing variables, and other potential problems —
load_data = load(filename);
then refer to the fields of ‘load_data’ (or whatever you choose to call it) to recover the various variables.
  4 件のコメント
Star Strider
Star Strider 2022 年 8 月 3 日
@Fabian Grodek — My pleasure!
@Walter Roberson — Thank you!


その他の回答 (2 件)

桂生 李
桂生 李 2022 年 8 月 3 日
[FileName, FilePath] = uigetfile;

Walter Roberson
Walter Roberson 2022 年 8 月 3 日
filename = '\Data_record_file_20220714T121615.mat';
[~, basename, ~] = fileparts(filename);
title( "Analysis of " + basename)
Note: we do not recommend that form of load(). We recommend that you assign the output of load() to a variable, which gives you a struct array with one field for each loaded variable; you would then pull fields out of the struct as needed. This is more robust and more efficient, and is needed if you later want to do code generation. There are situations in which MATLAB will ignore variables created by load() that does not have an output variable name.
  3 件のコメント
Fabian Grodek
Fabian Grodek 2022 年 8 月 3 日
Understood. Thanks.





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