title according to the file name
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Fercho_Sala
2021 年 4 月 15 日
Does anybody know how to (in a plot) put the ‘title’ as the name of the file where the X,Y,Z variables are included? the idea is to generate several and independent plots, based on ‘imagesc’, ‘contolchart’ and other plotting functions. Thanks.
4 件のコメント
John Ostrander
2022 年 3 月 24 日
@Rik I am looking for something similar. There is a python code for it (I am learning both and very much a newbie) but in my case and I suspect the qeustion is the same:
C:\path\morepath\.......\filename.csv some of my files are buried deep in the file structure.
Strip "filename" from this automatically and insert it as the graph title.
I work with many files, and many graphs look alike. This would save time.
Rik
2022 年 3 月 24 日
The fileparts function should do what you need:
[p,f,e]=fileparts('C:\path\morepath\filename.csv')
%(these results are on Linux, on Windows you should get this)
採用された回答
Constantino Carlos Reyes-Aldasoro
2021 年 4 月 15 日
Perhaps you want to add values to the titles of your figures, try something like this
for k=1:9
subplot(3,3,k)
title(strcat('Subplot number =',num2str(k)))
end
If this does not answer your question, we would need more information.
1 件のコメント
Adam Danz
2021 年 4 月 15 日
or,
title(['Subplot number = ', num2str(k)])
or,
title(sprintf('Subplot number = %d', k))
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