title according to the file name

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Fercho_Sala
Fercho_Sala 2021-4-15
评论： Rik 2022-3-24
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.
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John Ostrander
John Ostrander 2022-3-24
@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
Rik 2022-3-24
Ran in:
The fileparts function should do what you need:
[p,f,e]=fileparts('C:\path\morepath\filename.csv')
p = 0×0 empty char array
f = 'C:\path\morepath\filename'
e = '.csv'
%(these results are on Linux, on Windows you should get this)
p = 'C:\path\morepath'
f = 'filename'
e = '.csv'

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采纳的回答

Constantino Carlos Reyes-Aldasoro
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 个评论
Adam Danz
Adam Danz 2021-4-15
or,
title(['Subplot number = ', num2str(k)])
or,
title(sprintf('Subplot number = %d', k))

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更多回答（1 个）

Chunru
Chunru 2021-4-15
filename = "abc";
load(filename, "x", "y");
plot(x, y)
title(sprintf("File name: %s", filename));
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Rik
Rik 2021-4-15
You shouldn't encourage loading variables like this. Always load to a struct:
S=load(filename, "x", "y");x=S.x;y=S.y;

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