title according to the file name

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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 个评论
Rik
Rik 2022-3-24
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));
  1 个评论
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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