title according to the file name

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Fercho_Sala
Fercho_Sala el 15 de Abr. de 2021
Comentada: Rik el 24 de Mzo. de 2022
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 comentarios
John Ostrander
John Ostrander el 24 de Mzo. de 2022
@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 el 24 de Mzo. de 2022
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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Respuesta aceptada

Constantino Carlos Reyes-Aldasoro
Constantino Carlos Reyes-Aldasoro el 15 de Abr. de 2021
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 comentario
Adam Danz
Adam Danz el 15 de Abr. de 2021
or,
title(['Subplot number = ', num2str(k)])
or,
title(sprintf('Subplot number = %d', k))

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Más respuestas (1)

Chunru
Chunru el 15 de Abr. de 2021
filename = "abc";
load(filename, "x", "y");
plot(x, y)
title(sprintf("File name: %s", filename));
  1 comentario
Rik
Rik el 15 de Abr. de 2021
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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