title according to the file name

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.

[Rik]

If you have the file name available as a char or string, you can use it in the title function call. Otherwise I don't know what you would mean.

[Fercho_Sala]

for the case you upload a data file and need to title the chart as the original name of the file found in some specific folder i.e C:\Users\Mat\Desktop\matrix_A.dat , obviously without the '.dat' extencion.

[John Ostrander]

@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]

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'

Answer 1

[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.

[Adam Danz]

or,

title(['Subplot number = ', num2str(k)])  

or,

title(sprintf('Subplot number = %d', k))

Answer 2

[Chunru]

filename = "abc";
load(filename, "x", "y");  
plot(x, y)
title(sprintf("File name: %s", filename));

[Rik]

You shouldn't encourage loading variables like this. Always load to a struct:

S=load(filename, "x", "y");x=S.x;y=S.y;