How do I create a dynamic file name when printing to jpeg?

3 vues (au cours des 30 derniers jours)
Matt Brown
Matt Brown le 19 Jan 2018
Commenté : Matt Brown le 22 Jan 2018
I have a function which generates the same plot for multiple data sets. My goal is to save each plot under a unique filename. It looks something like this:
figure1=figure('visible', 'off')
axes1=axes('Parent',figure1)
plot(x,y)
print -djpeg test.jpeg
fclose(figure1)
This will overwrite the same jpeg each time. What I would like to have is test01.jpg, test02.jpg and so on corresponding to the string identifying each data set (i.e. test01, test02). I have tried dropping the string into the print command, but to no avail.
print -djpeg horzcat(idstring,'.jpeg')

Réponse acceptée

Walter Roberson
Walter Roberson le 19 Jan 2018
print('-djpeg', [idstring,'.jpeg'])

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