How can I create a Figure Title with a variable in it?

25 vues (au cours des 30 derniers jours)
onamaewa
onamaewa le 28 Mai 2019
Modifié(e) : Stephen23 le 29 Mai 2019
I have a figure:
DEPTH = 10; % Depth changes based on user input
figure('Numbers at a depth of %d meters)')
How can I make my figure title update depending on my input for depth?

Réponses (1)

Adam Danz
Adam Danz le 28 Mai 2019
Modifié(e) : Adam Danz le 29 Mai 2019
DEPTH = 10; % Depth changes based on user input
title(sprintf('Numbers at a depth of %.0f meters', DEPTH))
% For the figure name (not title):
figure('Name', sprintf('Numbers at a depth of %.0f meters', DEPTH))
I prefer %.0f rather than %d to convert integers because 1) if DEPTH isn't an integer, %d will print out scientific notation and 2) if you ever want to add decimal places you can just change the 0.

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