Creating matplotlib widget callback inside a function

Question

I am using python 2.7 on Windows. I have a function which creates a figure with a CheckButtons widget, and it also includes the definition of the button's callback. When I call the function once, everything is OK, but when I call it more than once, the buttons stops responding, as follows:

If the figure is created using plt.subplots(), none of the buttons respond.
If the figure was created using plt.figure(), the behavior is inconsistent; sometimes only the 1st created button responds, and sometimes both respond.

My guess is that is has to do with the scope of the callback, but I couldn't pinpoint the problem using trial-and-error.

Sample code:

import matplotlib.pyplot as plt
from matplotlib.widgets import CheckButtons

def create_button():
    plt.subplots() # or: plt.figure()
    rax = plt.axes([0.2, 0.2, 0.2, 0.2])
    check = CheckButtons(rax, ['on'], [True])
    def callback(label):
        check.labels[0].set_text('on' if check.lines[0][0].get_visible() else 'off')
        plt.draw()
    check.on_clicked(callback)

create_button()
#create_button() # uncomment to reproduce problem
plt.show()

Itamar Katz · Accepted Answer · 2014-08-16 17:34:20Z

3

It turns out the problem was that the CheckButtons instance created inside the function no longer exists after the function returns.

The solution I came up with was to keep a list in the scope where the function is called (I used a static variable in a class), and append the instance to this list from within the function. This way the CheckButtons instance still exists when the function exits. In order for that list to not grow more than needed, I also wrote a function which deletes the corresponding instance from the list, and registered this function as a callback for the event of the figure being closed by the user.

I will be happy to hear comments on my solution, or suggestions for more Pythonish solution, if such a solution exists.

answered Aug 16, 2014 at 17:34

Itamar Katz

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Comments

PRD · Accepted Answer · 2014-12-02 14:11:23Z

1

I think also if you return check at the end of the function this will work to keep the button alive on exit.

answered Dec 2, 2014 at 14:11

PRD

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