Python callback passing default argument

Question

I've been reading for the past half hour on various ways of constructing callbacks in Python and I haven't successfully passed a default parameter to be included when it "bounces back".

The code below demonstrates that I would like the integer 5 to be included when the callback is issued through exiting the Tkinter GUI.

class Heatmap(Updater):
""" Displays the Heatmap provided the data streaming in. """
def __init__(self, data_queue, slider_callback, closed_callback):
    """
    Initialization function for the Osc heatmap.
    :param data_queue: Data streamed in from the buffer for visualization.
    :type data_queue: Queue
    :param closed_callback: When the figure is closed, callback should be used to remove the figure.
    :type closed_callback: Function
    :param slider_callback: Callback function to return the state of the slider to parent caller.
    :type slider_callback: Function
    """
    super(Heatmap, self).__init__()

    self.data_queue = data_queue
    self.closed_callback = closed_callback
    self.slider_callback = slider_callback

    self.window = Tk()
    #self.window.protocol('WM_DELETE_WINDOW', closed_callback)
    atexit.register(self.closed_callback)
...
if __name__ == '__main__':
q = Queue()
for i in xrange(100):
    q.put(i)
def close_cb(idx):
    print 'idx {} window has been closed'.format(idx)
def slider_cb(val):
    print 'slidy {}'.format(val)
closely = lambda x: close_cb(5)
hm = Heatmap(q, slider_cb, closely)
hm.run()

ERROR
TypeError: <lambda>() takes exactly 1 argument (0 given)

I see the error is pointing towards my atexit.register(self.closed_callback) as it might expect some argument be passed through? How would I go about satisfying this while not passing fault info, all the while maintaining the fact that I receive 5 on a successful exit call back.

I've also been playing with functools.partial.

Are you asking how to set a default for a parameter of a lambda function? — MisterMiyagi
– MisterMiyagi, Commented Oct 21, 2019 at 19:14

chepner · Accepted Answer · 2019-10-21 19:14:10Z

3

A lambda expression doesn't require any formal parameters.

hm = Heatmap(q, slider_cb, lambda: close_cb(5))

answered Oct 21, 2019 at 19:14

chepner

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2 Comments

Kyle Over a year ago

Perfect thanks! I'll mark it after the time. Whats the difference between lambda x: ... and lambda: ?

chepner Over a year ago

One requires an argument, the other does not. The parameter list has the same format as that in a def statement.

blhsing · Accepted Answer · 2019-10-21 19:37:44Z

0

You mentioned in your question that you have tried partial but did not show us how you attempted it. Here is an example of how partial can be used in your case:

from functools import partial
hm = Heatmap(q, slider_cb, partial(close_cb, 5))

answered Oct 21, 2019 at 19:37

blhsing

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