refactor removeAllLayoutWidgets

def removeAllLayoutWidgets(layout):
    """Remove all existing widgets from QLayout."""
    for i in reversed(range(layout.count())):
        layout.itemAt(i).widget().setParent(None)

ChatGPT's suggestion:

The function removeAllLayoutWidgets that you've provided sets the parent of each widget in the layout to None, which effectively removes the widget from the layout. However, this does not necessarily delete the widget from memory. The widget still exists and takes up memory, but it is no longer visible and does not have a parent widget.

Using setParent(None) may be sufficient if you plan to reuse the widgets later or if you are managing their deletion elsewhere in your code. However, if you want to ensure that the widgets are deleted and their memory is freed, you should use deleteLater(), as it schedules the widget for deletion in a safe manner.

Here's how you could modify your removeAllLayoutWidgets function to use deleteLater():

 def removeAllLayoutWidgets(layout):
     """Remove all existing widgets from QLayout and schedule them for deletion."""
     for i in reversed(range(layout.count())):
         widget = layout.itemAt(i).widget()
         if widget is not None:  # Check if the layout item is a widget and not None
             widget.deleteLater()

This version of the function will schedule all widgets for deletion, and Qt will delete them the next time it processes events. This is generally the preferred way to remove and delete widgets in Qt when you no longer need them and want to ensure that resources are freed appropriately.

BCDA-APS / mdaviz

refactor removeAllLayoutWidgets #89