Clear all widgets in a layout in pyqt

This may be a bit too late but just wanted to add this for future reference:

def clearLayout(layout):
  while layout.count():
    child = layout.takeAt(0)
    if child.widget():
      child.widget().deleteLater()

Adapted from Qt docs http://doc.qt.io/qt-5/qlayout.html#takeAt. Remember that when you are removing children from the layout in a while or for loop, you are effectively modifying the index # of each child item in the layout. That's why you'll run into problems using a for i in range() loop.

After a lot of research (and this one took quite time, so I add it here for future reference), this is the way I found to really clear and delete the widgets in a layout:

for i in reversed(range(layout.count())): 
    layout.itemAt(i).widget().setParent(None)

What the documentation says about the QWidget is that:

The new widget is deleted when its parent is deleted.

Important note: You need to loop backwards because removing things from the beginning shifts items and changes the order of items in the layout.

To test and confirm that the layout is empty:

for i in range(layout.count()): print i

---

There seems to be another way to do it. Instead of using the setParent function, use the deleteLater() function like this:

for i in reversed(range(layout.count())): 
    layout.itemAt(i).widget().deleteLater()

The documentation says that QObject.deleteLater (self)

Schedules this object for deletion.

However, if you run the test code specified above, it prints some values. This indicates that the layout still has items, as opposed to the code with setParent.

The answer from PALEN works well if you do not need to put new widgets to your layout.

for i in reversed(range(layout.count())): 
    layout.itemAt(i).widget().setParent(None)

But you will get a "Segmentation fault (core dumped)" at some point if you empty and fill the layout many times or with many widgets. It seems that the layout keeps a list of widget and that this list is limited in size.

If you remove the widgets that way:

for i in reversed(range(layout.count())): 
    widgetToRemove = layout.itemAt(i).widget()
    # remove it from the layout list
    layout.removeWidget(widgetToRemove)
    # remove it from the gui
    widgetToRemove.setParent(None)

You won't get that problem.