Pfthb

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.

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.

You can use the close() method of widget:

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

That's how I clear a layout :

def clearLayout(layout):
 if layout != None:
 while layout.count():
 child = layout.takeAt(0)
 if child.widget() is not None:
 child.widget().deleteLater()
 elif child.layout() is not None:
 clearLayout(child.layout())

My solution to this problem is to override the setLayout method of QWidget. The following code updates the layout to the new layout which may or may not contain items that are already displayed. You can simply create a new layout object, add whatever you want to it, then call setLayout. Of course, you can also just call clearLayout to remove everything.

def setLayout(self, layout):
 self.clearLayout()
 QWidget.setLayout(self, layout)

def clearLayout(self):
 if self.layout() is not None:
 old_layout = self.layout()
 for i in reversed(range(old_layout.count())):
 old_layout.itemAt(i).widget().setParent(None)
 import sip
 sip.delete(old_layout)

From the docs:

To remove a widget from a layout, call removeWidget(). Calling QWidget.hide() on a widget also effectively removes the widget from the layout until QWidget.show() is called.

removeWidget is inherited from QLayout, that's why it's not listed among the QGridLayout methods.

I use:

 while layout.count() > 0: 
 layout.itemAt(0).setParent(None)

A couple of solutions, if you are swapping between known views using a stacked widget and just flipping the shown index might be a lot easier than adding and removing single widgets from a layout.

If you want to replace all the children of a widget then the QObject functions findChildren should get you there e.g. I don't know how the template functions are wrapped in pyqt though. But you could also search for the widgets by name if you know them.

this my first time actually answering a stack overflow question but I saw that all the answers here are slightly wrong. (yes, I know that the question was to delete all widgets) The problem with most of them is that they don't account for nested layouts, so I made a recursive function, that given a layout it will recursively delete everything inside it, and all the layouts inside of it. here it is:

def clearLayout(layout):
print("-- -- input layout: "+str(layout))
for i in reversed(range(layout.count())):
 layoutItem = layout.itemAt(i)
 if layoutItem.widget() is not None:
 widgetToRemove = layoutItem.widget()
 print("found widget: " + str(widgetToRemove))
 widgetToRemove.setParent(None)
 layout.removeWidget(widgetToRemove)
 elif layoutItem.spacerItem() is not None:
 print("found spacer: " + str(layoutItem.spacerItem()))
 else:
 layoutToRemove = layout.itemAt(i)
 print("-- found Layout: "+str(layoutToRemove))
 clearLayout(layoutToRemove)

I might not have accounted for all UI types, not sure. Hope this helps!

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

or

 for i in range(layout.count()):
 layout.itemAt(0).widget().close()
 layout.takeAt(0)