[Paraview] layout question, pvpython

[Sgouros, Thomas]

Alas, indeed. Just as I thought I was starting to understand what's going
on. Was I just misled because there is a trace output that has this in it?

# get layout
layout1 = GetLayoutByName("Layout #1")
# split cell
layout1.SplitHorizontal(0, 0.5)


Without layout support, can I use pvpython to present two render views and
link them with a camera link? Can you suggest what direction might succeed?

You may have suggested this already, but I'm not sure the hint about
view.GetRenderWindow().SetPosition(x, y) is enough for me. I don't really
get how that's a way forward. What are you suggesting I use that function
to do?

Thanks,

 -Tom

On Fri, May 11, 2018 at 3:04 PM, Utkarsh Ayachit <
utkarsh.ayachit at kitware.com> wrote:

> Tom,
>
> Alas, there's no layout support in pvpython. There indeed is a layout proxy
> (vtkSMLayoutProxy), but they rely on the Qt ParaView components to layout
> the windows. The proxy itself is the "model", if you will.  And it works
> when you execute the layout related updates in the Python shell in the
> ParaView UI.
>
> A clumpsy way to set a window's position in Python is:
> > view.GetRenderWindow().SetPosition(x, y)
>
> You can use that to lay out the windows, short of doing something more
> elaborate with a GUI library.
>
> Utkarsh
>
>
>
>
> On Fri, May 11, 2018 at 2:34 PM Sgouros, Thomas <thomas_sgouros at brown.edu>
> wrote:
>
> > Hi All:
>
> > I've become accustomed to throwing around render view objects in
> pvpython, swapping them in and out of the view with simple.Show() and
> simple.Hide(). Now I have to learn about layout objects. Can someone
> suggest the best way to think about these? Are they containers into which I
> can drop one or more render views and move them around, or are they more
> like molds into which I can pour a render view and then use the layout as
> handles with which to manipulate the views?
>
> > Where can I find a list of the methods of a layout object? Or would it be
> better to call it a layout proxy?
>
> > Thank you,
>
> >   -Tom
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