31 May
2013
31 May
'13
12:07 p.m.
On Thu, May 30, 2013 at 9:57 PM, Tudor Girba <tudor(a)tudorgirba.com> wrote:
Hi,
Not really. You could do it because you in your example you controlled the
objects corresponding to the roots of the hierarchy.
But by definition, the tree has always a root. In my opinion the
computation of the root is the necessary step in a layout involving trees.
You can compute the root by looking at the incoming connections and setting
as the root the one which does not have any, or has the least?. We had this
problem when we have to represent a graph as a tree, we ask the user to
provide the root.
Here is my challenge. Make this graph layout the trees it can and wrap the
lines afterwards:
view nodes: (1 to: 100).
view edgesFrom: [ :x | x atRandom ].
:)
For everyone else: the idea is to be able to compose layouts out of layout
pieces. For example, we should be able to apply a tree layout and then
arrange the resulting trees in a flow layout or in a compact rectangle
layout.
Cheers,
Doru
On May 30, 2013, at 5:41 PM, mathieubmddehouck(a)mailoo.org wrote:
Hi
We speak with Tudor about mixing layout, and here is something I try.
====================================
| view rawView |
rawView := ROView new.
view := ROMondrianViewBuilder view: rawView.
"-------------"
"-------------"
view shape rectangle borderColor: (Color white).
view nodes: ROObject withAllSubclasses forEach: [ :each |
view nodes: each withAllSubclasses .
view edgesFrom: #superclass.
view layout: (ROTreeLayout new horizontalGap: 4) .
].
view layout:( RORectanglePackingLayout new ratioWidth: 100
height: 100 ).
"-------------"
"-------------"
"Below is the initiation of the menu and opening the visualization"
ROEaselMorphic new populateMenuOn: view.
view open
=======================================
So, it seems it can be done with not too long a script.
Regards
Math
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