Justin,

thank you for your example. I never knew you can put a for loop right
inside a list.

I was looking for a faster alternative because it's such a big set
with up to a thousand geometries all imported into the scene. But for
my purposes, a normal loop would do too.

The set was provided by another studio and it's seriously not cleaned
up: tiny basic shapes with over 250k vertices kinda hidden inside
other models for no reason.


On Mon, Dec 26, 2011 at 11:33 PM, Justin Israel <[email protected]> wrote:
> As far as I know, maya's database doesn't index all geometry vertex counts
> in a way that you can filter select on it. The API selection mechanism only
> lets you select on name patterns and then filter on type. And the python
> commands wrap around that to add more type filtering.
> I believe your only option is to loop over all geometry and do a
> polyEvaluate(v=True) on each one to test the vertex count, and then append
> to a list. A list comprehension would be the same speed as a multi-line for
> loop:
>
> # list comp
> matches = [geom for geom in cmds.ls(dag=True, g=True) if
> cmds.polyEvaluate(geom, v=True) > 1000]
>
> # same as normal loop
> matches = []
> for geom in cmds.ls(dag=True, g=True):
>     if cmds.polyEvaluate(geom, v=True) > 1000:
>         matches.append(geom)
>
> This would also be slightly faster if you were to do it in the API because
> of the selection iterator.
>
> Is this something where you need extremely low latency searches for queries
> happening constantly? Or were you just looking for a simpler way to do the
> query? Because if you needed a solution faster than looping over every
> single one, in the way thats similar to your SQL style query, you would need
> to index all the geometry yourself by vertex count into a dictionary. And
> then either manage it yourself each some geometry is modified or attach some
> kind of scriptJob to keep updating it. But Im not sure if thats even your
> goal. The SQL query just sorta made me think you needed fast queries that
> are indexed.
>
>
> On Dec 26, 2011, at 4:32 AM, Panupat wrote:
>
> say I want to select geometries that have more than X amount of
> vertices. Using some kind of logic like this
>
> select geometry where vertex count > 10000
>
> Is this doable? I can think of a way to cycle through all geometries
> and check their vertices but wondering if there's a better way.
>
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