[Closed] Delete modifier by name?

there is semi-pro way:

for node in objects as array where (mods = join #() node.modifiers).count  > 0 do for m in mods where iskindof m MeshSmooth do deletemodifier node m

and there is a pro way:

for t in (getclassinstances MeshSmooth astrackviewpick:on) do for node in (refs.dependentnodes t.client) do deletemodifier node t.anim

PS. more PRO sometimes doesn’t mean BETTER (or faster)

Interesting, did not know you could use getClassInstances that way, but I never tried it. Very cool, I like the semi-pro way better then the pro just because its easier to read.

quick question: My method it works, but what are the redundancies in it?

there is nothing bad… but of course disable/enable view redraw has to be moved out the node loop… as:

with redraw off for node in nodes ... 

it’s better to replace do append with collect

  mods = for m in node.modifier where ... collect m
  for m in mods do deletemodifier node m
  

and… there is a trick. you can delete modifiers in backward order:

  for k=node.modifiers.count to 1 by -1 where iskindof (m = node.modifiers[k]) ... do deletemodifier node m 
  

last (backward loop) solution saves some memory. which is not bad sometimes.

Yes I noticed that after I submitted the post! thx man!

Pro tip!

(getclassinstances MeshSmooth astrackviewpick:on)

Thanks for that Denis!