Align pivot point to bottom back left of object

Question

Thank you for the great script.

I have tried to use this script, but I want a slightly different setup. Do you know how you can change this script so the pivot point is set to the bottom left back?

import bpy
from bpy import context as C

for z, co in [(o.bound_box[0][2], v.co) for o in C.selected_objects for v in o.data.vertices]:
co.z -= z 

Answer 1

Something like this should do the trick.

import bpy
import numpy as np
from mathutils import Vector, Matrix


obj = bpy.context.object
mesh = obj.data

# get the lowest point in all 3 axes for the object bounding box
#this should be the bottom-left

bb = obj.bound_box
bb_locs = np.array([v[:] for v in bb])
bot_left = []

for i, row in enumerate(bb_locs.transpose()):
    # go through each row in the bounding box axes
    # if the axis is "Y"
    if i == 1:
        # get the greatest value in that row:
        val = max(row)
    else:
        # otherwise get the lowest value
        val = min(row)
    bot_left.append(val)
    
bot_left = Vector(bot_left)
    

# make a translation matrix that will move the mesh in the opposite direction
# leaving the origin in place.
mat = Matrix.Translation(-bot_left)
if mesh.is_editmode:
    bm = bmesh.from_edit_mesh(mesh)
    bm.transform(mat)
    bmesh.update_edit_mesh(me, False, False)
else:
    mesh.transform(mat)

mesh.update()

# then move the entire object by the matrix itself. 
# Which puts the object back where it was with the origin in the new location.

obj.matrix_world.translation = bot_left

Result: