So I just learned about symmetrize, but don't want to start over on my mesh. Is there a way I can easily select the half I like less, delete it, and then use symmetrize and automerge doubles to have a perfectly even mesh?


3 Answers 3


Select one vertex in the middle of your model, then press Menu > Select > Side of Active and choose the axis and axis mode in the redo panel:

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JuhaW's answer is probably the way to go, but if you don't want to click through all those menus, often a quick border select (B) will do:

enter image description here


I wrote this script so that I could select all vertices that are along the same plane. I needed this because I had a mesh with millions of vertices, but actually describing a simple shape. Instructions:

  • Split the blender pane
  • Change the editor type to text
  • Paste this script
  • Select an object
  • Enter edit mode
  • Select 3 vertices (it will error if not exactly 3 selected)
  • Run the script with alt+p (while mouse is focused on text window)
  • A plane with the given tolerance will be selected

Here is the code:

import bpy
import bmesh
import numpy as np

#  Usage
#    Select an object
#    Enter edit mode
#    Select 3 vertices (error if not exactly 3 selected)
#    run the script with alt+p
#    A plane with the given tolerance will be selected

# user edit these
tol = 1e-3

def isclose(a, b, rel_tol=1e-09, abs_tol=0.0):
    return abs(a-b) <= max(rel_tol * max(abs(a), abs(b)), abs_tol)

# Setup stuff and unselect everything

obj = bpy.context.active_object

count_sel = 0
selected = []

if 1:
    count_sel = 0
    if obj.mode == 'EDIT':
        print("Object was in edit")
        for v in bm.verts:
            if v.select:
                count_sel += 1
        if count_sel != 3:
            raise Exception("must select 3 verts")
        raise Exception("Object is not in edit mode.")


p1 = np.array(selected[0])
p2 = np.array(selected[1])
p3 = np.array(selected[2])

# These two vectors are in the plane
v1 = p3 - p1
v2 = p2 - p1

# the cross product is a vector normal to the plane
cp = np.cross(v1, v2)
a, b, c = cp

# This evaluates a * x3 + b * y3 + c * z3 which equals d
d = np.dot(cp, p3)

print('The equation is {0}x + {1}y + {2}z = {3}'.format(a, b, c, d))

def pdist(p, a, b, c, d):
    x = p[0]
    y = p[1]
    z = p[2]

    return np.abs(a*x + b*y + c*z - d) / np.sqrt(a*a + b*b + c*c)

if 1:
    bpy.ops.object.mode_set(mode = 'EDIT') 
    bpy.ops.mesh.select_all(action = 'DESELECT')
    bpy.ops.object.mode_set(mode = 'OBJECT')

if 1:
    for i,v in enumerate(obj.data.vertices):
        pd = pdist(v.co, a, b, c, d)
        # print('p {0},{1},{2} has distance {3} from plane'.format(v.co[0], v.co[1], v.co[2], pd))

        if pd <= tol:
            v.select = True

    bpy.ops.object.mode_set(mode = 'EDIT')

In my use case, after the plane was selected, I click back to the 3d view and press f to join the vertices into a plane.

  • 2
    $\begingroup$ This appears to be selecting coplanar verts I believe the OP simply wants to select one half, which for for local x axis mirror would be v.select = v.co.x >= 0 For any plane there is a mathutils distance point to plane method, which will return positive or negative values depending which side of plane. (or 0 on plane) $\endgroup$
    – batFINGER
    Dec 29, 2018 at 13:03

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