How do I convert camera coordinates to world coordinates?

Question

Blender has a great function to convert world coordinates to camera coordinates. However, in my case I am trying to do the exact opposite and can't find an existing method of doing so. I have tried reversing the math on this but haven't had any success. I have attached the function here from bpy_extras/object_utils.py

def world_to_camera_view(scene, obj, coord):
"""
Returns the camera space coords for a 3d point.
(also known as: normalized device coordinates - NDC).

Where (0, 0) is the bottom left and (1, 1)
is the top right of the camera frame.
values outside 0-1 are also supported.
A negative 'z' value means the point is behind the camera.

Takes shift-x/y, lens angle and sensor size into account
as well as perspective/ortho projections.

:arg scene: Scene to use for frame size.
:type scene: :class:`bpy.types.Scene`
:arg obj: Camera object.
:type obj: :class:`bpy.types.Object`
:arg coord: World space location.
:type coord: :class:`mathutils.Vector`
:return: a vector where X and Y map to the view plane and
   Z is the depth on the view axis.
:rtype: :class:`mathutils.Vector`
"""
from mathutils import Vector

co_local = obj.matrix_world.normalized().inverted() @ coord
z = -co_local.z

camera = obj.data
frame = [-v for v in camera.view_frame(scene=scene)[:3]]
if camera.type != 'ORTHO':
    if z == 0.0:
        return Vector((0.5, 0.5, 0.0))
    else:
        frame = [(v / (v.z / z)) for v in frame]

min_x, max_x = frame[1].x, frame[2].x
min_y, max_y = frame[0].y, frame[1].y

x = (co_local.x - min_x) / (max_x - min_x)
y = (co_local.y - min_y) / (max_y - min_y)

return Vector((x, y, z))

Answer 1

Is a scripted version necessary to you? Because there is a very simple node setup to accomplish this:

In this image, I convert the vector 0 0 -1 from camera space into world space. If I look from the top, it will be blue, from the back it will be green, from the right (as here) it will be red. This way the color always corresponds to the orientation gizmo.