国内公司为什么很少用blender?

import bpy import math # Define the number of vertices and the size of the heart num_vertices = 100 size = 0.05 # Calculate the vertices of the heart shape vertices = [] for i in range(num_vertices): t = math.pi * 2 * i / num_vertices x = size * (16 * math.sin(t)**3) y = size * (13 * math.cos(t) - 5 * math.cos(2*t) - 2 * math.cos(3*t) - math.cos(4*t)) vertices.append((x, y, 0)) # Define the faces of the heart shape faces = [(i, (i+1) % num_vertices, num_vertices) for i in range(num_vertices)] # Add a point at the center of the heart vertices.append((0, 0, 0)) # Create a new mesh and a new object mesh = bpy.data.meshes.new(name="HeartMesh") obj = bpy.data.objects.new(name="HeartObject", object_data=mesh) # Link the object to the current collection bpy.context.collection.objects.link(obj) # Set the object to be the active object bpy.context.view_layer.objects.active = obj obj.select_set(True) # Create the mesh from the vertices and faces mesh.from_pydata(vertices, [], faces) mesh.update() # Switch to edit mode bpy.ops.object.mode_set(mode='EDIT') # Select all bpy.ops.mesh.select_all(action='SELECT') # Extrude along the z-axis to give some height to the heart bpy.ops.mesh.extrude_region_move( TRANSFORM_OT_translate={"value": (0, 0, 0.2)} ) # Switch back to object mode bpy.ops.object.mode_set(mode='OBJECT') # Create an arrow bpy.ops.mesh.primitive_cone_add(vertices=12, radius1=0.1, depth=4, location=(0, 0, 0.5)) arrow_shaft = bpy.context.object bpy.ops.mesh.primitive_cone_add(vertices=12, radius1=0.2, depth=0.8, location=(0, 0, 2.4)) arrow_head = bpy.context.object # Combine the parts of the arrow into one object bpy.ops.object.select_all(action='DESELECT') arrow_shaft.select_set(True) arrow_head.select_set(True) bpy.context.view_layer.objects.active = arrow_shaft bpy.ops.object.join() # Position the arrow at the heart's center arrow_shaft.location = (0, 0, 0) # Rotate the arrow arrow_shaft.rotation_euler = (math.pi/1.5, math.pi/4, -math.pi/6) # Scale the arrow to fit the heart arrow_shaft.scale = (0.5, 0.5, 0.5)