Python 在pyglet/opengl中渲染纹理立方体的最有效方法

目前，我有这个代码，它渲染一个纹理立方体。它只需要几个立方体就可以了，但当我添加数百个立方体时，创建它需要花费很长时间

def get_tex(file):
    tex = pyglet.image.load(file).get_texture()
    glTexParameterf(GL_TEXTURE_2D, GL_TEXTURE_MIN_FILTER, GL_NEAREST)
    glTexParameterf(GL_TEXTURE_2D, GL_TEXTURE_MAG_FILTER, GL_NEAREST)
    return pyglet.graphics.TextureGroup(tex)


def make_textures(textures):
    result = {}
    base = "imgs/"
    faces = ("left", "right", "top", "bottom", "front", "back")
    for face in faces:
        file_name = textures.get(face, textures.get("side"))
        result[face] = get_tex(base + file_name)
    return result

def get_vertices(x, y, z):
    dx, dy, dz = x + 1, y + 1, z + 1
    return [
        (x, y, z, x, y, dz, x, dy, dz, x, dy, z),  # side
        (dx, y, dz, dx, y, z, dx, dy, z, dx, dy, dz),  # side
        (x, dy, dz, dx, dy, dz, dx, dy, z, x, dy, z),  # top
        (x, y, z, dx, y, z, dx, y, dz, x, y, dz),  # bottom
        (dx, y, z, x, y, z, x, dy, z, dx, dy, z),  # side
        (x, y, dz, dx, y, dz, dx, dy, dz, x, dy, dz),  # side
    ]
class Grass:
    textures = make_textures({"side": "grass_side.png", "top": "grass_top.png", "bottom": "dirt.png"})


   
def draw_block(self, block):

    position = (block.x,block.y,block.z)
    if block._type == "grass":
        block = Grass
    else:
        block = Grass
    vertices = get_vertices(*position)
    faces = ("left", "right", "top", "bottom", "front", "back")
    for vertex, face in zip(vertices, faces):
        self.batch.add(
            4,
            gl.GL_QUADS,
            block.textures[face],
            ("v3f/static", vertex),
            ("t2f/static", (0, 0, 1, 0, 1, 1, 0, 1)),
        )

---

但这需要很长的时间才能创建，我不能使用其他线程来完成，因为我正在调用gl函数。

不确定函数名是否与它的功能相符。但不要在每个渲染帧上绘制块。我希望/假设您已经知道这一点。不确定函数名是否与它的功能相符。但不要在每个渲染帧上绘制块。我希望/假设你已经知道这一点。