Python,ruby或LUA中的游戏开发?

时间:2013-04-09 08:01:12

标签: python ruby lua game-engine

我在Action Script 3和C ++的某些游戏引擎中有游戏开发经验。 但是,我想提高工作效率,所以我想用Python,ruby或LUA开发一个新项目。 这是个好主意吗?如果是,您会建议哪一个?什么是杀手游戏开发工具集或引擎?

1 个答案:

答案 0 :(得分:1)

如果你有任何好处,请选择Pyglet 它是针对OpenGL的跨平台Python版本独立钩子,具有出色的性能。这有点棘手,但它比Python世界中的其他任何东西做得更好。

如果你是初学者,我会选择Pygame 它对系统有点负担,但是现代计算机不是问题..而且,它有预先打包的API用于游戏开发(因此得名):)

Python游戏/图形引擎的“官方”列表: http://wiki.python.org/moin/PythonGames

一些好的:

  • 的Panda3D
  • Pyglet
  • pygame的
  • Blender3D

示例Pyglet代码:

#!/usr/bin/python
import pyglet
from time import time, sleep

class Window(pyglet.window.Window):
    def __init__(self, refreshrate):
        super(Window, self).__init__(vsync = False)
        self.frames = 0
        self.framerate = pyglet.text.Label(text='Unknown', font_name='Verdana', font_size=8, x=10, y=10, color=(255,255,255,255))
        self.last = time()
        self.alive = 1
        self.refreshrate = refreshrate
        self.click = None
        self.drag = False

    def on_draw(self):
        self.render()

    def on_mouse_press(self, x, y, button, modifiers):
        self.click = x,y

    def on_mouse_drag(self, x, y, dx, dy, buttons, modifiers):
        if self.click:
            self.drag = True
            print 'Drag offset:',(dx,dy)

    def on_mouse_release(self, x, y, button, modifiers):
        if not self.drag and self.click:
            print 'You clicked here', self.click, 'Relese point:',(x,y)
        else:
            print 'You draged from', self.click, 'to:',(x,y)
        self.click = None
        self.drag = False

    def render(self):
        self.clear()
        if time() - self.last >= 1:
            self.framerate.text = str(self.frames)
            self.frames = 0
            self.last = time()
        else:
            self.frames += 1
        self.framerate.draw()
        self.flip()

    def on_close(self):
        self.alive = 0

    def run(self):
        while self.alive:
            self.render()
            # ----> Note: <----
            #  Without self.dispatc_events() the screen will freeze
            #  due to the fact that i don't call pyglet.app.run(),
            #  because i like to have the control when and what locks
            #  the application, since pyglet.app.run() is a locking call.
            event = self.dispatch_events()
            sleep(1.0/self.refreshrate)

win = Window(23) # set the fps
win.run()




使用Python 3.X注意Pyglet:

你必须下载 1.2alpha1 ,否则会抱怨你没有安装Python3.X.)