Question

我在Python 3.5中使用Kivy。我正在尝试建立一个应用程序，在该应用程序中，按下按钮即可安排功能调用。一个前提：这是我使用Kivy的第一个应用程序，而我对Python的了解不是很好。这是我的简化代码：

class MyApp(App):
    def build(self):
        layout = GridLayout(cols=2)
        TextStream = Label(text = 'Something will be written here: ')
        StartButton = Button(text = 'Start writing')
        StartButton.bind(on_press=lambda x:self.start_program(freq=10))
        layout.add_widget(TextStream)
        layout.add_widget(StartButton)
        return layout

    def start_program(self, freq):
        Clock.schedule_interval(self.write_something, 1.0/freq)
        # Also tried Clock.schedule_interval(self.write_something(), 1.0/freq)

    def write_something(self):
        TextStream.text =+ 'Something '

if __name__ == '__main__':
    MyApp().run()

运行脚本时得到的是：

TypeError: write_something() takes 1 positional argument but 2 were given

在阅读关于this question的答案时，我了解到Python中的方法实际上与在代码中的显示方式不同。

更准确地说，从this comment可以理解，正确地声明方法write_something为参数self是正确的方法。另外，从this answer中，我得到Python计数的传递参数的数量应为1 + N，其中在调用中传递了N。我没有在电话中传递任何参数。

那么，是否有Kivy传递了一些沉默的论点？我以错误的方式调用该方法吗？

编辑-错误堆栈

如评论所建议，我包括了整个错误堆栈：

Traceback (most recent call last):

  File "<ipython-input-1-5d17b5569154>", line 1, in <module>
    runfile('/home/raggot/Projects/MyApp/scripts/_test_kivy.py', wdir='/home/raggot/Projects/MyApp/scripts')

  File "/usr/local/lib/python3.5/dist-packages/spyder/utils/site/sitecustomize.py", line 705, in runfile
    execfile(filename, namespace)

  File "/usr/local/lib/python3.5/dist-packages/spyder/utils/site/sitecustomize.py", line 102, in execfile
    exec(compile(f.read(), filename, 'exec'), namespace)

  File "/home/raggot/Projects/MyApp/scripts/_test_kivy_camera.py", line 78, in <module>
    MyApp().run()

  File "/usr/local/lib/python3.5/dist-packages/kivy/app.py", line 826, in run
    runTouchApp()

  File "/usr/local/lib/python3.5/dist-packages/kivy/base.py", line 502, in runTouchApp
    EventLoop.window.mainloop()

  File "/usr/local/lib/python3.5/dist-packages/kivy/core/window/window_pygame.py", line 403, in mainloop
    self._mainloop()

  File "/usr/local/lib/python3.5/dist-packages/kivy/core/window/window_pygame.py", line 289, in _mainloop
    EventLoop.idle()

  File "/usr/local/lib/python3.5/dist-packages/kivy/base.py", line 337, in idle
    Clock.tick()

  File "/usr/local/lib/python3.5/dist-packages/kivy/clock.py", line 581, in tick
    self._process_events()

  File "kivy/_clock.pyx", line 384, in kivy._clock.CyClockBase._process_events

  File "kivy/_clock.pyx", line 414, in kivy._clock.CyClockBase._process_events

  File "kivy/_clock.pyx", line 412, in kivy._clock.CyClockBase._process_events

  File "kivy/_clock.pyx", line 167, in kivy._clock.ClockEvent.tick

TypeError: write_something() takes 1 positional argument but 2 were given

Answer 1

当它连接任何事件时，在这种情况下，schedule_interval Kivy会向其传递其他参数，因此它会向您抛出该错误，在您的情况下，它仅使用args而您没有任何问题。对于Clock，它将通过dt，这是准确的通话时间。

def start_program(self, freq):
    Clock.schedule_interval(self.write_something, 1.0/freq)

def write_something(self, *args):
    print(args)

使用Kivy进行函数调用时发生意外的TypeError

1 个答案: