wxpython如何用代码控制工具栏开关按钮按下和弹起？

如题，急，谢谢！
问题补充：
kind=wx.ITEM_CHECK设置了，“调用ToggleTool设置按钮的状态
”，请问如何设置？

找了很久都没相关的方法。

好吧，这是文档，看wxWidget文档，下面有一段我写得代码
wxToolBar::ToggleTool
void ToggleTool(int toolId, const bool toggle)

Toggles a tool on or off. This does not cause any event to get emitted.

Parameters

toolId

Tool in question.
toggle

If true, toggles the tool on, otherwise toggles it off.
Remarks

Only applies to a tool that has been specified as a toggle tool.

#-----------------------------------------------

import wx

ID_ToggleTest = wx.NewId()

class MyFrame(wx.Frame):
def init(self):
wx.Frame.__init__(self, parent=None)
self.toolbar = self.CreateToolBar(wx.TB_HORIZONTAL|wx.NO_BORDER|wx.TB_FLAT)
self.toolbar.AddCheckLabelTool(ID_ToggleTest, 'ToggleTest', wx.ArtProvider.GetBitmap(wx.ART_NEW), wx.NullBitmap, '')
self.toolbar.Realize()

    button = wx.Button(self, -1, 'Toggle ToolButton')
    self.Bind(wx.EVT_BUTTON, self.OnToggleToolButton, button)

def OnToggleToolButton(self, event):
    self.toolbar.ToggleTool(ID_ToggleTest, (not self.toolbar.GetToolState(ID_ToggleTest)))

if name == '__main__':
app = wx.PySimpleApp()
frame = MyFrame()
frame.Show()
app.MainLoop()

首先AddTool时设置kind=wx.ITEM_CHECK，调用ToggleTool设置按钮的状态