如何通过 Python 访问数位板笔数据？

Question

我需要通过 Python 访问 Windows 数位板笔数据（例如表面）。我主要需要位置、压力和倾斜值。

我知道如何访问 Wacom 笔数据，但 windows 笔不同。

有一个名为 Kivy 的 Python 库可以处理多点触控，但它会将我的笔识别为手指 (WM_TOUCH) 而不是笔 (WM_PEN)。

这是我的 Kivy 代码（不报告压力和倾斜）：

 from kivy.app import App
 from kivy.uix.widget import Widget

class TouchInput(Widget):

def on_touch_down(self, touch):
    print(touch)
def on_touch_move(self, touch):
    print(touch)
def on_touch_up(self, touch):
    print("RELEASED!",touch)

class SimpleKivy4(App):

def build(self):
    return TouchInput()

有一个很棒的 processing 库，名为 Tablet，它只适用于具有简单 API 的 Wacom 数位板（例如，tablet.getPressure()）

我需要这样的东西。

Answer 1

如果您想查看设备：

import pyglet
pyglet.input.get_devices()

如果您想查看设备控件：

tablets = pyglet.input.get_devices() #tablets is a list of input devices
tablets[0].get_controls() #get_controls gives a list of possible controls of the device  

现在来获取数据。我有一台 xp-pen g640 平板电脑，但没有倾斜传感器，但如果你有，修改代码很容易：

if tablets:
    print('Tablets:')
    for i, tablet in enumerate(tablets):
        print('  (%d) %s' % (i , tablet.name))
i = int(input('type the index of the tablet.'))

device = tablets[i]
controls = device.get_controls()
df = pd.DataFrame()
window = pyglet.window.Window(1020, 576)

# Here you will have to write a line like "control_tilt_x = controls[j]" where j is
# the controls list index of the tilt control.
control_presion = controls[7]
Button = controls[3]
control_x =controls[5]
control_y =controls[6]
control_punta = controls[4]
control_alcance = controls [0]
name = tablets[9].name

try:
    canvas = device.open(window)
except pyglet.input.DeviceException:
    print('Failed to open tablet %d on window' % index)

print('Opened %s' % name)

    
@control_presion.event
def on_change(presion):
    global df   
    df_temp = pd.DataFrame({'x':[control_x.value/(2**15)],
                           'y':[control_y.value/(2**16)],
                           'p':[control_presion.value/8],
                           't':[time.time()]})
    df = pd.concat([df,df_temp])
          
pyglet.app.run()

Answer 2

此解决方案适用于我，适用于来自 here 的 Python 3。

有用的东西：笔、橡皮擦、笔按钮、两个压力传感器。

首先安装 pyglet 库：pip install pyglet。然后使用代码：

import pyglet

window = pyglet.window.Window()
tablets = pyglet.input.get_tablets()
canvases = []

if tablets:
    print('Tablets:')
    for i, tablet in enumerate(tablets):
        print('  (%d) %s' % (i + 1, tablet.name))
    print('Press number key to open corresponding tablet device.')
else:
    print('No tablets found.')

@window.event
def on_text(text):
    try:
        index = int(text) - 1
    except ValueError:
        return

    if not (0 <= index < len(tablets)):
        return

    name = tablets[i].name

    try:
        canvas = tablets[i].open(window)
    except pyglet.input.DeviceException:
        print('Failed to open tablet %d on window' % index)

    print('Opened %s' % name)

    @canvas.event
    def on_enter(cursor):
        print('%s: on_enter(%r)' % (name, cursor))

    @canvas.event
    def on_leave(cursor):
        print('%s: on_leave(%r)' % (name, cursor))

    @canvas.event
    def on_motion(cursor, x, y, pressure, a, b):  # if you know what "a" and "b" are tell me (tilt?)
        print('%s: on_motion(%r, x=%r, y=%r, pressure=%r, %s, %s)' % (name, cursor, x, y, pressure, a, b))

@window.event
def on_mouse_press(x, y, button, modifiers):
    print('on_mouse_press(%r, %r, %r, %r' % (x, y, button, modifiers))

@window.event
def on_mouse_release(x, y, button, modifiers):
    print('on_mouse_release(%r, %r, %r, %r' % (x, y, button, modifiers))

pyglet.app.run()