如何通过Python访问数位板笔数据?

时间:2019-01-02 08:20:08

标签: python tablet surface stylus-pen

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笔却不同。

有一个名为Kivy的Python库,可以处理多点触摸,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(不报告压力和倾斜度):

 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,该库仅适用于具有简单API(例如tablet.getPressure())的Wacom平板电脑

我需要这样的东西。

2 个答案:

答案 0 :(得分: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。我有一个 xp-pen g640 平板电脑,没有倾斜传感器,但如果你有它,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()

答案 1 :(得分:0)

此解决方案对我有效,适用于here的Python 3。

有效的方法:笔,橡皮,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()