有没有办法让 pynput.keyboard 计算每次按下（短按和长按）一次，同时可以同时按住多个键？

Question

我试图在 python 中制作简单的密钥计数器，但我发现了一个问题。当我按住一个键一段时间后，它开始快速计算，因为我按住了这个键。

from pynput.keyboard import Listener
n = 0
def on_press(key):
    global n
    n += 1
def on_release(key):
    pass
with Listener(on_press=on_press, on_release=on_release) as listener:
    listener.join()

所以我做了一个锁，当我按住一个键时，它不能计数。

from pynput.keyboard import Listener
pressed = False
n = 0
def on_press(key):
    global n, pressed
    if pressed == False:
        n += 1
        pressed = True
def on_release(key):
    global pressed
    pressed = False
with Listener(on_press=on_press, on_release=on_release) as listener:
        listener.join()

但现在当我开始打字或玩游戏时，它不会计算每一次按下，因为我有时会同时按住 2 个键，所以 1 个键不被计算在内。

我尝试在按住键 0.5 秒后锁定，所以它会在我打字时计算每次按下，但当我按住多个键超过 0.5 秒时（当我按住多个键更长时间时）仍然存在问题超过 0.5 秒，只算 1).

有没有办法让它对每次按下（无论是否按下）只计算一次，同时可以同时按住多个键？

Answer 1

您的问题可以通过跟踪数组中当前按下的按钮来解决。

from pynput.keyboard import Key, Listener

keys_currently_pressed = []
n = 0

def on_press(key):
    global n
    if key not in keys_currently_pressed:
        keys_currently_pressed.append(key)
        n += 1
        print(key, 'pressed')

def on_release(key):
    global n
    if key in keys_currently_pressed:
        keys_currently_pressed.remove(key)
    print(key, 'released')
    if key == Key.esc:
        print("You pressed", n, "keys")
        return False

with Listener(on_press=on_press, on_release=on_release) as listener:
        listener.join()

有没有办法让 pynput.keyboard 计算每次按下（短按和长按）一次，同时可以同时按住多个键？

Is there a way to make pynput.keyboard count every press (short and long) just once with possibility holding multiple keys at once?

python

pynput