python - 如何处理 pygame.get_pressed() 来接收输入

Question

pygame.get_keypressed() 为每个按下的按键返回一个由 0 和 1 组成的长列表，可以由 pygame 映射。下面的示例，是否有直接的方法来提取按下的按键的字母表示？

我试图避免使用很长的多个 if 语句来测试 K_a、K_b... 等是否被单击，有没有办法处理 1 和下面有 0 吗？

(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
 0, 0, 0)

Answer 1

它看起来像二进制表示形式的数字，因此您可以将其转换为整数并使用按位“AND”将其与某些“掩码”(表示您需要的键)进行比较。我不知道这是否值得做。

<小时/>

要测试更多键(例如 h,e,l,o )，您可以使用

pressed = pygame.get_keypressed()

if all( (pressed[x] for x in (K_h, K_e, K_l, K_o)) ):
    print "all keys are pressed: h, e, l, o"

if any( (pressed[x] for x in (K_h, K_e, K_l, K_o)) ):
    print "at least one key is pressed: h, e, l, o"

你可以把它变成函数

def test_all_keys( list_of_keys, pressed ):
    return all( (pressed[x] for x in list_of_keys) )

if test_all_keys((K_h, K_e, K_l, K_o), pressed):
    print "all keys are pressed: h, e, l, o"

如果您需要按键列表:

list_of_pressed = [ i for i in range(len(pressed)) if pressed[i] ]

if K_a in list_of_pressed:
    print "key 'a' was pressed"