Question

我从名为sounds的列表中播放声音。它播放声音，在start中存储播放声音的时间，等待6秒，然后播放列表中的下一个声音。现在，我想通过按下按键来捕获这6秒之间的反应时间。如果条件成立，那么我单击按钮，它将捕获时间并将其存储在end中。然后，end和start之间的差异应该给我结果。问题是，它无法正确地衡量时间。即使单击之前我走了更长的路，它也总是给我千分之一。我想知道我在这里做错了吗？

start = time.time()

    for i in range(len(arr)):
        pygame.mixer.music.load(sounds[i])
        pygame.mixer.music.play()

            for e in pygame.event.get():
                if e.type == pygame.KEYDOWN:
                    if e.key == pygame.K_RIGHT:

                        if condition:
                            end = time.time()
                            diff = end - start

            while pygame.mixer.music.get_busy():
                  time.sleep(6)

Answer 1

我认为最简单的解决方案是在下一个声音开始播放时重置开始时间。

import time
import pygame as pg


pg.init()
screen = pg.display.set_mode((640, 480))
clock = pg.time.Clock()
BG_COLOR = pg.Color('gray12')
SOUND = pg.mixer.Sound('a_sound.wav')

start = time.time()

done = False
while not done:
    for event in pg.event.get():
        if event.type == pg.QUIT:
            done = True
        elif event.type == pg.KEYDOWN:
            if event.key == pg.K_RIGHT:
                diff = time.time() - start
                print(diff)

    passed_time = time.time() - start
    pg.display.set_caption(str(passed_time))
    if passed_time > 6:
        start = time.time()  # Reset the start time.
        SOUND.play()

    screen.fill(BG_COLOR)
    pg.display.flip()
    clock.tick(60)

pg.quit()

使用time.time（）和Keydown测量时间

1 个答案: