Pygame 空格触发计时器

我最近在学习pygame，准备写一个小游戏，游戏内容大概是：两个人，p1操作q键， p2操作p键，相应的人按下按键，就开始计时最后再按一下相应的按键，停止计时，最后看谁按下相应按键的间隔最接近10秒

import pygame as pg
import sys

# Initialization
pg.init()
pg.font.init()
# Variables
scr_width = 800
scr_height = 700
scr_color = (230, 230, 230)
# Get font from the "Fonts" folder
inkfree = pg.font.Font('C:\\Windows\\Fonts\\tt1255m_.ttf', 80)
# Create the main screen
screen = pg.display.set_mode((scr_width, scr_height))
pg.display.set_caption("COUNT TO TEN")
# Defined the clock
clock = pg.time.Clock()
# Defined font
text = inkfree.render("COUNT TO TEN", True, (10, 10, 10))

# Game loop
while True:
    for event in pg.event.get():
        if event.type == pg.QUIT:
            pg.quit()
            sys.exit()

    screen.blit(text, (62, 50))
    pg.display.flip()
    screen.fill(scr_color)
    pg.display.update()
    clock.tick(60)

我看了看网站里的其他博客，因为本人刚开始学习，都看不懂，请朋友们给点简单的方法谢谢

导入time库
你参考一下，下面接你程序 ,按k键表示时间间隔
text = inkfree.render("COUNT TO TEN", True, (10, 10, 10))
screen.fill(scr_color)
# Game loop
pre = [[0, 0], [0, 0]]
while True:
    for event in pg.event.get():
        if event.type == pg.QUIT:
            pg.quit()
            sys.exit()
        elif event.type == pg.KEYDOWN:
            if event.key == pg.K_k:                
                if pre[0][0] == 0:
                    pre[0][0] = time.time()
                else:
                    pre[0][1] = time.time()
                    print(pre, pre[0][1] - pre[0][0])
    screen.blit(text, (62, 50))
    pg.display.flip()
    
    pg.display.update()
    clock.tick(60)