倒数计时器以延迟方式递减

Question

我一直在做一个游戏;其中一个主要组件是倒数计时器 - 但是这个计时器被延迟了，我无法推断出原因。我希望它每秒递减一次，但它似乎每 6 秒递减一次。

以下是我设置计时器的方法：

loops = 0
minute = 1
tens = 0
ones = 0 
#Timer Calculation
    screen.blit(cloudSky, (0,0))
    if go == True:
        loops = loops + 1
        if (loops % 60)== 0:
            if ones == 0 and tens == 0 and minute != 0:
                tens = 6
                ones = 0
                minute = minute - 1
            if ones == 0 and tens != 0:
                ones = 9
                tens = tens - 1
            elif ones != 0:
                ones = ones - 1
            elif tens == 0 and ones == 0:
                tens = 5
                minute = minute - 1
            elif ones == 0 and tens != 0:
                tens = tens - 1

            if minute <= 0 and tens == 0 and ones == 0:
                go = False

我用下面的代码在屏幕上打印它：

#Draw Clock Time
time = timeFont.render(str (minute)+ ":" + str (tens) + str (ones), True, WHITE)
screen.blit(time, (750,10))

任何帮助将不胜感激！

Answer 1

这可能是由于您依赖于通过此计时器计算恰好 60 次/秒的事实引起的。如果您每秒更新 10 次，它将为每 1 秒实时增加 6 秒。您可能应该使用类似模块time来更精确地计时。

import time
# To remember current time
timer = time.time() # timer = 1497737106.913825
# To read time since the clock started
seconds_passed = time.time() - timer # seconds_passed = 11.117798089981079
# To get a tuple containing calculated time (minutes, seconds, etc.)
time_passed = time.gmtime(seconds_passed)
# time_passed = time.struct_time(tm_year=1970, tm_mon=1, tm_mday=1,
#               tm_hour=0, tm_min=0, tm_sec=11, tm_wday=3, tm_yday=1, tm_isdst=0)
# As you can see year returned is 1970 because 0 means start of Unix Epoch time
# but you don't use it anyway

由于它被命名为元组，因此您可以使用索引：

time_passed[0] # 1970

或名称：

time_passed.tm_year # 1970

https://docs.python.org/3/library/time.html#time.time