Don*_*one 4 python windows time clock performancecounter
我想用Python编写自己的时钟对象.我希望它非常非常准确.我在Windows上看到,我可以使用QueryPerformanceCounter().但是怎么样?我不知道任何C; 只有Python 2.x.
有人可以给我一个提示,如何在Python中使用它来在Win上制作一个准确的时钟吗?
我已经使用模块移植了你给Python 的C++示例ctypes:
C++
LARGE_INTEGER StartingTime, EndingTime, ElapsedMicroseconds;
LARGE_INTEGER Frequency;
QueryPerformanceFrequency(&Frequency);
QueryPerformanceCounter(&StartingTime);
// Activity to be timed
QueryPerformanceCounter(&EndingTime);
ElapsedMicroseconds.QuadPart = EndingTime.QuadPart - StartingTime.QuadPart;
ElapsedMicroseconds.QuadPart *= 1000000;
ElapsedMicroseconds.QuadPart /= Frequency.QuadPart;
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蟒蛇
import ctypes
import ctypes.wintypes
import time
kernel32 = ctypes.WinDLL('kernel32', use_last_error=True)
starting_time = ctypes.wintypes.LARGE_INTEGER()
ending_time = ctypes.wintypes.LARGE_INTEGER()
elapsed_microseconds = ctypes.wintypes.LARGE_INTEGER()
frequency = ctypes.wintypes.LARGE_INTEGER()
kernel32.QueryPerformanceFrequency(ctypes.byref(frequency))
kernel32.QueryPerformanceCounter(ctypes.byref(starting_time))
# Activity to be timed, e.g.
time.sleep(2)
kernel32.QueryPerformanceCounter(ctypes.byref(ending_time))
elapsed_microseconds = ending_time.value - starting_time.value
elapsed_microseconds *= 1000000
elapsed_microseconds /= frequency.value
print(elapsed_microseconds)
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我真的很感激@eryksun的有用提示!
上面的代码应该打印附近的东西2000000(例如2000248.7442040185,值可能不时地不同).您也可以使用round()或int()函数来删除小数.
正如@eryksun所评论的那样,您也可以使用time.clock(),它在C中实现并且也可以使用QueryPerformanceCounter().
示例与使用的示例完全相同ctypes:
import time
starting_time = time.clock()
# Activity to be timed, e.g.
time.sleep(2)
ending_time = time.clock()
elapsed_microseconds = ending_time - starting_time
elapsed_microseconds *= 1000000
print(elapsed_microseconds)
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希望这可以帮助!
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