我想测量执行一个函数所花费的时间。我没时间工作:
import timeit
start = timeit.timeit()
print("hello")
end = timeit.timeit()
print(end - start)
我想测量执行一个函数所花费的时间。我没时间工作:
import timeit
start = timeit.timeit()
print("hello")
end = timeit.timeit()
print(end - start)
当前回答
这里有一个很好的文档记录和完全类型提示的装饰器,我将其用作通用工具:
from functools import wraps
from time import perf_counter
from typing import Any, Callable, Optional, TypeVar, cast
F = TypeVar("F", bound=Callable[..., Any])
def timer(prefix: Optional[str] = None, precision: int = 6) -> Callable[[F], F]:
"""Use as a decorator to time the execution of any function.
Args:
prefix: String to print before the time taken.
Default is the name of the function.
precision: How many decimals to include in the seconds value.
Examples:
>>> @timer()
... def foo(x):
... return x
>>> foo(123)
foo: 0.000...s
123
>>> @timer("Time taken: ", 2)
... def foo(x):
... return x
>>> foo(123)
Time taken: 0.00s
123
"""
def decorator(func: F) -> F:
@wraps(func)
def wrapper(*args: Any, **kwargs: Any) -> Any:
nonlocal prefix
prefix = prefix if prefix is not None else f"{func.__name__}: "
start = perf_counter()
result = func(*args, **kwargs)
end = perf_counter()
print(f"{prefix}{end - start:.{precision}f}s")
return result
return cast(F, wrapper)
return decorator
示例用法:
from timer import timer
@timer(precision=9)
def takes_long(x: int) -> bool:
return x in (i for i in range(x + 1))
result = takes_long(10**8)
print(result)
输出:耗时:4.942629056秒真的
可以通过以下方式检查doctest:
$ python3 -m doctest --verbose -o=ELLIPSIS timer.py
类型提示:
$ mypy timer.py
其他回答
计算操作持续时间的最简单方法:
import time
start_time = time.monotonic()
<operations, programs>
print('seconds: ', time.monotonic() - start_time)
这里有官方文件。
对于Python 3
如果使用时间模块,则可以获取当前时间戳,然后执行代码,然后再次获取时间戳。现在,所用时间将是第一个时间戳减去第二个时间戳:
import time
first_stamp = int(round(time.time() * 1000))
# YOUR CODE GOES HERE
time.sleep(5)
second_stamp = int(round(time.time() * 1000))
# Calculate the time taken in milliseconds
time_taken = second_stamp - first_stamp
# To get time in seconds:
time_taken_seconds = round(time_taken / 1000)
print(f'{time_taken_seconds} seconds or {time_taken} milliseconds')
还有一种使用timeit的方法:
from timeit import timeit
def func():
return 1 + 1
time = timeit(func, number=1)
print(time)
使用time.time()测量两点之间经过的墙上时钟时间:
import time
start = time.time()
print("hello")
end = time.time()
print(end - start)
这给出了以秒为单位的执行时间。
Python 3.3之后的另一个选项可能是使用perf_counter或process_time,具体取决于您的需求。在3.3之前,建议使用time.clock(感谢Amber)。但是,它目前已被弃用:
在Unix上,将当前处理器时间作为浮点数返回以秒表示。准确度,事实上就是定义“处理器时间”的含义取决于C函数的含义具有相同名称。在Windows上,此函数返回自该函数的第一次调用,作为浮点数,基于Win32函数QueryPerformanceCounter()。分辨率通常为优于一微秒。自3.3版起已弃用:此函数的行为取决于在平台上:改用perf_counter()或process_time(),根据您的要求,要有明确的行为。
这里有一个很好的文档记录和完全类型提示的装饰器,我将其用作通用工具:
from functools import wraps
from time import perf_counter
from typing import Any, Callable, Optional, TypeVar, cast
F = TypeVar("F", bound=Callable[..., Any])
def timer(prefix: Optional[str] = None, precision: int = 6) -> Callable[[F], F]:
"""Use as a decorator to time the execution of any function.
Args:
prefix: String to print before the time taken.
Default is the name of the function.
precision: How many decimals to include in the seconds value.
Examples:
>>> @timer()
... def foo(x):
... return x
>>> foo(123)
foo: 0.000...s
123
>>> @timer("Time taken: ", 2)
... def foo(x):
... return x
>>> foo(123)
Time taken: 0.00s
123
"""
def decorator(func: F) -> F:
@wraps(func)
def wrapper(*args: Any, **kwargs: Any) -> Any:
nonlocal prefix
prefix = prefix if prefix is not None else f"{func.__name__}: "
start = perf_counter()
result = func(*args, **kwargs)
end = perf_counter()
print(f"{prefix}{end - start:.{precision}f}s")
return result
return cast(F, wrapper)
return decorator
示例用法:
from timer import timer
@timer(precision=9)
def takes_long(x: int) -> bool:
return x in (i for i in range(x + 1))
result = takes_long(10**8)
print(result)
输出:耗时:4.942629056秒真的
可以通过以下方式检查doctest:
$ python3 -m doctest --verbose -o=ELLIPSIS timer.py
类型提示:
$ mypy timer.py