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

这是一种很晚的反应,但也许对某人来说是有目的的。这是一种我认为非常干净的方法。

import time

def timed(fun, *args):
    s = time.time()
    r = fun(*args)
    print('{} execution took {} seconds.'.format(fun.__name__, time.time()-s))
    return(r)

timed(print, "Hello")

请记住,“print”是Python 3中的函数,而不是Python 2.7中的函数。但是,它可以与任何其他功能一起使用。干杯

其他回答

print_elapsed_time函数如下

def print_elapsed_time(prefix=''):
    e_time = time.time()
    if not hasattr(print_elapsed_time, 's_time'):
        print_elapsed_time.s_time = e_time
    else:
        print(f'{prefix} elapsed time: {e_time - print_elapsed_time.s_time:.2f} sec')
        print_elapsed_time.s_time = e_time

用这种方式

print_elapsed_time()
.... heavy jobs ...
print_elapsed_time('after heavy jobs')
.... tons of jobs ...
print_elapsed_time('after tons of jobs')

结果是

after heavy jobs elapsed time: 0.39 sec
after tons of jobs elapsed time: 0.60 sec  

这个函数的优点和缺点是你不需要经过开始时间

如果您想方便地对函数计时,可以使用一个简单的修饰符:

import time

def timing_decorator(func):
    def wrapper(*args, **kwargs):
        start = time.perf_counter()
        original_return_val = func(*args, **kwargs)
        end = time.perf_counter()
        print("time elapsed in ", func.__name__, ": ", end - start, sep='')
        return original_return_val

    return wrapper

您可以在您希望计时的函数上使用它,如下所示:

@timing_decorator
def function_to_time():
    time.sleep(1)

function_to_time()

无论何时调用function_to_time,它都会打印所用的时间和正在计时的函数的名称。

测量时间(秒):

from timeit import default_timer as timer
from datetime import timedelta

start = timer()

# ....
# (your code runs here)
# ...

end = timer()
print(timedelta(seconds=end-start))

输出:

0:00:01.946339

虽然问题中没有严格要求,但通常情况下,您需要一种简单、统一的方法来递增地测量几行代码之间的经过时间。

如果您使用的是Python 3.8或更高版本,则可以使用赋值表达式(也称为walrus运算符)以相当优雅的方式实现这一点:

import time

start, times = time.perf_counter(), {}

print("hello")
times["print"] = -start + (start := time.perf_counter())

time.sleep(1.42)
times["sleep"] = -start + (start := time.perf_counter())

a = [n**2 for n in range(10000)]
times["pow"] = -start + (start := time.perf_counter())

print(times)

=>

{'print': 2.193450927734375e-05, 'sleep': 1.4210970401763916, 'power': 0.005671024322509766}

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