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
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。
其他回答
下面是一个返回“hh:mm:ss”字符串的小型计时器类:
class Timer:
def __init__(self):
self.start = time.time()
def restart(self):
self.start = time.time()
def get_time_hhmmss(self):
end = time.time()
m, s = divmod(end - self.start, 60)
h, m = divmod(m, 60)
time_str = "%02d:%02d:%02d" % (h, m, s)
return time_str
用法:
# Start timer
my_timer = Timer()
# ... do something
# Get time string:
time_hhmmss = my_timer.get_time_hhmmss()
print("Time elapsed: %s" % time_hhmmss )
# ... use the timer again
my_timer.restart()
# ... do something
# Get time:
time_hhmmss = my_timer.get_time_hhmmss()
# ... etc
我参加聚会已经很晚了,但这种方法以前没有涉及过。当我们想要手动对某段代码进行基准测试时,我们可能需要首先找出哪些类方法占用了执行时间,这有时并不明显。我构建了以下元类来解决这个问题:
from __future__ import annotations
from functools import wraps
from time import time
from typing import Any, Callable, TypeVar, cast
F = TypeVar('F', bound=Callable[..., Any])
def timed_method(func: F, prefix: str | None = None) -> F:
prefix = (prefix + ' ') if prefix else ''
@wraps(func)
def inner(*args, **kwargs): # type: ignore
start = time()
try:
ret = func(*args, **kwargs)
except BaseException:
print(f'[ERROR] {prefix}{func.__qualname__}: {time() - start}')
raise
print(f'{prefix}{func.__qualname__}: {time() - start}')
return ret
return cast(F, inner)
class TimedClass(type):
def __new__(
cls: type[TimedClass],
name: str,
bases: tuple[type[type], ...],
attrs: dict[str, Any],
**kwargs: Any,
) -> TimedClass:
for name, attr in attrs.items():
if isinstance(attr, (classmethod, staticmethod)):
attrs[name] = type(attr)(timed_method(attr.__func__))
elif isinstance(attr, property):
attrs[name] = property(
timed_method(attr.fget, 'get') if attr.fget is not None else None,
timed_method(attr.fset, 'set') if attr.fset is not None else None,
timed_method(attr.fdel, 'del') if attr.fdel is not None else None,
)
elif callable(attr):
attrs[name] = timed_method(attr)
return super().__new__(cls, name, bases, attrs)
它允许如下使用:
class MyClass(metaclass=TimedClass):
def foo(self):
print('foo')
@classmethod
def bar(cls):
print('bar')
@staticmethod
def baz():
print('baz')
@property
def prop(self):
print('prop')
@prop.setter
def prop(self, v):
print('fset')
@prop.deleter
def prop(self):
print('fdel')
c = MyClass()
c.foo()
c.bar()
c.baz()
c.prop
c.prop = 2
del c.prop
MyClass.bar()
MyClass.baz()
它打印:
foo
MyClass.foo: 1.621246337890625e-05
bar
MyClass.bar: 4.5299530029296875e-06
baz
MyClass.baz: 4.291534423828125e-06
prop
get MyClass.prop: 3.814697265625e-06
fset
set MyClass.prop: 3.5762786865234375e-06
fdel
del MyClass.prop: 3.5762786865234375e-06
bar
MyClass.bar: 3.814697265625e-06
baz
MyClass.baz: 4.0531158447265625e-06
它可以与其他答案相结合,以更精确的方式代替time.time。
使用一个上下文管理器可以很有趣地做到这一点,它可以自动记住进入with块时的开始时间,然后在块退出时冻结结束时间。通过一些小技巧,您甚至可以从同一个上下文管理器函数获得块内的运行时间计数。
核心库没有这个(但可能应该有)。一旦就位,您可以执行以下操作:
with elapsed_timer() as elapsed:
# some lengthy code
print( "midpoint at %.2f seconds" % elapsed() ) # time so far
# other lengthy code
print( "all done at %.2f seconds" % elapsed() )
以下是足以完成此任务的contextmanager代码:
from contextlib import contextmanager
from timeit import default_timer
@contextmanager
def elapsed_timer():
start = default_timer()
elapser = lambda: default_timer() - start
yield lambda: elapser()
end = default_timer()
elapser = lambda: end-start
以及一些可运行的演示代码:
import time
with elapsed_timer() as elapsed:
time.sleep(1)
print(elapsed())
time.sleep(2)
print(elapsed())
time.sleep(3)
注意,通过设计此函数,elapsed()的返回值在块退出时被冻结,并且进一步的调用返回相同的持续时间(在这个玩具示例中大约为6秒)。
import time
def getElapsedTime(startTime, units):
elapsedInSeconds = time.time() - startTime
if units == 'sec':
return elapsedInSeconds
if units == 'min':
return elapsedInSeconds/60
if units == 'hour':
return elapsedInSeconds/(60*60)
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