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
作为lambda,获取经过的时间和时间戳:
import datetime
t_set = lambda: datetime.datetime.now().astimezone().replace(microsecond=0)
t_diff = lambda t: str(t_set() - t)
t_stamp = lambda t=None: str(t) if t else str(t_set())
在实践中:
>>>
>>> t_set()
datetime.datetime(2021, 3, 21, 1, 25, 17, tzinfo=datetime.timezone(datetime.timedelta(days=-1, seconds=61200), 'PDT'))
>>> t = t_set()
>>> t_diff(t)
'0:00:14'
>>> t_diff(t)
'0:00:23'
>>> t_stamp()
'2021-03-21 01:25:57-07:00'
>>> t_stamp(t)
'2021-03-21 01:25:22-07:00'
>>>
其他回答
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
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)
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
使用探查器模块。它提供了非常详细的概况。
import profile
profile.run('main()')
它输出类似于:
5 function calls in 0.047 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 0.000 0.000 :0(exec)
1 0.047 0.047 0.047 0.047 :0(setprofile)
1 0.000 0.000 0.000 0.000 <string>:1(<module>)
0 0.000 0.000 profile:0(profiler)
1 0.000 0.000 0.047 0.047 profile:0(main())
1 0.000 0.000 0.000 0.000 two_sum.py:2(twoSum)
我发现它很有启发性。
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