我有一个Python命令行程序,需要一段时间才能完成。我想知道完成跑步所需的确切时间。
我看过timeit模块,但它似乎只适用于小代码片段。我想给整个节目计时。
我有一个Python命令行程序,需要一段时间才能完成。我想知道完成跑步所需的确切时间。
我看过timeit模块,但它似乎只适用于小代码片段。我想给整个节目计时。
当前回答
我喜欢datetime模块提供的输出,其中时间增量对象以人类可读的方式显示天、小时、分钟等。
例如:
from datetime import datetime
start_time = datetime.now()
# do your work here
end_time = datetime.now()
print('Duration: {}'.format(end_time - start_time))
样本输出,例如。
Duration: 0:00:08.309267
or
Duration: 1 day, 1:51:24.269711
正如J.F.Sebastian所提到的,这种方法在当地时间可能会遇到一些棘手的情况,因此使用更安全:
import time
from datetime import timedelta
start_time = time.monotonic()
end_time = time.monotonic()
print(timedelta(seconds=end_time - start_time))
其他回答
我在很多地方都遇到了同样的问题,所以我创建了一个方便的套装占星术。你可以用pip安装钟表,然后以优雅的方式安装:
from horology import Timing
with Timing(name='Important calculations: '):
prepare()
do_your_stuff()
finish_sth()
将输出:
Important calculations: 12.43 ms
或者更简单(如果你有一个功能):
from horology import timed
@timed
def main():
...
将输出:
main: 7.12 h
它负责单位和舍入。它适用于python 3.6或更高版本。
from time import time
start_time = time()
...
end_time = time()
time_taken = end_time - start_time # time_taken is in seconds
hours, rest = divmod(time_taken,3600)
minutes, seconds = divmod(rest, 60)
我很喜欢保罗·麦奎尔的答案,但我使用的是Python 3。因此,对于感兴趣的人来说:这里是他在*nix上使用Python 3的答案的修改(我想,在Windows下,应该使用clock()而不是time()):
#python3
import atexit
from time import time, strftime, localtime
from datetime import timedelta
def secondsToStr(elapsed=None):
if elapsed is None:
return strftime("%Y-%m-%d %H:%M:%S", localtime())
else:
return str(timedelta(seconds=elapsed))
def log(s, elapsed=None):
line = "="*40
print(line)
print(secondsToStr(), '-', s)
if elapsed:
print("Elapsed time:", elapsed)
print(line)
print()
def endlog():
end = time()
elapsed = end-start
log("End Program", secondsToStr(elapsed))
start = time()
atexit.register(endlog)
log("Start Program")
如果你觉得这很有用,你仍然应该投票给他的答案,而不是这一个,因为他做了大部分工作;)。
我在查找两种不同方法的运行时间时遇到的问题,这两种方法用于查找所有<=一个数的素数。当在程序中进行用户输入时。
错误的方法
#Sample input for a number 20
#Sample output [2, 3, 5, 7, 11, 13, 17, 19]
#Total Running time = 0.634 seconds
import time
start_time = time.time()
#Method 1 to find all the prime numbers <= a Number
# Function to check whether a number is prime or not.
def prime_no(num):
if num<2:
return False
else:
for i in range(2, num//2+1):
if num % i == 0:
return False
return True
#To print all the values <= n
def Prime_under_num(n):
a = [2]
if n <2:
print("None")
elif n==2:
print(2)
else:
"Neglecting all even numbers as even numbers won't be prime in order to reduce the time complexity."
for i in range(3, n+1, 2):
if prime_no(i):
a.append(i)
print(a)
"When Method 1 is only used outputs of running time for different inputs"
#Total Running time = 2.73761 seconds #n = 100
#Total Running time = 3.14781 seconds #n = 1000
#Total Running time = 8.69278 seconds #n = 10000
#Total Running time = 18.73701 seconds #n = 100000
#Method 2 to find all the prime numbers <= a Number
def Prime_under_num(n):
a = [2]
if n <2:
print("None")
elif n==2:
print(2)
else:
for i in range(3, n+1, 2):
if n%i ==0:
pass
else:
a.append(i)
print(a)
"When Method 2 is only used outputs of running time for different inputs"
# Total Running time = 2.75935 seconds #n = 100
# Total Running time = 2.86332 seconds #n = 1000
# Total Running time = 4.59884 seconds #n = 10000
# Total Running time = 8.55057 seconds #n = 100000
if __name__ == "__main__" :
n = int(input())
Prime_under_num(n)
print("Total Running time = {:.5f} seconds".format(time.time() - start_time))
上述所有情况下获得的不同运行时间都是错误的。对于我们正在接受输入的问题,我们必须在接受输入后才开始计时。这里,用户键入输入所花费的时间也与运行时间一起计算。
正确的方法
我们必须从开头删除start_time=time.time()并将其添加到主块中。
if __name__ == "__main__" :
n = int(input())
start_time = time.time()
Prime_under_num(n)
print("Total Running time = {:.3f} seconds".format(time.time() - start_time))
因此,两种方法单独使用时的输出如下:-
# Method 1
# Total Running time = 0.00159 seconds #n = 100
# Total Running time = 0.00506 seconds #n = 1000
# Total Running time = 0.22987 seconds #n = 10000
# Total Running time = 18.55819 seconds #n = 100000
# Method 2
# Total Running time = 0.00011 seconds #n = 100
# Total Running time = 0.00118 seconds #n = 1000
# Total Running time = 0.00302 seconds #n = 10000
# Total Running time = 0.01450 seconds #n = 100000
现在我们可以看到,与错误方法相比,总运行时间有显著差异。即使方法2在两种方法中的性能优于方法1,但第一种方法(错误方法)是错误的。
使用line_profiler。
line_profiler将描述单个代码行执行所需的时间。分析器通过Cython在C语言中实现,以减少分析开销。
from line_profiler import LineProfiler
import random
def do_stuff(numbers):
s = sum(numbers)
l = [numbers[i]/43 for i in range(len(numbers))]
m = ['hello'+str(numbers[i]) for i in range(len(numbers))]
numbers = [random.randint(1,100) for i in range(1000)]
lp = LineProfiler()
lp_wrapper = lp(do_stuff)
lp_wrapper(numbers)
lp.print_stats()
结果将是:
Timer unit: 1e-06 s
Total time: 0.000649 s
File: <ipython-input-2-2e060b054fea>
Function: do_stuff at line 4
Line # Hits Time Per Hit % Time Line Contents
==============================================================
4 def do_stuff(numbers):
5 1 10 10.0 1.5 s = sum(numbers)
6 1 186 186.0 28.7 l = [numbers[i]/43 for i in range(len(numbers))]
7 1 453 453.0 69.8 m = ['hello'+str(numbers[i]) for i in range(len(numbers))]