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
Python中最简单的方法:
import time
start_time = time.time()
main()
print("--- %s seconds ---" % (time.time() - start_time))
这假设程序运行至少需要十分之一秒。
打印:
--- 0.764891862869 seconds ---
其他回答
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))]
您可以使用Python分析器cProfile来测量CPU时间,以及每个函数内部花费的时间以及每个函数被调用的次数。如果您想在不知道从哪里开始的情况下提高脚本的性能,这非常有用。对另一个堆栈溢出问题的回答很好。查看文档总是很好的。
以下是如何从命令行使用cProfile评测脚本的示例:
$ python -m cProfile euler048.py
1007 function calls in 0.061 CPU seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 0.061 0.061 <string>:1(<module>)
1000 0.051 0.000 0.051 0.000 euler048.py:2(<lambda>)
1 0.005 0.005 0.061 0.061 euler048.py:2(<module>)
1 0.000 0.000 0.061 0.061 {execfile}
1 0.002 0.002 0.053 0.053 {map}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler objects}
1 0.000 0.000 0.000 0.000 {range}
1 0.003 0.003 0.003 0.003 {sum}
Timeit是Python中的一个类,用于计算小代码块的执行时间。
Default_timer是此类中的一个方法,用于测量墙上时钟计时,而不是CPU执行时间。因此,其他进程执行可能会对此产生干扰。因此,它对小代码块很有用。
代码示例如下:
from timeit import default_timer as timer
start= timer()
# Some logic
end = timer()
print("Time taken:", end-start)
这是保罗·麦奎尔的回答,对我来说很有用。以防有人在运行这个问题时遇到问题。
import atexit
from time import clock
def reduce(function, iterable, initializer=None):
it = iter(iterable)
if initializer is None:
value = next(it)
else:
value = initializer
for element in it:
value = function(value, element)
return value
def secondsToStr(t):
return "%d:%02d:%02d.%03d" % \
reduce(lambda ll,b : divmod(ll[0],b) + ll[1:],
[(t*1000,),1000,60,60])
line = "="*40
def log(s, elapsed=None):
print (line)
print (secondsToStr(clock()), '-', s)
if elapsed:
print ("Elapsed time:", elapsed)
print (line)
def endlog():
end = clock()
elapsed = end-start
log("End Program", secondsToStr(elapsed))
def now():
return secondsToStr(clock())
def main():
start = clock()
atexit.register(endlog)
log("Start Program")
导入文件后,从程序中调用timing.main()。
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。