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

根据这个答案,创建了一个简单但方便的工具。

import time
from datetime import timedelta

def start_time_measure(message=None):
    if message:
        print(message)
    return time.monotonic()

def end_time_measure(start_time, print_prefix=None):
    end_time = time.monotonic()
    if print_prefix:
        print(print_prefix + str(timedelta(seconds=end_time - start_time)))
    return end_time

用法:

total_start_time = start_time_measure()    
start_time = start_time_measure('Doing something...')
# Do something
end_time_measure(start_time, 'Done in: ')
start_time = start_time_measure('Doing something else...')
# Do something else
end_time_measure(start_time, 'Done in: ')
end_time_measure(total_start_time, 'Total time: ')

输出:

Doing something...
Done in: 0:00:01.218000
Doing something else...
Done in: 0:00:01.313000
Total time: 0:00:02.672000

其他回答

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。

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}

我也喜欢Paul McGuire的回答,并提出了一个更符合我需求的上下文管理器表单。

import datetime as dt
import timeit

class TimingManager(object):
    """Context Manager used with the statement 'with' to time some execution.

    Example:

    with TimingManager() as t:
       # Code to time
    """

    clock = timeit.default_timer

    def __enter__(self):
        """
        """
        self.start = self.clock()
        self.log('\n=> Start Timing: {}')

        return self

    def __exit__(self, exc_type, exc_val, exc_tb):
        """
        """
        self.endlog()

        return False

    def log(self, s, elapsed=None):
        """Log current time and elapsed time if present.
        :param s: Text to display, use '{}' to format the text with
            the current time.
        :param elapsed: Elapsed time to display. Dafault: None, no display.
        """
        print s.format(self._secondsToStr(self.clock()))

        if(elapsed is not None):
            print 'Elapsed time: {}\n'.format(elapsed)

    def endlog(self):
        """Log time for the end of execution with elapsed time.
        """
        self.log('=> End Timing: {}', self.now())

    def now(self):
        """Return current elapsed time as hh:mm:ss string.
        :return: String.
        """
        return str(dt.timedelta(seconds = self.clock() - self.start))

    def _secondsToStr(self, sec):
        """Convert timestamp to h:mm:ss string.
        :param sec: Timestamp.
        """
        return str(dt.datetime.fromtimestamp(sec))

在Linux或Unix中:

$ time python yourprogram.py

在Windows中,请参阅StackOverflow问题:如何在Windows命令行上测量命令的执行时间?

对于更详细的输出,

$ time -v python yourprogram.py
    Command being timed: "python3 yourprogram.py"
    User time (seconds): 0.08
    System time (seconds): 0.02
    Percent of CPU this job got: 98%
    Elapsed (wall clock) time (h:mm:ss or m:ss): 0:00.10
    Average shared text size (kbytes): 0
    Average unshared data size (kbytes): 0
    Average stack size (kbytes): 0
    Average total size (kbytes): 0
    Maximum resident set size (kbytes): 9480
    Average resident set size (kbytes): 0
    Major (requiring I/O) page faults: 0
    Minor (reclaiming a frame) page faults: 1114
    Voluntary context switches: 0
    Involuntary context switches: 22
    Swaps: 0
    File system inputs: 0
    File system outputs: 0
    Socket messages sent: 0
    Socket messages received: 0
    Signals delivered: 0
    Page size (bytes): 4096
    Exit status: 0
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)