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))

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 ---

要使用metakermit对Python 2.7的更新答案，您需要单调包。

代码如下：

from datetime import timedelta
from monotonic import monotonic

start_time = monotonic()
end_time = monotonic()
print(timedelta(seconds=end_time - start_time))

有一个timeit模块，可用于对Python代码的执行时间进行计时。

它在Python文档26.6中有详细的文档和示例。timeit-测量小代码片段的执行时间。

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。