与@rogeriopvl的响应类似，我添加了一个轻微的修改，使用相同的库将长时间运行的作业转换为小时-分钟-秒。

import time
start_time = time.time()
main()
seconds = time.time() - start_time
print('Time Taken:', time.strftime("%H:%M:%S",time.gmtime(seconds)))

样本输出

Time Taken: 00:00:08

首先，以管理员身份打开命令提示符（CMD）并在那里键入，安装人性化的软件包-pip安装人性化

代码：

from humanfriendly import format_timespan
import time
begin_time = time.time()
# Put your code here
end_time = time.time() - begin_time
print("Total execution time: ", format_timespan(end_time))

输出：

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；）。

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))

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。

我将这个timing.py模块放入我自己的站点包目录中，并在模块顶部插入导入计时：

import atexit
from time import clock

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
    print

def endlog():
    end = clock()
    elapsed = end-start
    log("End Program", secondsToStr(elapsed))

def now():
    return secondsToStr(clock())

start = clock()
atexit.register(endlog)
log("Start Program")

如果程序中有重要的阶段，我也可以在程序中调用timing.log。但仅包括导入计时就可以打印开始和结束时间，以及总运行时间。（请原谅我晦涩难懂的secondsToStr函数，它只是将浮点秒数设置为hh:mm:ss.sss格式。）

注意：上述代码的Python3版本可以在这里找到。