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。

Python程序执行度量的时间可能不一致，具体取决于：

可以使用不同的算法评估相同的程序运行时间因算法而异运行时间因实现而异运行时间因计算机而异基于小输入，运行时间不可预测

这是因为最有效的方法是使用“增长顺序”，并学习大“O”符号来正确地执行。

无论如何，您可以尝试使用以下简单算法来评估任何Python程序在每秒特定机器计数步骤中的性能：使其适应您想要评估的计划

import time

now = time.time()
future = now + 10
step = 4 # Why 4 steps? Because until here already four operations executed
while time.time() < future:
    step += 3 # Why 3 again? Because a while loop executes one comparison and one plus equal statement
step += 4 # Why 3 more? Because one comparison starting while when time is over plus the final assignment of step + 1 and print statement
print(str(int(step / 10)) + " steps per second")

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；）。

我尝试使用以下脚本找到时间差。

import time

start_time = time.perf_counter()
[main code here]
print (time.perf_counter() - start_time, "seconds")

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。