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")

只需使用timeit模块。它同时适用于Python 2和Python 3。

import timeit

start = timeit.default_timer()

# All the program statements
stop = timeit.default_timer()
execution_time = stop - start

print("Program Executed in "+str(execution_time)) # It returns time in seconds

它在几秒钟内返回，您可以获得执行时间。这很简单，但您应该在启动程序执行的主函数中编写这些。如果您想获得执行时间，即使在出现错误时，也可以将参数“Start”设置为它，并在那里进行如下计算：

def sample_function(start,**kwargs):
     try:
         # Your statements
     except:
         # except statements run when your statements raise an exception
         stop = timeit.default_timer()
         execution_time = stop - start
         print("Program executed in " + str(execution_time))

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

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))]

我使用来自ttictoc的tic和toc。

pip install ttictoc

然后可以在脚本中使用：

from ttictoc import tic,toc
tic()

# foo()

print(toc())

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。