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。

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

我在很多地方都遇到了同样的问题，所以我创建了一个方便的套装占星术。你可以用pip安装钟表，然后以优雅的方式安装：

from horology import Timing

with Timing(name='Important calculations: '):
    prepare()
    do_your_stuff()
    finish_sth()

将输出：

Important calculations: 12.43 ms

或者更简单（如果你有一个功能）：

from horology import timed

@timed
def main():
    ...

将输出：

main: 7.12 h

它负责单位和舍入。它适用于python 3.6或更高版本。

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")

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；）。