python 优化利器（二）运用line_profile查看耗时函数中的行耗时_line profile里面用到决策树吗-优快云博客

本文链接：https://blog.youkuaiyun.com/budong282712018/article/details/80281267

安装

pip install line_profiler

如果我从上一遍找到primes这函数最耗时，而想看看里面的代码行耗时，可以在此函数上面添加@profile 符。接着运行

kernprof -l cprofile_test.py
得到一个文件cprofile_test.py.lprof
python -m line_profiler cprofile_test.py.lprof


[root@k8s-node-1 alex]# python -m line_profiler cprofile_test.py.lprof 
Timer unit: 1e-06 s

Total time: 88.2254 s
File: cprofile_test.py
Function: primes at line 10

Line #      Hits         Time  Per Hit   % Time  Line Contents
==============================================================
    10                                           @profile
    11                                           def primes(n):
    12                                           
    13        40         64.0      1.6      0.0      if n == 2:
    14                                           
    15                                                   return [2]
    16                                           
    17        40         34.0      0.8      0.0      elif n < 2:
    18                                           
    19                                                   return []
    20                                           
    21        40         21.0      0.5      0.0      s = []
    22                                           
    23  39999960   18282042.0      0.5     20.7      for i in range(3, n+1):
    24                                           
    25  39999920   14411959.0      0.4     16.3          if i % 2 != 0:
    26                                           
    27  19999960    7159780.0      0.4      8.1              s.append(i)
    28                                           
    29        40        212.0      5.3      0.0      mroot = n ** 0.5
    30                                           
    31        40         82.0      2.0      0.0      half = (n + 1) / 2 - 1
    32                                           
    33        40         24.0      0.6      0.0      i = 0
    34                                           
    35        40         23.0      0.6      0.0      m = 3
    36                                           
    37     20000       7466.0      0.4      0.0      while m <= mroot:
    38                                           
    39     19960       7213.0      0.4      0.0          if s[i]:
    40                                           
    41      6680       2801.0      0.4      0.0              j = (m * m - 3) / 2
    42                                           
    43      6680       3029.0      0.5      0.0              s[j] = 0
    44                                           
    45  32449400   10795395.0      0.3     12.2              while j < half:
    46                                           
    47  32442720   12643545.0      0.4     14.3                  s[j] = 0
    48                                           
    49  32442720   10744559.0      0.3     12.2                  j += m
    50                                           
    51     19960       6664.0      0.3      0.0          i = i + 1
    52                                           
    53     19960       7618.0      0.4      0.0          m = 2 * i + 3
    54                                           
    55        40         53.0      1.3      0.0      l = [2]
    56                                           
    57  20000000    6688249.0      0.3      7.6      for x in s:
    58                                           
    59  19999960    6286709.0      0.3      7.1          if x:
    60                                           
    61   3139880    1177810.0      0.4      1.3              l.append(x)
    62                                           
    63        40         14.0      0.3      0.0      return l