我的代码的瓶颈目前是使用ctypes从Python列表转换为C数组,如in this question所述。
一个小实验表明,与其他Python指令相比,它确实非常慢:
1.790962941000089
0.0911122129996329
0.3200237319997541
给出:
perf
我用CPython 3.4.2获得了这些结果。我在CPython 2.7.9和Pypy 2.4.0上得到了类似的时间。
我尝试使用timeit
运行上述代码,评论 Performance counter stats for 'python3 perf.py':
1807,891637 task-clock (msec) # 1,000 CPUs utilized
8 context-switches # 0,004 K/sec
0 cpu-migrations # 0,000 K/sec
59 523 page-faults # 0,033 M/sec
5 755 704 178 cycles # 3,184 GHz
13 552 506 138 instructions # 2,35 insn per cycle
3 217 289 822 branches # 1779,581 M/sec
748 614 branch-misses # 0,02% of all branches
1,808349671 seconds time elapsed
指令,一次只运行一个。我得到了这些结果:
ctypes的
Performance counter stats for 'python3 perf.py':
144,678718 task-clock (msec) # 0,998 CPUs utilized
0 context-switches # 0,000 K/sec
0 cpu-migrations # 0,000 K/sec
12 913 page-faults # 0,089 M/sec
458 284 661 cycles # 3,168 GHz
1 253 747 066 instructions # 2,74 insn per cycle
325 528 639 branches # 2250,011 M/sec
708 280 branch-misses # 0,22% of all branches
0,144966969 seconds time elapsed
阵列
Performance counter stats for 'python3 perf.py':
369,786395 task-clock (msec) # 0,999 CPUs utilized
0 context-switches # 0,000 K/sec
0 cpu-migrations # 0,000 K/sec
108 584 page-faults # 0,294 M/sec
1 175 946 161 cycles # 3,180 GHz
2 086 554 968 instructions # 1,77 insn per cycle
422 531 402 branches # 1142,636 M/sec
768 338 branch-misses # 0,18% of all branches
0,370103043 seconds time elapsed
设置
ctypes
set
的代码具有比具有 +------+
| |
| A1 | Class A1
| |
+--+---+ Class A2 - child
^ Class A3 - child
|
| A2->B1 (Class B1)
+--------+--------+ A3->B2 (Class B2)
| |
| |
+-----+ +--+---+ +--+--+ +-----+
| | | | | | | |
| B1 +---+ A2 | | A3 +---+ B2 |
| | | | | | | |
+-----+ +------+ +-----+ +-----+
的代码更少的页面错误,并且具有与其他两个相同的分支未命中数。我唯一看到的是有更多的指令和分支(但我仍然不知道为什么)和更多的上下文切换(但它肯定是更长的运行时间而不是原因的结果)。
因此我有两个问题:
答案 0 :(得分:5)
虽然这不是一个明确的答案,但问题似乎是*t
的构造函数调用。相反,执行以下操作可显着降低开销:
array = (ctypes.c_uint32 * len(t))()
array[:] = t
测试:
import timeit
setup="from array import array; import ctypes; t = [i for i in range(1000000)];"
print(timeit.timeit(stmt='(ctypes.c_uint32 * len(t))(*t)',setup=setup,number=10))
print(timeit.timeit(stmt='a = (ctypes.c_uint32 * len(t))(); a[:] = t',setup=setup,number=10))
print(timeit.timeit(stmt='array("I",t)',setup=setup,number=10))
print(timeit.timeit(stmt='set(t)',setup=setup,number=10))
输出:
1.7090932869978133
0.3084979929990368
0.08278547400186653
0.2775516299989249
答案 1 :(得分:4)
解决方案是使用array
模块并转换地址或使用from_buffer方法......
import timeit
setup="from array import array; import ctypes; t = [i for i in range(1000000)];"
print(timeit.timeit(stmt="v = array('I',t);assert v.itemsize == 4; addr, count = v.buffer_info();p = ctypes.cast(addr,ctypes.POINTER(ctypes.c_uint32))",setup=setup,number=10))
print(timeit.timeit(stmt="v = array('I',t);a = (ctypes.c_uint32 * len(v)).from_buffer(v)",setup=setup,number=10))
print(timeit.timeit(stmt='(ctypes.c_uint32 * len(t))(*t)',setup=setup,number=10))
print(timeit.timeit(stmt='set(t)',setup=setup,number=10))
使用Python 3时速度提高了许多倍:
$ python3 convert.py
0.08303386811167002
0.08139665238559246
1.5630637975409627
0.3013848252594471