Python的itertools - 慢？

Question

我正在嘗試使用Python的itertools模塊來加速三重嵌套for循環。下面的測試碼進行比較以itertools'產品的方法，並輸出一個標準的三重嵌套循環：

嵌套循環時間= 2.35023秒

Itertools循環時間= 2.67766秒

我缺少的東西？

import numpy 
import itertools 
import time 

n = 128 
a = numpy.arange(n**3).reshape((n,n,n)) 
b = numpy.zeros((n,n,n)) 
c = numpy.zeros((n,n,n)) 

t = time.time() 
for i in range(n): 
    for j in range(n): 
     for k in range(n): 
      b[i,j,k] = a[i,j,k] 
print 'Nested loop time = %g secs' % (time.time() - t) 

t = time.time() 
for (i,j,k) in itertools.product(range(n), repeat=3): 
    c[i,j,k] = a[i,j,k] 
print 'Itertools loop time = %g secs' % (time.time() - t) 

Answer 1

它似乎像itertools.product是的n大值慢：

 
In [24]: print _23 
from itertools import product 

def nested_loops(n): 
    for i in range(n): 
     for j in range(n): 
      for k in range(n): 
       pass 

def itertools_product(n): 
    for (i,j,k) in product(range(n), repeat=3): 
     pass 


In [25]: %timeit nested_loops(128) 
10 loops, best of 3: 68.6 ms per loop 

In [26]: %timeit itertools_product(128) 
10 loops, best of 3: 162 ms per loop 

In [27]: %timeit nested_loops(10) 
10000 loops, best of 3: 84.5 us per loop 

In [28]: %timeit itertools_product(10) 
10000 loops, best of 3: 79.8 us per loop 

In [30]: %timeit nested_loops(300) 
1 loops, best of 3: 833 ms per loop 

In [31]: %timeit itertools_product(300) 
1 loops, best of 3: 2.07 s per loop 

而且沒有元組拆包：

 
In [40]: print _39 
from itertools import product 

def itertools_product(n): 
    for ijk in product(range(n), repeat=3): 
     pass 

In [41]: %timeit itertools_product(128) 
10 loops, best of 3: 115 ms per loop 

In [42]: %timeit itertools_product(10) 
10000 loops, best of 3: 59.2 us per loop 

In [43]: %timeit itertools_product(300) 
1 loops, best of 3: 1.47 s per loop 

此外，爲了好玩，列表內涵和發電機表達式：

 
def list_comprehension_product(n): 
    range_n = range(n) 
    for (i,j,k) in [ (i, j, k) for i in range_n for j in range_n for k in range_n ]: 
     pass 

def generator_expression_product(n): 
    range_n = range(n) 
    for (i,j,k) in ((i, j, k) for i in range_n for j in range_n for k in range_n): 
     pass 

In [51]: %timeit list_comprehension_product(128) 
1 loops, best of 3: 583 ms per loop 

In [52]: %timeit generator_expression_product(128) 
1 loops, best of 3: 480 ms per loop 

個

這些基準測試與python --version運行：

2.6.7 (r267:88850, Jul 31 2011, 19:30:54) 
[GCC 4.2.1 (Based on Apple Inc. build 5658) (LLVM build 2335.15.00)]

Answer 2

它似乎是第二個循環是也許是因爲元組拆包比第一速度較慢。你沒有做到這一點，我覺得它使第二循環更快地做到這一點是這樣的：

for ijk in itertools.product(range(n), repeat=3): 
    c[ijk] = a[ijk] 

當然，與numpy的，你要避免遍歷所有的元素，而是使用立即對整個陣列進行numpy操作。這樣，所有的循環等，都在C中完成，而且你會得到巨大的加速。