如何迭代使用itertools在python中具有不同长度的迭代？

Question

我有不同长度的两个iterables如下

range(5) 
    numpy.arange(0,0.3,0.1) 

我想有对如下

(0,0.) 
    (1,0.) 
    (2,0.) 
    (3,0.) 
    (4,0.) 

    (0,0.1) 
    (1,0.1) 
    (2,0.1) 
    (3,0.1) 
    (4,0.1) 

    (0,0.2) 
    (1,0.2) 
    (2,0.2) 
    (3,0.2) 
    (4,0.2) 

如何一个可以用itertools做呢？

Answer 1

一般为itertools.product工作：

>>> from itertools import product 
>>> for x in product(range(5), numpy.arange(0, 0.3, 0.1)): 
    print x 
...  
(0, 0.0) 
(0, 0.10000000000000001) 
(0, 0.20000000000000001) 
(1, 0.0) 
(1, 0.10000000000000001) 
(1, 0.20000000000000001) 
(2, 0.0) 
(2, 0.10000000000000001) 
(2, 0.20000000000000001) 
(3, 0.0) 
(3, 0.10000000000000001) 
(3, 0.20000000000000001) 
(4, 0.0) 
(4, 0.10000000000000001) 
(4, 0.20000000000000001) 

既然你想要的“其他”命令，你可以使用一个理解：

>>> [(x,y) for y in numpy.arange(0, 0.3, 0.1) for x in range(5)] 

[(0, 0.0), 
(1, 0.0), 
(2, 0.0), 
(3, 0.0), 
(4, 0.0), 
(0, 0.10000000000000001), 
(1, 0.10000000000000001), 
(2, 0.10000000000000001), 
(3, 0.10000000000000001), 
(4, 0.10000000000000001), 
(0, 0.20000000000000001), 
(1, 0.20000000000000001), 
(2, 0.20000000000000001), 
(3, 0.20000000000000001), 
(4, 0.20000000000000001)] 

或者你可以逆转的参数，然后反向每个元组itertools.product吐出（他们总是循环最快的最右边的元素）。

>>> [x[::-1] for x in product(numpy.arange(0, 0.3, 0.1), range(5))] 

[(0, 0.0), 
(1, 0.0), 
(2, 0.0), 
(3, 0.0), 
(4, 0.0), 
(0, 0.10000000000000001), 
(1, 0.10000000000000001), 
(2, 0.10000000000000001), 
(3, 0.10000000000000001), 
(4, 0.10000000000000001), 
(0, 0.20000000000000001), 
(1, 0.20000000000000001), 
(2, 0.20000000000000001), 
(3, 0.20000000000000001), 
(4, 0.20000000000000001)]