我怎样才能简化“for x in a for y in b for z in c ...”与无序？

Question

#!/usr/bin/python
#
# Description: I try to simplify the implementation of the thing below.
# Sets, such as (a,b,c), with irrelavant order are given. The goal is to
# simplify the messy "assignment", not sure of the term, below.
#
#
# QUESTION: How can you simplify it? 
#
# >>> a=['1','2','3']
# >>> b=['bc','b']
# >>> c=['#']
# >>> print([x+y+z for x in a for y in b for z in c])
# ['1bc#', '1b#', '2bc#', '2b#', '3bc#', '3b#']
#
# The same works with sets as well
# >>> a
# set(['a', 'c', 'b'])
# >>> b
# set(['1', '2'])
# >>> c
# set(['#'])
#
# >>> print([x+y+z for x in a for y in b for z in c])
# ['a1#', 'a2#', 'c1#', 'c2#', 'b1#', 'b2#']


#BROKEN TRIALS
d = [a,b,c]

# TRIAL 2: trying to simplify the "assignments", not sure of the term
# but see the change to the abve 
# print([x+y+z for x, y, z in zip([x,y,z], d)])

# TRIAL 3: simplifying TRIAL 2
# print([x+y+z for x, y, z in zip([x,y,z], [a,b,c])])

[更新]缺少一件事，如果你真的有for x in a for y in b for z in c ...，即任意数量的结构，写作product(a,b,c,...) 很麻烦。假设您有一个列表列表，例如上面示例中的 d。你能简单点吗？ Python 让我们使用 *a 对列表进行解包，并使用 **b 进行字典评估，但这只是表示法。任意长度的嵌套 for 循环和此类怪物的简化超出了 SO，需要进一步研究 此处。我想强调的是，标题中的问题是开放式的，所以如果我接受问题，请不要被误导！

Answer 1

试试这个

>>> import itertools
>>> a=['1','2','3']
>>> b=['bc','b']
>>> c=['#'] 
>>> print [ "".join(res) for res in itertools.product(a,b,c) ]
['1bc#', '1b#', '2bc#', '2b#', '3bc#', '3b#']

Answer 2

>>> from itertools import product
>>> a=['1','2','3']
>>> b=['bc','b']
>>> c=['#']
>>> map("".join, product(a,b,c))
['1bc#', '1b#', '2bc#', '2b#', '3bc#', '3b#']

编辑：

您可以在很多事情上使用产品，就像您也想要的那样

>>> list_of_things = [a,b,c]
>>> map("".join, product(*list_of_things))