Python 我怎样才能简化；对于a中的x表示y表示b中的y表示c中的z……”；无序的？

[更新]缺少一件事，如果你真的有

为x在A中为y在b中为z在c中…

，也就是说，任意数量的结构，编写

产品（A，b，c…）是很麻烦的。假设您有一个列表列表，如上例中的
d
。你能简单点吗？Python让我们用
*a
对列表进行
解包
，用
**b
对字典进行评估，但这只是表示法。嵌套的任意长度的循环和简化这样的怪物是超越如此，为进一步的研究。我想强调，标题中的问题是开放的，所以如果我接受一个问题，请不要被误导
 试试这个

#!/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])])

编辑：

你可以在很多事情上使用这个产品，就像你想做的一样

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

@HH，我在我的回答中添加了这样的内容：
产品（*d）
等同于
产品（a、b、c）
>>> from itertools import product
>>> a=['1','2','3']
>>> b=['bc','b']
>>> c=['#']
>>> map("".join, product(a,b,c))
['1bc#', '1b#', '2bc#', '2b#', '3bc#', '3b#']

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