```
#!/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])])
```

**[Update]** A thing missing, what about if you really have `for x in a for y in b for z in c ...`

, i.e. arbirtary amount of structures, writing `product(a,b,c,...)`

is cumbersome. Suppose you have a list of lists such as the `d`

in the above example. Can you get it simpler? Python let us do `unpacking`

with `*a`

for lists and the dictionary evaluation with `**b`

but it is just the notation. Nested for-loops of arbitrary-length and simplification of such monsters is beyond SO, for further research here. I want to stress that the problem in the title is open-ended, so do not be misguided if I accept a question!