如何将2个短列表中的元素追加到另一个大元组列表中？

Question

Say I have the following lists:

l1 = [('a','b','c'),('d','e','f'),('g','h','i'),('j','k','l')]
l2 = ['x','y','z']
l3 = ['m','n']

I want to extract elements from l2 and l3, then append l2[i]( i in range(len(l2)) ) as the first element inside every tuple, and l3[i]( i in range(len(l2)) ) as the last element inside every tuple.

so the result will look like the followings:

l1 = [('x','a','b','c','m'),('x','a','b','c','n'),('y','a','b','c','m'),('y','a','b','c','n'), ('z','a','b','c','m'),('z','a','b','c','n')]

and yes, the len of l1 will be increased.

Answer 1

You can do that with the help of itertools.chain.from_iterable and itertools.product , and get the cartesian product, like this

>>> from itertools import chain, product
>>> from pprint import pprint
>>> pprint([tuple(chain.from_iterable(i)) for i in product(l2, [l1[0]], l3)])
[('x', 'a', 'b', 'c', 'm'),
 ('x', 'a', 'b', 'c', 'n'),
 ('y', 'a', 'b', 'c', 'm'),
 ('y', 'a', 'b', 'c', 'n'),
 ('z', 'a', 'b', 'c', 'm'),
 ('z', 'a', 'b', 'c', 'n')]

You are finding the cartesian product between l2 , the first element of l1 and l3 . Since the result will be a tuple with the element from l2 (a string) and the first element of l1 (a tuple) and an element from l3 (a string), we flatten it with chain.from_iterable .

Let's say we don't flatten the tuples, then this is what you will get

>>> pprint([tuple(items) for items in product(l2, [l1[0]], l3)])
[('x', ('a', 'b', 'c'), 'm'),
 ('x', ('a', 'b', 'c'), 'n'),
 ('y', ('a', 'b', 'c'), 'm'),
 ('y', ('a', 'b', 'c'), 'n'),
 ('z', ('a', 'b', 'c'), 'm'),
 ('z', ('a', 'b', 'c'), 'n')]

This is why we use chain.from_iterable and flatten the tuples.

Answer 2

What about playing with zip and list comprehension :

>>> [zip(*i) for i in zip(zip(l2,l2),zip(l1,l1),(l3 for _ in range(2*len(l1))))]
[[('x', ('a', 'b', 'c'), 'm'), ('x', ('a', 'b', 'c'), 'n')], [('y', ('d', 'e', 'f'), 'm'), ('y', ('d', 'e', 'f'), 'n')], [('z', ('g', 'h', 'i'), 'm'), ('z', ('g', 'h', 'i'), 'n')]]