python - 如何以 Pythonic 方式将多个列表分组为单个列表？

Question

我有一个从列表列表中返回列表的函数，其中返回列表按索引号对每个列表的成员进行分组。代码和示例:

def listjoinervar(*lists: list) -> list:
    """returns list of grouped values from each list 
        keyword arguments:
        lists: list of input lists
    """ 
    assert(len(lists) > 0) and (all(len(i) == len(lists[0]) for i in lists))
    joinedlist = [None] * len(lists) * len(lists[0])
    for i in range(0, len(joinedlist), len(lists)):
        for j in range(0, len(lists[0])):
            joinedlist[i//len(lists[0]) + j*len(lists[0])] = lists[i//len(lists[0])][j]
    return joinedlist

a = ['a', 'b', 'c']
b = [1, 2, 3]
c = [True, False, False]
listjoinervar(a, b, c)
# ['a', 1, True, 'b', 2, False, 'c', 3, False]

有没有办法使用 itertools、generators 等使它更像 Pythonic？我看过像 this 这样的例子但在我的代码中，没有交互 b/w 各个列表的元素。谢谢

Answer 1

使用itertools.chain.from_iterable + zip :

from itertools import chain

def listjoinervar(*a):
    return list(chain.from_iterable(zip(*a)))

用法:

>>> a = ['a', 'b', 'c']
>>> b = [1, 2, 3]
>>> c = [True, False, False]
>>> listjoinervar(a, b, c)
['a', 1, True, 'b', 2, False, 'c', 3, False]