交错元组

Question

定义一个函数 interleaved_tuple_two 接受两个参数，tuple_a 和 tuple_b return 是一个交错值的元组，它交错 tuple_b 的一个值对于 tuple_a 的每两个值。假设 tuple_a 总是 tuple_b 的两倍。

例如：

tuple_a = (1,3,5,7,9,11,13,15)
tuple_b = (2,4,6,8)

interleaved_tuple_two(tuple_a, tuple_b) 将 return:

(1, 3, 2, 5, 7, 4, 9, 11, 6, 13, 15, 8)

目前，我有这个，但它不正确。不太确定如何对此进行改进

def interleaved_tuple_two(tuple_a, tuple_b):
    return [val for pair in zip(tuple_a, tuple_b) for val in pair]

Answer 1

您可以对三个列表使用 zip，前两个 tuple_a 一次跳过一个值但起始位置不同，然后第三个是 tuple_b.

tuple_a = (1, 3, 5, 7, 9, 11, 13, 15)
tuple_b = (2, 4, 6, 8)

tuple_c = tuple(
    v
    for triple in zip(tuple_a[::2], tuple_a[1::2], tuple_b)
    for v in triple
)
print(tuple_c)

另一种选择是使用索引并将其映射到适当的查找：

tuple_c = tuple(
    v
    for i in range(len(tuple_b))
    for v in (tuple_a[i*2], tuple_a[i*2+1], tuple_b[i])
)

您也可以使用 itertools.chain 来避免理解：

from itertools import chain

tuple_a = (1, 3, 5, 7, 9, 11, 13, 15)
tuple_b = (2, 4, 6, 8)

tuple_c = tuple(chain(*zip(tuple_a[::2], tuple_a[1::2], tuple_b)))
print(tuple_c)

交错元组

Interleaving tuple

python

tuples