如何将较大的数组分布在较小的数组上

Question

我的问题有点复杂，但是这个问题可以用一个例子来写得相当笼统：我有一个池列表（pools）需要有一个子列表（children）均匀分布在列表中pools。该children列表已经排序，因此可以安全地假设它可以按pools当前顺序分布在 上。例如，如果我有并且[pool1, pool2]我[child1, child2, child3]希望pool1被分配child1并且将被分配：child3pool2child2pools = ['pool1', 'pool2']children = ['child1', 'child2', 'child3']def print_assignment(pool, child)  print('{} assigned to {}'.format(child, pool)# The expectation is that distribute would perform the core logic and # call print_assignment during each assignmentdistribute(pools, children, print_assignment)预期输出为：child1 assigned to pool1child2 assigned to pool2child3 assigned to pool1期望pools和的数量children可以是任意大小，但是，以下情况始终为真：len(pools) < len(children)。

Answer 1

from itertools import cycle

pools = ['pool1', 'pool2']

children = ['child1', 'child2', 'child3']

c = cycle(pools)

for child in children:

    print('{} assigned to {}'.format(child, next(c)))

印刷：

child1 assigned to pool1

child2 assigned to pool2

child3 assigned to pool1

Answer 2

我认为它更具可读性：

from itertools import cycle

pools = ['pool1', 'pool2']

children = ['child1', 'child2', 'child3']

for child, pool in zip(children, cycle(pools)):

    print(f'{child} assigned to {pool}')

输出：

child1 assigned to pool1

child2 assigned to pool2

child3 assigned to pool1

Answer 3

你可以这样做：

for elem in children:

    if children.index(elem) % 2 == 0:

        print(f"{elem} to {pools[0]}")

    else:

        print(f"{elem} to {pools[1]}")

考虑到你只有两个池，如果他的索引是奇数，你可以将孩子分配给 pool1。