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) 在 pools 的列表中均匀分布。

children 列表已经排序，因此可以安全地假设它可以按当前顺序分布在 pools 中。

例如，如果我有 [pool1, pool2] 和 [child1, child2, child3] 我希望 pool1 被分配 child1 和 child3 和 pool2将被分配 child2:

pools = ['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 assignment
distribute(pools, children, print_assignment)

预期输出为：

child1 assigned to pool1
child2 assigned to pool2
child3 assigned to pool1

期望 pools 和 children 的计数可以是任意大小，但是，以下情况始终为真：len(pools) < len(children).

Answer 1

您可以使用 itertools.cycle 完成任务：

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)))

打印：

child1 assigned to pool1
child2 assigned to pool2
child3 assigned to pool1

Answer 2

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]}")

考虑到您只有两个池，如果他的索引是奇数，您可以将 children 分配给 pool1。

Answer 3

这是对的轻微修改，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

如何将较大的数组分布在较小的数组上

How to distribute a larger array over a smaller array

python

algorithm

round-robin