我的問題有點復雜，但是這個問題可以用一個例子來寫得相當籠統：我有一個池列表（pools）需要有一個子列表（children）均勻分布在列表中pools。該children列表已經排序，因此可以安全地假設它可以按pools當前順序分布在 上。例如，如果我有并且[pool1, pool2]我[child1, child2, child3]希望pool1被分配child1并且將被分配：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)預期輸出為：child1 assigned to pool1child2 assigned to pool2child3 assigned to pool1期望pools和的數量children可以是任意大小，但是，以下情況始終為真：len(pools) < len(children)。
3 回答
守著一只汪
TA貢獻1872條經驗 獲得超4個贊
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
瀟瀟雨雨
TA貢獻1833條經驗 獲得超4個贊
我認為它更具可讀性：
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}')
輸出：
child1 assigned to pool1
child2 assigned to pool2
child3 assigned to pool1
子衿沉夜
TA貢獻1828條經驗 獲得超3個贊
你可以這樣做：
for elem in children:
if children.index(elem) % 2 == 0:
print(f"{elem} to {pools[0]}")
else:
print(f"{elem} to {pools[1]}")
考慮到你只有兩個池，如果他的索引是奇數，你可以將孩子分配給 pool1。
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