我的問(wèn)題有點(diǎn)復(fù)雜，但是這個(gè)問(wèn)題可以用一個(gè)例子來(lái)寫得相當(dāng)籠統(tǒng)：我有一個(gè)池列表（pools）需要有一個(gè)子列表（children）均勻分布在列表中pools。該children列表已經(jīng)排序，因此可以安全地假設(shè)它可以按pools當(dāng)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(yù)期輸出為：child1 assigned to pool1child2 assigned to pool2child3 assigned to pool1期望pools和的數(shù)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(xiàn)1872條經(jīng)驗(yàn) 獲得超4個(gè)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(xiàn)1833條經(jīng)驗(yàn) 獲得超4個(gè)贊
我認(rèn)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(xiàn)1828條經(jīng)驗(yàn) 獲得超3個(gè)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(gè)池，如果他的索引是奇數(shù)，你可以將孩子分配給 pool1。
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