Question

我有这个无限生成器：

def infiniList():
    count = 0
    ls = []
    while True:
        yield ls
        count += 1
        ls.append(count)

有没有办法采用前n个元素？我的意思是，我做到了：

n = 5
ls = infiniList()
for i in range(n):
  rs = next(ls)

输出：

print(rs)
[1, 2, 3, 4]

Answer 1

itertools.islice确实做到了这一点，尽管在您的示例中，您需要注意不要重复产生对不断修改的同一对象的引用：

def infiniList():
    count = 0
    ls = []
    while True:
        yield ls[:]  # here I added copying
        count += 1
        ls.append(count)  # or you could write "ls = ls + [count]" and not need to make a copy above

import itertools
print(list(itertools.islice(infiniList(), 5)))

Answer 2

仅需介绍@NPE的良好答案，我需要做的就是：

import itertools

def infiniList():
    count = 0
    ls = []
    while True:
        yield ls
        count += 1
        ls = ls + [count]


print(list(itertools.islice(infiniList(), 5, 6))[0])

输出：

[1, 2, 3, 4, 5]

如何从无限生成器列表中获取前n个元素？

2 个答案: