python - itertools tee() 迭代器分割

Question

我很困惑 tee() 的实际工作原理。

l = [1, 2, 3, 4, 5, 6]
iterators3 = itertools.tee(l, 3)
for i in iterators3:
    print (list(i))

输出:

[1, 2, 3, 4, 5, 6]
[1, 2, 3, 4, 5, 6]
[1, 2, 3, 4, 5, 6]

这没关系。但如果我尝试:

a, b, c = itertools.tee(l)

我收到此错误:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: not enough values to unpack (expected 3, got 2)

为什么？

Answer 1

tee 接受 2 个参数，一个迭代器和一个数字，它将复制实际迭代器(及其上下文)作为参数传递的次数，因此您实际上无法解压更多生成器比 tee 创建的:

a,b = tee(l) #this is ok, since it just duplicate it so you got 2 
a,b,c = tee(l, 3) #this is also ok, you will get 3 so you can unpack 3
a,b = tee(l, 3) #this will fail, tee is generating 3 but you are trying to unpack just 2 so he dont know how to unpack them

在 python 3 中你可以像这样解压:

a, *b = tee(l, 3)

其中 a 将保存 tee 中的第一个迭代器，b 将保存列表中的其余迭代器。