zip(* [iter(s)] * n)如何在Python中运行? [英] How does zip(*[iter(s)]*n) work in Python?
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问题描述
s = [1,2,3,4,5,6,7,8,9]
n = 3
zip(*[iter(s)]*n) # returns [(1,2,3),(4,5,6),(7,8,9)]
zip(* [iter(s)] * n)
如何工作?如果用更详细的代码编写它会是什么样子?
How does zip(*[iter(s)]*n)
work? What would it look like if it was written with more verbose code?
推荐答案
iter()
是序列上的迭代器。 [x] * n
生成一个包含 n
数量 x $ c $的列表c>,即长度
n
的列表,其中每个元素都是 x
。 * arg
将序列解包为函数调用的参数。因此,您将相同的迭代器传递3次到 zip( )
,每次都从迭代器中提取一个项目。
iter()
is an iterator over a sequence. [x] * n
produces a list containing n
quantity of x
, i.e. a list of length n
, where each element is x
. *arg
unpacks a sequence into arguments for a function call. Therefore you're passing the same iterator 3 times to zip()
, and it pulls an item from the iterator each time.
x = iter([1,2,3,4,5,6,7,8,9])
print zip(x, x, x)
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