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)
是如何工作的?如果用更冗长的代码编写会是什么样子?
How does zip(*[iter(s)]*n)
work? What would it look like if it was written with more verbose code?
iter()
是一个序列的迭代器.[x] * n
生成一个包含 n
个 x
数量的列表,即长度为 n
的列表,其中每个元素都是x
.*arg
将序列解压缩为函数调用的参数.因此,您将相同的迭代器 3 次传递给 zip()
,每次都会从迭代器中拉取一个item.
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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