zip(*[iter(s)]*n) 在 Python 中是如何工作的?
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) 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?
解决方案
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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