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)工作怎么样?如果用更详细的代码编写它会是什么样子?
zip(*[iter(s)]*n)
python iterator
iterator ×1
python ×1