将列表的每个元素传递给在Python中带有多个参数的函数？

Question

例如,如果我有

a=[['a','b','c'],[1,2,3],['d','e','f'],[4,5,6]]

如何在不必键入的情况下将每个元素a作为一个论证zip

zip(a[0],a[1],a[2],a[3])？

Answer 1

Using sequence unpacking (thanks to delnan for the name):

zip(*a)