针对多列表嵌套,如何获取所有元素?...second_layer:
print third_layer
else:
print first_layer
但是,若遇到列表有.../usr/bin/env python
#coding:utf-8
fruit=['a','b',123,['c',345,'d',['e','f',90],22,'cc'],'po',34]
def...list):
layer(first_layer)
else:
print first_layer
layer(fruit)
这样,不管多少个列表嵌套