python学习笔记--打印嵌套list中每个数据（遍历列表）

遍历new_list列表中元素

new_list = ["H1","H2",1999]

for each_list in new_list:
    print (each_list);

若列表中包含嵌套列表,怎样处理？

笨方法：判断列表中元素是不是列表;并继续使用for来循环打印, 缺点：多个嵌套列表时会使代码过长过重复 难读

new_list = ["H1","H2",1999,["hello","day"]]

for each_list in new_list:
    if isinstance(each_list,list):
        for new_each in each_list:
            print (new_each)
    else:
        print (each_list);

简单方法：编写递归函数

def each_list(list_name):
    for yuansu in list_name:
        if isinstance(yuansu,list):
            each_list(yuansu)
        else:
            print (yuansu)


new_list = ["H1","H2",1999,["hello","day"],["one","two"]]

each_list(new_list)

如果想遇到列表就缩进一次怎么办？

增加一个形参呗；

def each_list(list_name,level=0):
    for yuansu in list_name:
        if isinstance(yuansu,list):   #判断当前元素是不是列表
            each_list(yuansu,level+1) #如是,则递归调用,并且标记当前元素是列表
        else:
            for tab in range(level):#固定次数
                print ("\t",end='')
            print (yuansu)


new_list = ["H1","H2",1999,["hello","day",["one","two"]]]

each_list(new_list)

如果加入一个开启机制,不想加缩进&想加缩进 怎么办；

再次加入一个形参控制

def each_list(list_name,count=False,level=0):  #加入控制形参 count 默认为不开启缩进
    for yuansu in list_name:
        if isinstance(yuansu,list):   #判断当前元素是不是列表
            each_list(yuansu,count,level+1) #如是,则递归调用,并且标记当前元素是列表
        else:
            if count:  #判断是否开启缩进
                for tab in range(level):#固定次数
                    print ("\t",end='')
                print (yuansu)
            else:
                print (yuansu)


new_list = ["H1","H2",1999,["hello","day",["one","two"]]]

each_list(new_list)

来源：CSDN

作者：lclcy306

链接：https://blog.csdn.net/lclcy306/article/details/38733215