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?spa
笨方法:判断列表中元素是否是列表;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太重复 难读.net
1 new_list = ["H1","H2",1999,["hello","day"]] 2 3 for each_list in new_list: 4 if isinstance(each_list,list): 5 for new_each in each_list: 6 print (new_each) 7 else: 8 print (each_list);
若是想遇到列表就缩进一次怎么办?code
增长一个形参呗;blog
1 def each_list(list_name,level=0): 2 for yuansu in list_name: 3 if isinstance(yuansu,list): #判断当前元素是否是列表 4 each_list(yuansu,level+1) #如是,则递归调用,而且标记当前元素是列表 5 else: 6 for tab in range(level):#固定次数 7 print ("\t",end='') 8 print (yuansu) 9 10 11 new_list = ["H1","H2",1999,["hello","day",["one","two"]]] 12 13 each_list(new_list)
若是加入一个开启机制,不想加缩进&想加缩进 怎么办;递归
再次加入一个形参控制get
1 def each_list(list_name,count=False,level=0): #加入控制形参 count 默认为不开启缩进 2 for yuansu in list_name: 3 if isinstance(yuansu,list): #判断当前元素是否是列表 4 each_list(yuansu,count,level+1) #如是,则递归调用,而且标记当前元素是列表 5 else: 6 if count: #判断是否开启缩进 7 for tab in range(level):#固定次数 8 print ("\t",end='') 9 print (yuansu) 10 else: 11 print (yuansu) 12 13 14 new_list = ["H1","H2",1999,["hello","day",["one","two"]]] 15 16 each_list(new_list)