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
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);
如果想遇到列表就缩进一次怎么办?
增加一个形参呗;
1 def each_list(list_name,level=0):
2 for yuansu in list_name:
3 if isinstance(yuansu,list): #判断当前元素是不是列表
4 each_list(yuansu,level+1) #如是,则递归调用,并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)
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
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): #判断当前元素是不是列表
4 each_list(yuansu,count,level+1) #如是,则递归调用,并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)