python 列表元组字符串字典集合return等梳理

Posted 2020-09-19

 

 

有必要对这些数据类型及操作做下梳理：

1.列表：增删改查　　

a.查找：

 1 >>> names=["zhang","wang","li","zhao"]
 2 #列表天生具有下标，基于下标0,1,2,...进行查找
 3 >>> names[1] 
 4 ‘wang‘
 5 #列表的切片，即也是基于下标进行操作
 6 >>> names[1:3]
 7 [‘wang‘, ‘li‘]
 8 >>> names[-1:]
 9 [‘zhao‘]
10 >>> names[:3]
11 [‘zhang‘, ‘wang‘, ‘li‘]
12 >>> names[1:]
13 [‘wang‘, ‘li‘, ‘zhao‘]
14 #基于列表中的值找下标
15 >>> print(names.index(‘li‘))
16 2
17 >>> print(names[names.index(‘li‘)])
18 li
19 #基于for循环查找列表值
20 >>> for i in names:
21 ...     print(i)
22 ...
23 zhang
24 wang
25 li
26 zhao

View Code

b.修改

 1 >>> names=["zhang","wang","li","zhao"]
 2 >>> names.append("Sun")
 3 >>> names
 4 [‘zhang‘, ‘wang‘, ‘li‘, ‘zhao‘, ‘Sun‘]
 5 >>> names.insert(1,‘qian‘)
 6 >>> names
 7 [‘zhang‘, ‘qian‘, ‘wang‘, ‘li‘, ‘zhao‘, ‘Sun‘]
 8 >>> names[2] = ‘gao‘
 9 >>> names
10 [‘zhang‘, ‘qian‘, ‘gao‘, ‘li‘, ‘zhao‘, ‘Sun‘]

View Code

c.删除

 1 >>> names
 2 [‘zhang‘, ‘qian‘, ‘gao‘, ‘li‘, ‘zhao‘, ‘Sun‘]
 3 >>> names.remove(‘zhang‘)
 4 >>> names
 5 [‘qian‘, ‘gao‘, ‘li‘, ‘zhao‘, ‘Sun‘]
 6 >>> del names[1]
 7 >>> names
 8 [‘qian‘, ‘li‘, ‘zhao‘, ‘Sun‘]
 9 >>> names.pop()
10 ‘Sun‘
11 >>> names.pop(1)
12 ‘li‘
13 >>> names
14 [‘qian‘, ‘zhao‘]

View Code

d.增加

 1 >>> names
 2 [‘qian‘, ‘zhao‘]
 3 >>> names.append(‘Sun‘) #追加
 4 >>> names
 5 [‘qian‘, ‘zhao‘, ‘Sun‘]
 6 >>> names.insert(1,‘zhang‘)  #插入
 7 >>> names
 8 [‘qian‘, ‘zhang‘, ‘zhao‘, ‘Sun‘]
 9 >>> names2=[1,2,3,4]
10 >>> names.extend(names2) #合并
11 >>> names
12 [‘qian‘, ‘zhang‘, ‘zhao‘, ‘Sun‘, 1, 2, 3, 4]
13 #深浅copy
14 >>> names3 = names.copy()#浅copy，指向同一块内存地址
15 >>> names
16 [‘qian‘, ‘zhang‘, ‘zhao‘, ‘Sun‘, 1, 2, 3, 4]
17 >>> names3
18 [‘qian‘, ‘zhang‘, ‘zhao‘, ‘Sun‘, 1, 2, 3, 4]
19 #深copy，新增一块内存地址，把数据完全copy一份到新内存
20 >>> import copy
21 >>> names
22 [‘qian‘, ‘zhang‘, ‘zhao‘, ‘Sun‘, 1, 2, 3, 4]
23 >>> names4 = copy.deepcopy(names)
24 >>> names4
25 [‘qian‘, ‘zhang‘, ‘zhao‘, ‘Sun‘, 1, 2, 3, 4]

View Code

2.元组：只能查询和切片，也叫只读列表，只有count和index方法

 1 >>> names = (‘alex‘,‘wang‘,‘eric‘)
 2 >>> names.append(‘li‘)
 3 Traceback (most recent call last):
 4   File "<stdin>", line 1, in <module>
 5 AttributeError: ‘tuple‘ object has no attribute ‘append‘
 6 >>> names[1]
 7 ‘wang‘
 8 
 9 >>> print(names.index(‘alex‘))
10 0
11 >>> print(names[names.index(‘alex‘)])
12 alex

View Code

3.字符串：文件中只能存储字符串，且不可修改

 1 name="my name is alex"
 2 print(name.center(50,"-")) 打印50个字符，不够用-补上，并把name放中间
 3 -----------------my name is alex------------------
 4 >>> print(name.startswith(‘my‘)) #判断是否以某字符开头
 5 True
 6 print(name.endswith("ex")) 判断name字符串是否以ex结尾
 7 My name is alex
 8 
 9 name="my\\tnameis{name}andmyageis{year}old"
10 print(name.format(name=‘alex‘,year=23))
11 my name is alex and my age is 23 old
12 print(name.index(‘a‘))
13 5
14 print(‘123‘.isdigit())是否为整数
15 True
16 print(‘+‘.join([‘1‘,‘2‘,‘3‘,‘4‘]))  经常用
17 1+2+3+4
18 print(‘alexli‘.replace(‘l‘,‘L‘)) 把小写l替换成大写的L
19 aLex Li
20 print(‘alex li‘.split())把字符串按空格组成一个列表
21 [‘alex‘, ‘li‘]

View Code

4.字典：key：value的数据类型，无序

>>> info = {‘stu1101‘:‘wang‘,‘stu1102‘:‘zhang‘,‘stu1103‘:‘li‘}

a.增加

 1 >>> info = {‘stu1101‘:‘wang‘,‘stu1102‘:‘zhang‘,‘stu1103‘:‘li‘}
 2 >>> info[‘stu1104‘] = ‘zhao‘
 3 >>> info
 4 {‘stu1101‘: ‘wang‘, ‘stu1102‘: ‘zhang‘, ‘stu1103‘: ‘li‘, ‘stu1104‘: ‘zhao‘}
 5 
 6 #合并
 7 >>> info2 = {1:3,2:5}
 8 >>> info.update(info2)
 9 >>> info
10 {‘stu1101‘: ‘wang‘, ‘stu1102‘: ‘zhang‘, ‘stu1103‘: ‘li‘, ‘stu1104‘: ‘zhao‘, 1:
11 , 2: 5}

View Code

b.修改

1 >>> info = {‘stu1101‘:‘wang‘,‘stu1102‘:‘zhang‘,‘stu1103‘:‘li‘}
2 >>> info[‘stu1101‘] = ‘lu‘
3 >>> info
4 {‘stu1101‘: ‘lu‘, ‘stu1102‘: ‘zhang‘, ‘stu1103‘: ‘li‘}

View Code

c.删除

1 >>> info
2 {‘stu1101‘: ‘lu‘, ‘stu1102‘: ‘zhang‘, ‘stu1103‘: ‘li‘}
3 >>> del info[‘stu1101‘]
4 >>> info
5 {‘stu1102‘: ‘zhang‘, ‘stu1103‘: ‘li‘}
6 
7 info.pop("stu1101")  #pop删除
8 info.popitem()随机删除

View Code

d.查询

 1 >>> info
 2 {‘stu1102‘: ‘zhang‘, ‘stu1103‘: ‘li‘}
 3 >>> info[‘stu1102‘]
 4 ‘zhang‘
 5 #判断是否存在字典中
 6 print(info.get(‘stu1104‘))  安全获取的方法
 7 None
 8 >>> print(‘stu1103‘ in info)
 9 True
10 
11 >>> info
12 {‘stu1102‘: ‘zhang‘, ‘stu1103‘: ‘li‘}
13 >>> info[‘stu1102‘]
14 ‘zhang‘
15 >>> for i in info:  #最建议的查询方式
16 ...     print(i,info[i])
17 ...
18 stu1102 zhang
19 stu1103 li
20 >>> for k,v in info.items():
21 ...     print(k,v)
22 ...
23 stu1102 zhang
24 stu1103 li
25 
26 >>> print(info.items())
27 dict_items([(‘stu1102‘, ‘zhang‘), (‘stu1103‘, ‘li‘)])

View Code

e.多级字典的查询方式

5.集合：无序，天生去重

1 >>> list = [1,2,3,4]
2 >>> list_1 = set(list)
3 >>> print(type(list_1),list_1)
4 <class ‘set‘> {1, 2, 3, 4}

View Code

关系型测试：

交差并子父对称集操作

a.增加

 1 >>> list = [1,2,3,4]
 2 >>> list_1 = set(list)
 3 >>> print(type(list_1),list_1)
 4 <class ‘set‘> {1, 2, 3, 4}
 5 >>> list_1.add(999)
 6 >>> list_1
 7 {1, 2, 3, 4, 999}
 8 >>> list_1.update([777,888])
 9 >>> list_1
10 {1, 2, 3, 4, 999, 777, 888}

View Code

b.删除

1 >>> list_1.remove(999)
2 >>> list_1
3 {1, 2, 3, 4, 777, 888}
4 >>> list_1.discard(888)
5 >>> list_1
6 {1, 2, 3, 4, 777}
7 >>> list_1.pop()
8 1

View Code

c.深浅复制，跟列表一样

d.判断

len(list_1)  #测试长度

判断是否存在集合中，大于小于等

6.return

两个作用：

1.得到程序的执行结果；

2.程序结束，return后的代码不再执行；