一、数据类型
1、列表
lis = [11, 22, 33, 44, 55] for i in range(len(lis)): print(i) # i = 0 i = 1 i = 2 del lis[i] print(lis) # [11,22,33,44,55] [22, 44, 55] [22, 44] 循环按照列表的索引0、1、2...循环删除,最后3 的时候元素不够报错
--->赋值
l1 = [] l2 = l1 l3 = l1 l3.append('a') print(l1,l2,l3) #['a'] ['a'] ['a']
-->隔一个元素进行删除
lis = [11, 22, 33, 44, 55] lis = lis[::2] print(lis)
--->或者
lis = [11, 22, 33, 44, 55] l1 = [] for i in lis: if lis.index(i) % 2 == 0: l1.append(i) lis = l1 print(lis)
--->又或者
lis = [11,22,33,44,55] for i in range(len(lis)-1,-1,-1): if i % 2 == 1: del lis[i] print(lis)
2、字典
dic = dict.fromkeys([1,2,3],'春哥') #列表赋值 print(dic) dic = dict.fromkeys([1,2,3],[]) print(dic) # {1: [], 2: [], 3: []} dic[1].append('袁姐') print(dic) #{1: ['袁姐'], 2: ['袁姐'], 3: ['袁姐']} dic[2].extend('二哥') print(dic) #{1: ['二', '哥'], 2: ['二', '哥'], 3: ['二', '哥']}
3、删除字典内指定键值组
--->第1种
dic = {'k1':'v1','k2':'v2','a3':'v3'} dic1 = {} for i in dic: if 'k' not in i: dic1.setdefault(i,dic[i]) dic = dic1 print(dic)
--->第2种
dic = {'k1':'v1','k2':'v2','a3':'v3'} l = [] for i in dic: if 'k' in i: l.append(i) for i in l: del dic[i] print(dic)
4、元组 (如果元祖里面只有一个元素且不加,那此元素是什么类型,就是什么类型)
tu1 = (1) tu2 = (1,) print(tu1,type(tu1)) # 1 <class 'int'> print(tu2,type(tu2)) # (1,) <class 'tuple'> tu1 = ([1]) tu2 = ([1],) print(tu1,type(tu1)) # [1] <class 'list'> print(tu2,type(tu2)) # ([1],) <class 'tuple'> dic = dict.fromkeys([1,2,3,],3) dic[1] = 4 print(dic) # {1: 4, 2: 3, 3: 3}
二、集合 (可变的数据类型,他里边的元素必须是不可变的数据类型,无序,不重复)
set1 = {1,2,3} print(set1) set2 = {1,2,3,[1,2,3],{'name':'alex'}} #错误表达方式,含有列表跟字典等可变数据类型 print(set2)
1、集合的增
set1 = {'peter','alex','jimmy','xiaoming'} set1.add('老王') #第一种,直接增加元素 set1.update('abc') #第二种,拆分为最小元素增加 print(set1)
2、集合的删
set1 = {'peter','alex','jimmy','xiaoming'} set1.pop() # 随机产出集合中某个元素,有返回值 print(set1.pop()) # 查看返回值 set1.remove('peter') #按照元素进行删除 set1.clear() # 清空集合 set() del(set1) #删除整个集合 print(set1)
3、集合的查
set1 = {'peter','alex','jimmy','xiaoming'} for i in set1: #顺序是变化的 print(i)
4、集合的交集
set1 = {1,2,3,4,5} set2 = {4,5,6,7,8} print(set1 & set2) #求set1 与 set2 的交集 {4, 5} print(set1.intersection(set2)) #求set1 与 set2 的交集 {4, 5} set3 = set2 & set1 #赋值set3 print(set3)
5、集合的并集
set1 = {1,2,3,4,5} set2 = {4,5,6,7,8} print(set1 | set2) #求set1 与 set2 的并集 {1, 2, 3, 4, 5, 6, 7, 8} print(set1.union(set2)) #求set1 与 set2 的并集 {1, 2, 3, 4, 5, 6, 7, 8}
6、集合的反交集
set1 = {1,2,3,4,5} set2 = {4,5,6,7,8} print(set1 ^ set2) #求set1 与 set2 的反交集 {1, 2, 3, 6, 7, 8} print(set1.symmetric_difference(set2)) #求set1 与 set2 的反交集 {1, 2, 3, 6, 7, 8}
7、集合的差集
set1 = {1,2,3,4,5} set2 = {4,5,6,7,8} print(set1 - set2) #set1独有的 {1, 2, 3} print(set1.difference(set2)) #set1独有的 {1, 2, 3} print(set2 - set1) # set2 独有的 {8, 6, 7} print(set2.difference(set1)) # set2 独有的 {8, 6, 7}
8、集合的子集与超集
set1 = {1,2,3} set2 = {1,2,3,4,5,6} print(set1 < set2) # set1 是set2 的子集 True print(set1.issubset(set2)) # set1 是set2 的子集 True print(set2 > set1) # set2 是set1 的超集 True print(set2.issuperset(set1)) # set2 是set1 的超集 True
9、面试题---去掉列表重复的元素li = [1,22,22,33,45,66,66,90]
li = [1,22,22,33,45,66,66,90] set1 = set(li) li = list(set1) print(li)
10、冻结功能
s = frozenset('peter') print(s,type(s)) # 冻结 frozenset({'e', 'r', 'p', 't'}) <class 'frozenset'> for i in s: print(i)
三、深浅copy
1、赋值运算
l1 = [1,2,3] l2 = l1 l1.append('a') print(l1,l2) print(l1 is l2) # [1, 2, 3, 'a'] [1, 2, 3, 'a'] l1 跟l2 是同一个地址的列表
2、copy
l1 = [1,2,3] l2 = l1.copy() print(l1,l2) # [1, 2, 3] [1, 2, 3] print(l1,id(l1)) # [1, 2, 3] 18895432 print(l2,id(l2)) # [1, 2, 3] 18800456 print(l1 is l2) # False l1 跟l2 不是同一个地址的列表 l2.append('a') print(l1,l2) # [1, 2, 3] [1, 2, 3, 'a']
3、浅度copy
l1 = [1,2,[4,5,6],3] l2 = l1.copy() print(l1,id(l1)) # [1, 2, [4, 5, 6], 3] 18960648 print(l2,id(l2)) # [1, 2, [4, 5, 6], 3] 18960584 print(l1 is l2) # False l1 跟l2 不是同一个地址的列表 l1.append('a') print(l1,l2) #[1, 2, [4, 5, 6], 3, 'a'] [1, 2, [4, 5, 6], 3] l1[2].append('a') print(l1,l2) # [1, 2, [4, 5, 6, 'a'], 3] [1, 2, [4, 5, 6, 'a'], 3] print(id(l1[2])) # 18865736 print(id(l2[2])) # 18865736 l1、l2 列表中的子列表存放地址是相同的 print(l1[2] is l2[2]) # True 浅copy 时,列表中的子列表是相同地址的
4、深度copy
import copy l1 = [1,2,[4,5,6],3] l2 = copy.deepcopy(l1) print(l1,id(l1)) # [1, 2, [4, 5, 6], 3] 19324552 print(l2,id(l2)) # [1, 2, [4, 5, 6], 3] 19419208 l1[2].append('a') print(l1,l2) # [1, 2, [4, 5, 6, 'a'], 3] [1, 2, [4, 5, 6], 3] 深copy 时,列表中的子列表是不同地址的
5、小知识点
li = ['alex','taibai','wusir','egon'] for i in li: print(li.index(i),i) # 索引从0 开始 for index,i in enumerate(li,1): # 索引加1 作为序号 print(index,i)