#!/usr/bin/env python __author__ = "lrtao2010" #Python 3.7.0 集合常用方法 #集合是无序的，元素不能重复，元素只能是数字、字符串、元祖 # a = set('abc') # print(a) # {'c', 'a', 'b'} # a = {1,2,3,4,4,'abc',('a')} # print(a) # {1, 'abc', 2, 3, 4, 'a'} # a = set(['hello','hello','tom','tom']) # print(a) # {'tom', 'hello'} # add(self, *args, **kwargs) # 向集合中添加一个元素，如果元素在集合中已经存在，可以执行，但不会再次添加 # a = {'a','b','c'} # a.add(123) # a.add('a') # print(a) # {'c', 123, 'b', 'a'} #clear(self, *args, **kwargs)清空集合 # a = {'a','b','c'} # a.clear() # print(a) # set() #copy(self, *args, **kwargs) 浅复制 # a = {'a','b','c'} # b = a.copy() # print(b) # a.add('d') # print(a) # print(b) #pop(self, *args, **kwargs) 随机删除一个元素，并返回该元素，如果集合是空集合，报异常 # a = {'1','a',3} # v = a.pop() # print(v) # print(a) # a # {3, '1'} # a = {} # a.pop() # print(a) # a.pop() # TypeError: pop expected at least 1 arguments, got 0 #remove(self, *args, **kwargs) 删除一个给定的元素，如果集合中没有该元素，报异常 # a = {'a',1,3} # a.remove('a') # print(a) # {1, 3} # a.remove('b') # a.remove('b') # KeyError: 'b' #discard(self, *args, **kwargs) 删除一个给定的元素，如果该元素不存在，什么也不做 # a = {'a',1,3} # a.discard(1) # a.discard(4) # print(a) # {'a', 3} #定义不可变集合 # a = frozenset('Nochange') # print(a) # frozenset({'n', 'e', 'c', 'N', 'a', 'g', 'h', 'o'}) #列表去重，不考虑原来顺序 # name = ['tom','lilei','tom','jon','zdk'] # name = list(set(name)) # print(name) # ['tom', 'jon', 'zdk', 'lilei'] ##########集合关系运算######## # name_1 = {'tom','lilei','jim','jon'} # name_2 = {'tom','bob','jhon','jon'} #intersection(self, *args, **kwargs)求交集，并返回一个新集合，集合交换位置，结果一样 # name_3 = name_1.intersection(name_2) # name_3 = name_1 & name_2 # print(name_1,name_2,name_3) # {'jon', 'tom', 'jim', 'lilei'} {'bob', 'tom', 'jon', 'jhon'} {'jon', 'tom'} #intersection_update(self, *args, **kwargs) #将原集合修改为两个集合的交集 #name_1.intersection_update(name_2) # name_1 = name_1 & name_2 # print(name_1,name_2) # {'jon', 'tom'} {'jon', 'bob', 'jhon', 'tom'} #union(self, *args, **kwargs) 求并集，并返回一个新集合,集合交换位置，结果一样 # name_3 = name_1.union(name_2) # name_3 = name_1 | name_2 # print(name_1,name_2) # print(name_3) # {'tom', 'jon', 'jim', 'lilei'} {'tom', 'bob', 'jon', 'jhon'} # {'jon', 'jim', 'tom', 'lilei', 'bob', 'jhon'} #update(self, *args, **kwargs) # 原集合更新为并集, # name_1.update(name_2) # print(name_1) # print(name_2) # {'lilei', 'jon', 'jim', 'bob', 'tom', 'jhon'} # {'jhon', 'jon', 'bob', 'tom'} # 也可以用作同时向集合中传入多个值（可以是元祖，列表，字符串（是可迭代对象即可）） #add 方法一次只能加入一个值 # a = {1} # b = {'a'} # c = {1,'b'} # a.update((1,2,3,)) # b.update('abcd') # c.update(['adb',123,('123',)]) # print(a) # print(b) # print(c) # {1, 2, 3} # {'c', 'a', 'd', 'b'} # {1, 'adb', ('123',), 'b', 123} #difference(self, *args, **kwargs)求差集，并返回一个新集合，集合交换位置，结果不一样 # name_3 = name_1.difference(name_2) # name_4 = name_2.difference(name_1) # name_3 = name_1 - name_2 # name_4 = name_2 - name_1 # print(name_1,name_2) # print(name_3) # print(name_4) # {'jon', 'lilei', 'jim', 'tom'} {'jon', 'bob', 'jhon', 'tom'} # {'jim', 'lilei'} # {'bob', 'jhon'} #difference_update(self, *args, **kwargs) # 本集合中除去两个集合的交集，原集合被修改 # name_1.difference_update(name_2) # name_1 = name_1 - name_2 # print(name_1,name_2) # {'lilei', 'jim'} {'jon', 'jhon', 'bob', 'tom'} #symmetric_difference(self, *args, **kwargs) # 交叉补集=并集-交集，返回一个新集合，集合交换位置，结果一样 # name_3 = name_1.symmetric_difference(name_2) # name_4 = name_2.symmetric_difference(name_1) # name_3 = name_1 ^ name_2 # name_4 = name_2 ^ name_1 # print(name_1,name_2) # print(name_3) # print(name_4) # {'jim', 'lilei', 'jon', 'tom'} {'jhon', 'jon', 'bob', 'tom'} # {'jim', 'lilei', 'bob', 'jhon'} # {'jim', 'lilei', 'bob', 'jhon'} #symmetric_difference_update(self, *args, **kwargs) #原集合修改为交叉补集 # name_1.symmetric_difference_update(name_2) # print(name_1) # print(name_2) # {'jhon', 'lilei', 'jim', 'bob'} # {'tom', 'jhon', 'bob', 'jon'} #isdisjoint(self, *args, **kwargs) # 如果两个集合没有交集(交集为空)返回True,集合交换位置，结果一样 # a = {1,2} # b = {1} # c = {3} # a_b = a.isdisjoint(b) # b_a = b.isdisjoint(a) # a_c = a.isdisjoint(c) # c_a = c.isdisjoint(a) # print(a_b,b_a,a_c,c_a) # False False True True # a = {1,2,3} # b = {1} # c = {1,3,5} #issubset(self, *args, **kwargs)是否为子集 # v1 = a.issubset(b) # v2 = b.issubset(a) # v3 = a.issubset(c) # v4 = b <= a # print(v1,v2,v3,v4) # False True False True #issuperset(self, *args, **kwargs)是否为父集 # v1 = a.issuperset(b) # v2 = b.issuperset(a) # v3 = a.issuperset(c) # v4 = a >= b # print(v1,v2,v3,v4) # True False False True