汇合（set）是一种无序的不反复元素序列，能够应用大括号  { }  或者 set()  函数创立汇合。

它是Python中一个十分重要，且频繁用到的概念。无论是在日常开发过程中，还是在面试过程中都会常常遇到，明天就来11「鲜为人知」的汇合用法。

difference(set)

set_1.difference(set_2) ：这个办法帮忙你取得两个汇合之间的差别，换句话说，它让你取得存在于set_1中而不存在于给定汇合（set_2）中的元素。

# example 1recepie_requirements = {'orange', 'chocolate', 'salt', 'pepper'}what_I_have = {'apple', 'banana','salt'}# I have to buy orange chocolate pepperprint('I have to buy', *recepie_requirements.difference(what_I_have))# example2all_subscribers = {"aya", "john", "smith", "sparf", "kyle"}admins = {"aya", "sparf"}users = all_subscribers.difference(admins)# {'kyle', 'smith', 'john'}print(users)

union(set)

set_1.union(set_2) ：(set_1 U set_2) 这个set办法返回一个蕴含set_1的元素和set_2的元素的汇合，此外，返回的汇合只蕴含惟一的元素。

admins = {'aya', 'sparf'}users = {'aya','kyle', 'smith', 'john'}all_subscribers = admins.union(users)# {'smith', 'aya', 'sparf', 'kyle', 'john'}print(all_subscribers)

intersection(set)

set_1.intersection(set_2) ：取两个汇合的交加，只返回同时存在于set_1和set_2中的元素。

shop = {'orange', 'pepper', 'banana', 'sugar'}what_I_have = {'orange', 'sugar'}# I should not buy {'orange', 'sugar'} because I have them!print(f'I should not buy {shop.intersection(what_I_have)} because I have them!')

issubset()

set_1.issubset(set_2) ：查看set_1的所有元素是否存在于set_2中。

nearest_library_books = {"the power of now", 'why we sleep', 'rich dad poor dad'}necessary_books = {'atomic habits','the 48 laws of power', 'why we sleep'}if necessary_books.issubset(nearest_library_books):  print('yes, you can buy these books from your nearest library')else:  print('unfortunately, you have to go to another library')# unfortunately, you have to go to another library

issuperset()

set_1.issuperset(set_2) : 查看set_2的所有元素是否存在于set_1中。

nearest_library_books = {"the power of now", 'why we sleep', 'rich dad poor dad'}necessary_books = {'atomic habits','the 48 laws of power', 'why we sleep'}if nearest_library_books.issuperset(necessary_books):  print('yes, you can buy these books from your nearest library')else:  print('unfortunately, you have to go to another library')# unfortunately, you have to go to another library

isdisjoint(set)

isdisjoint(set) : 查看这两个汇合是否不蕴含独特的元素。

set_1 = {12, 38, 36}set_2 = {4, 40, 12}# means can set_1 element - set_2 element == 0 ?can_substruction_be_zero = set_1.isdisjoint(set_2)print(can_substruction_be_zero) # False

discard(value), remove(value), pop()

pop() : 从一个汇合中删除一个随机元素。

discard(value) : 删除一个汇合中的指定元素，如果该元素不存在，不会引发谬误。

remove(value) : 删除一个汇合中的指定元素，如果该元素不存在，则引发谬误。

users = {"Aya Bouchiha", "John Doe", "Kyle Smith", "Nabo Snay"}deleted_account = 'Aya Bouchiha'users.discard(deleted_account)users.discard('Hi!')print(users) # {'Kyle Smith', 'John Doe', 'Nabo Snay'}users.remove('Kyle Smith') print(users) # {'Nabo Snay', 'John Doe'}users.pop()print(users) # {'John Doe'}users.remove('Hello!') # KeyError

clear()

clear() : 删除汇合中所有元素。

countries = {'Morocco', 'UK', 'Spain', 'USA', 'UK'}print(len(countries)) # 4countries.clear()print(countries) # set()print(len(countries)) # 0

copy

copy() : 这个办法让你失去一个指定元素集的正本

countries = {'Morocco', 'UK', 'Spain', 'USA', 'UK'}print(countries) # {'UK', 'Morocco', 'Spain', 'USA'}print(countries.copy()) # {'UK', 'Morocco', 'Spain', 'USA'}

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