集合（set）是一种无序的不重复元素序列，可以使用大括号 { } 或者 set() 函数创建集合。
它是 Python 中一个非常重要，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经常遇到，今天就来 11「不为人知」的集合用法。
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difference(set)
set_1.difference(set_2)：这个方法帮助你获得两个集合之间的差异，换句话说，它让你获得存在于 set_1 中而不存在于给定集合（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)
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union(set)
set_1.union(set_2)：(set_1 U set_2) 这个 set 方法返回一个包含 set_1 的元素和 set_2 的元素的集合，此外，返回的集合只包含唯一的元素。
admins = {'aya', 'sparf'}users = {'aya','kyle', 'smith', 'john'}
all_subscribers = admins.union(users)
# {'smith', 'aya', 'sparf', 'kyle', 'john'}print(all_subscribers)
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intersection(set)
set_1.intersection(set_2)：取两个集合的交集，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!')
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issubset()
set_1.issubset(set_2)：检查 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
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issuperset()
set_1.issuperset(set_2): 检查 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
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isdisjoint(set)
isdisjoint(set): 检查这两个集合是否不包含共同的元素。
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
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discard(value), remove(value), pop()
pop(): 从一个集合中删除一个随机元素。
discard(value): 删除一个集合中的指定元素，如果该元素不存在，不会引发错误。
remove(value): 删除一个集合中的指定元素，如果该元素不存在，则引发错误。
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
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clear()
clear(): 删除集合中所有元素。
countries = {'Morocco', 'UK', 'Spain', 'USA', 'UK'}
print(len(countries)) # 4
countries.clear()
print(countries) # set()print(len(countries)) # 0
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copy
copy(): 这个方法让你得到一个指定元素集的副本
countries = {'Morocco', 'UK', 'Spain', 'USA', 'UK'}
print(countries) # {'UK', 'Morocco', 'Spain', 'USA'}print(countries.copy()) # {'UK', 'Morocco', 'Spain', 'USA'}
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