总算彻底搞懂Python集合了

　　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。
　　程序员宝藏库 ：https://github.com/Jackpopc/CS-Books-Store 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 1 recepie_requirements = {＂orange＂, ＂chocolate＂, ＂salt＂, ＂pepper＂} what_I_have = {＂apple＂, ＂banana＂,＂salt＂} # I have to buy orange chocolate pepper print(＂I have to buy＂, *recepie_requirements.difference(what_I_have))  # example2 all_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)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!＂)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 libraryissuperset()
　　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 libraryisdisjoint(set)
　　isdisjoint(set) : 检查这两个集合是否包含共同的元素。 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) # Falsediscard(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，则会引发错误。 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!＂) # KeyErrorclear()
　　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)) # 0copy
　　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＂}