关于python:总算把集合玩明白了

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。

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 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)

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)

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!')

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

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) : 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

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() : 删除汇合中所有元素。

countries = {'Morocco', 'UK', 'Spain', 'USA', 'UK'}

print(len(countries)) # 4

countries.clear()

print(countries) # set()
print(len(countries)) # 0

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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关于python:总算把集合玩明白了

difference(set)

union(set)

intersection(set)

issubset()

issuperset()

isdisjoint(set)

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

clear()

copy