Python集合，,定义：由不同元素组成

Python集合，,定义：由不同元素组成

定义：由不同元素组成的集合，集合中是一组无须排列的可hash值，可以作为字典的Key

特点：

　　a.不同元素组成

　　b.无序

　　c.集合中元素必须是不可变类型（数字、字符串、元祖）

方法：

　　-add()

1 arr_Num = {1, 2, 3, 4, 5}2 arr_Num.add(‘alex‘)3 print(arr_Num)

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　　-clear()

1 arr_Num = {1, 2, 3, 4, 5}2 arr_Num.clear()3 print(arr_Num)

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　　-copy()

1 arr_Num = {1, 2, 3, 4, 5}2 arr_Num1 = arr_Num.copy()3 print(arr_Num1)

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　　-pop() 删除，不确定从左边删，还是右边删，因为集合是无序的

1 arr_Num = {1, 2, 3, 4, 5}2 arr_Num.pop()3 print(arr_Num)

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　　-remove() 指定元素删，若元素不存在，会报错

1 arr_Num = {1, 2, 3, 4, 5}2 arr_Num.remove(3)3 print(arr_Num)

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　　-discard() 指定元素删，若元素不存在，不会报错

1 arr_Num = {1, 2, 3, 4, 5}2 arr_Num.discard(33)3 print(arr_Num)

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集合关系：

　　-intersection() 找2个集合中的交集，即共同的部分

1 arr_Num1 = {1, 2, 3, 4, 5}2 arr_Num2 = {2, 4, 5}3 print(arr_Num1.intersection(arr_Num2))

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　　- “&” 同上一个方法

1 arr_Num1 = {1, 2, 3, 4, 5}2 arr_Num2 = {2, 4, 5}3 print(arr_Num1 & arr_Num2)

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　　-union() 并集

1 arr_Num1 = {1, 2, 3, 4, 5}2 arr_Num2 = {2, 4, 5, 7}3 print(arr_Num1.union(arr_Num2))

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　　- “|” 同上一个方法

1 arr_Num1 = {1, 2, 3, 4, 5}2 arr_Num2 = {2, 4, 5, 7}3 print(arr_Num1 | arr_Num2)

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　　- difference() 求差集

1 arr_Num1 = {1, 2, 3, 4, 5}2 arr_Num2 = {2, 4, 5, 7}3 print(‘差集:‘, arr_Num1.difference(arr_Num2))

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　　- “-” 同上一个方法

1 arr_Num1 = {1, 2, 3, 4, 5}2 arr_Num2 = {2, 4, 5, 7}3 print(‘差集:‘, arr_Num1 - arr_Num2)

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　　- symmetric_difference() 求交叉补集

1 arr_Num1 = {1, 2, 3, 4, 5}2 arr_Num2 = {2, 4, 5, 7}3 print(‘交叉补集:‘, arr_Num1.symmetric_difference(arr_Num2))

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　　- “^” 同上一个方法

1 arr_Num1 = {1, 2, 3, 4, 5}2 arr_Num2 = {2, 4, 5, 7}3 print(‘交叉补集:‘, arr_Num1 ^ arr_Num2)

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　　- difference_update() 交叉完之后，更新

1 arr_Num1 = {1, 2, 3, 4, 5}2 arr_Num2 = {2, 4, 5, 7}3 arr_Num1.difference_update(arr_Num2)4 print(‘交叉完之后，赋值给arr_Num1:‘, arr_Num1)

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　　-isdisjoint() 返回一个布尔值，如何二个集合没有交集，返回一个True；反之则，False

1 arr_Num1 = {1, 2, 3, 4, 5}2 arr_Num2 = {7}3 accept_result = arr_Num1.isdisjoint(arr_Num2)4 print(accept_result)

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　　- issubset() 返回一个布尔值，判断一个集合是否是第二个集合的子集

1 arr_Num1 = {1, 2, 3, 4, 5}2 arr_Num2 = {2, 3}3 accept_result = arr_Num2.issubset(arr_Num1)4 print(accept_result)

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　　-issuperset() 返回一个布尔值，判断一个集合是否是第二个集合的父集

1 arr_Num1 = {1, 2, 3, 4, 5}2 arr_Num2 = {2, 3}3 accept_result = arr_Num2.issuperset(arr_Num1)4 print(accept_result)

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　　-update() 一个集合更新到另一个集合

1 arr_Num1 = {1, 2, 3, 4, 5}2 arr_Num2 = {6, 7}3 arr_Num1.update(arr_Num2)4 print(arr_Num1)

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Python集合