Python 中set集合类型(去重、成员运算)

集合类型的作用:去重以及成员运算

# 第一种、创建集合,使用花括号{},打印默认去重
one_set = {10, 20, 33, 10, 22, 33, 20}
print(one_set)

# 第二种、创建集合,使用花括号set({值}),打印默认去重
two_set = set([10, 10, 20, 44, 10, 66, 44, 20])
print(two_set)

# 创建空集合
# 不能使用{}来创建空集合,默认创建的是空字典
empty_set = set()
print(type(empty_set))

执行结果:
{33, 10, 20, 22}
{10, 20, 66, 44}
<class 'set'>

示例:

one_set1 = set([10, 20, 33, 10, 22, 33, 20])
two_set1 = set({10, 10, 20, 44, 10, 66, 44, 20})

# 求交集,去重之后公共元素, & 及 .intersection(译:因特塞克神)
result_set1 = one_set1 & two_set1
result_set11 = one_set1.intersection(two_set1)
print(result_set1)      # 结果:{10, 20}
print(result_set11)     # 结果:{10, 20}

# 求并集,合并在一起,
result_set2 = one_set1 | two_set1
result_set22 = one_set1.union(two_set1)                   # (译:有你恩呢)
print(result_set2)      # 结果:{33, 66, 10, 44, 20, 22}
print(result_set22)     # 结果:{33, 66, 10, 44, 20, 22}


# 求差集,只在其中一个,不在另一个
result_set3 = one_set1 - two_set1
result_set33 = one_set1.difference(two_set1)             # (译:地府云词)
print(result_set3)          # 结果:{33, 22}
print(result_set33)        # 结果:{33, 22}


# 对称差集,要么在A,要么在B,没有公共元素
result_set4 = one_set1 ^ two_set1
result_set44 = one_set1.symmetric_difference(two_set1)
print(result_set4)          # 结果:{33, 66, 44, 22}
print(result_set44)         # 结果:{33, 66, 44, 22}

# 如何判断一个元素是否在其中
# 判断33是否在one_set1
a = 33 in one_set1

 

  

 

 

 

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posted @ 2020-04-20 22:36  守护往昔  阅读(500)  评论(0编辑  收藏  举报