set性质
可变的、无序的、不重复的元素的集合
set定义和初始化
语法：
- set() -> 空集合
- set(iterable) -> set对象
In [2]: s1=set() In [3]: s1 Out[3]: set() In [4]: s2=set(range(5)) In [5]: s2 Out[5]: {0, 1, 2, 3, 4}
In [6]: s1=set(enumerate(range(5))) In [7]: s1 Out[7]: {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4)} In [9]: s2={(1,2),'b',None,True,b'abc'}
set中的元素必须可hash（可变类型基本都不可哈希）
In [12]: s3={bytearray(b'ab'),{1},'a'} --------------------------------------------------------------------------- TypeError Traceback (most recent call last) <ipython-input-12-c756f10834fb> in <module> ----> 1 s3={bytearray(b'ab'),{1},'a'} TypeError: unhashable type: 'bytearray' In [8]: s2={(1,2),[1,'a',(4,1)],'b',None,True,b'abc'} --------------------------------------------------------------------------- TypeError Traceback (most recent call last) <ipython-input-8-984c42c3f9da> in <module> ----> 1 s2={(1,2),[1,'a',(4,1)],'b',None,True,b'abc'} TypeError: unhashable type: 'list' In [11]: s3={{1},'a'} --------------------------------------------------------------------------- TypeError Traceback (most recent call last) <ipython-input-11-18225384901d> in <module> ----> 1 s3={{1},'a'} TypeError: unhashable type: 'set'
set方法
set增加元素
add(element)
In [19]: s.add(range(5)) In [20]: s Out[20]: {1, range(0, 5)} In [22]: s.add(tuple(range(5))) In [23]: s Out[23]: {(0, 1, 2, 3, 4), 1, range(0, 5)}
update(*others) 将其他可迭代对象的元素到本集合中，就地修改
In [25]: s Out[25]: {(0, 1, 2, 3, 4), 1, range(0, 5)} In [26]: s.update({1,2}) In [27]: s Out[27]: {(0, 1, 2, 3, 4), 1, 2, range(0, 5)} In [28]: s.update({1,2},{2,1,3}) In [29]: s Out[29]: {(0, 1, 2, 3, 4), 1, 2, 3, range(0, 5)}
set删除元素
remove(elem) 移除一个元素，如果元素不存在，抛出KeyError异常
In [32]: s.remove(1) In [33]: s Out[33]: {(0, 1, 2, 3, 4), 2, 3, 5, range(0, 5)} In [34]: s.remove(10) --------------------------------------------------------------------------- KeyError Traceback (most recent call last) <ipython-input-34-23b5f1bb77c8> in <module> ----> 1 s.remove(10) KeyError: 10
discard(elem) 移除一个元素，如果元素不存在，不抛异常
In [35]: s Out[35]: {(0, 1, 2, 3, 4), 2, 3, 5, range(0, 5)} In [36]: s.discard(10) In [37]: s Out[37]: {(0, 1, 2, 3, 4), 2, 3, 5, range(0, 5)} In [38]: s.discard(2) In [39]: s Out[39]: {(0, 1, 2, 3, 4), 3, 5, range(0, 5)}
pop() ->item 移除任意一个元素，并返回该元素，如果没有元素，返回KeyError
In [39]: s Out[39]: {(0, 1, 2, 3, 4), 3, 5, range(0, 5)} In [40]: s.pop() Out[40]: range(0, 5) In [41]: s.pop() Out[41]: 3 In [42]: s.pop() Out[42]: (0, 1, 2, 3, 4) In [43]: s.pop() Out[43]: 5 In [44]: s.pop() --------------------------------------------------------------------------- KeyError Traceback (most recent call last) <ipython-input-44-c88c8c48122b> in <module> ----> 1 s.pop() KeyError: 'pop from an empty set'
clear() 移除所有元素
In [45]: s2 Out[45]: {(1, 2), None, True, 'b', b'abc'} In [46]: s2.clear() In [47]: s2 Out[47]: set()
set没有修改操作，要么增加，要么删除，set可以遍历，但不能被索引
set成员运算符比较和效率问题
列表的成员运算
In [53]: %%timeit lst1=list(range(100)) ...: -1 in lst1 ...: ...: 812 ns ± 31.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each) In [54]: %%timeit lst2=list(range(10000)) ...: -1 in lst2 ...: ...: 78.3 µs ± 624 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
集合的成员运算
In [51]: %%timeit s1=set(range(100)) ...: -1 in s1 ...: ...: 24 ns ± 0.159 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each) In [52]: %%timeit s2=set(range(10000)) ...: -1 in s2 ...: ...: 24.1 ns ± 0.247 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
集合运算
并集：将多个集合的元素合并到一起
union(*other) 返回多个可迭代对象合并后的新集合
In [55]: s1 Out[55]: {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4)} In [56]: s2 Out[56]: set() In [57]: s2={1,2,'abc'} In [58]: s1.union(s2) Out[58]: {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), 1, 2, 'abc'} In [60]: s1.union((3,b'abc'),{'a','b'}) Out[60]: {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), 3, 'a', 'b', b'abc'}
| 运算符重载 和 union相似
In [64]: s1|s2 Out[64]: {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), 1, 2, 'abc'}
update(*others) 前面写过了，就地修改
|= 和update相似
In [65]: s1|=s2 In [66]: s1 Out[66]: {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), 1, 2, 'abc'} In [67]: s2 Out[67]: {1, 2, 'abc'}
交集：求两个集合中相同的元素
intersection(*others) 返回多个集合的交集
In [25]: s1 Out[25]: {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), 1, '1', 2, '2', '3', 'abc'} In [26]: s2 Out[26]: {(2, 2), 1, 2, 'abc'} In [27]: s3 Out[27]: {'abc'} In [28]: s1.intersection(s2,s3) Out[28]: {'abc'}
intersection_update(*others)
In [29]: s1.intersection_update(s2) In [30]: s1 Out[30]: {(2, 2), 1, 2, 'abc'}
& 等价intersection
In [31]: s1 & s2 Out[31]: {(2, 2), 1, 2, 'abc'}
&= 等价intersection_update
In [35]: s1 &=s3 In [37]: s3 Out[37]: {'abc'} In [36]: s1 Out[36]: {'abc'}
差集：获取多个集合中不同的元素组成的集合
difference(*ohters)
In [38]: s1={(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), 1, '1', 2, '2', '3', 'abc'} In [39]: s2 Out[39]: {(2, 2), 1, 2, 'abc'} In [40]: s3 Out[40]: {'abc'} In [41]: s1.difference(s3) Out[41]: {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), 1, '1', 2, '2', '3'}
difference_update(*others)
In [45]: s1.difference_update(s3) In [46]: s1 Out[46]: {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), 1, '1', 2, '2', '3'}
"-" 等价于difference
In [49]: s1 Out[49]: {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), 1, '1', 2, '2', '3'} In [50]: s2 Out[50]: {(2, 2), 1, 2, 'abc'} In [51]: s1 - s2 Out[51]: {(0, 0), (1, 1), (3, 3), (4, 4), '1', '2', '3'}
"-=" 等价于difference_update
In [49]: s1 Out[49]: {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), 1, '1', 2, '2', '3'} In [50]: s2 Out[50]: {(2, 2), 1, 2, 'abc'} In [51]: s1 - s2 Out[51]: {(0, 0), (1, 1), (3, 3), (4, 4), '1', '2', '3'} In [52]: s1.difference_update(s2) In [53]: s1 Out[53]: {(0, 0), (1, 1), (3, 3), (4, 4), '1', '2', '3'}
对称差集：获取两个集合的差集，等价于(s1 -s2) ∪（s2-s1）
symmetric_difference(other)
In [55]: s1 Out[55]: {(0, 0), (1, 1), (3, 3), (4, 4), '1', '2', '3'} In [56]: s2 Out[56]: {(2, 2), 1, 2, 'abc'} In [57]: s1.symmetric_difference(s2) Out[57]: {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), 1, '1', 2, '2', '3', 'abc'}
symmetric_difference_update(other)
In [62]: s1={(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), 1, '1', 2, '2', '3', 'abc'} In [63]: s2 Out[63]: {(2, 2), 1, 2, 'abc'} In [64]: s1.symmetric_difference_update(s2) Out[64]: {(0, 0), (1, 1), (3, 3), (4, 4), '1', '2', '3'}
"^" 等价于symmetric_difference(other)
In [67]: s1={(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), 1, '1', 2, '2', '3', 'abc'} In [68]: s2 Out[68]: {(2, 2), 1, 2, 'abc'} In [69]: s1 ^ s2 Out[69]: {(0, 0), (1, 1), (3, 3), (4, 4), '1', '2', '3'}
"^=" 等价于symmetric_difference_update(other)
In [70]: s1 ^= s2 In [71]: s1 Out[71]: {(0, 0), (1, 1), (3, 3), (4, 4), '1', '2', '3'}
集合判断
issubset(other) , <= 判断某个集合是否是另一个集合的子集
In [75]: s1={(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), 1, '1', 2, '2', '3', 'abc'} In [76]: s2 Out[76]: {(2, 2), 1, 2, 'abc'} In [77]: s1 <= s2 Out[77]: False In [78]: s1.issubset(s2) Out[78]: False In [79]: s2.issubset(s1) Out[79]: True
set1 < set2 判断set1是否是set2的真子集
In [80]: s2 < s1 Out[80]: True In [81]: s1 < s2 Out[81]: False
issuperset(other)、>= 判断集合是否是other集合的超集
In [3]: s1.issuperset(s2) Out[3]: True In [4]: s2>=s1 Out[4]: False
set1 > set2 判断set1是否是set2的真超集
In [5]: s2 > s1 Out[5]: False
isdisjoint(other) 当前集合和other集合是否有交集，没有则返回True
In [5]: s2 > s1 Out[5]: False