集合类型内置方法

集合类型内置方法

一、 集合类型内置方法(set)

集合可以理解成一个集合体，学习Python的学生可以是一个集合体；学习Linux的学生可以是一个集合体。

pythoners = ['lh', 'nick', 'hello', 'word']
linuxers = ['nick', 'handsome', 'hi']

# 即报名pythoners又报名lniuxers学生
py_li_list = []
for stu in pythoners:
    if stu in linuxers:
        py_li_list.append(stu)
print(f"pythoners and linuxers: {py_li_list}")

pythoners and linuxers: ['nick']

上述的列表方式求两个集合体的关系运算非常复杂，因此有了我们的集合数据类型。

1.用途：用于关系运算的集合体，由于集合内的元素无序且集合元素不可重复，因此集合可以去重，但是去重后的集合会打乱原来元素的顺序。

2.定义：{}内用逗号分隔开多个元素，每个元素必须是不可变类型。

s = {1, 2, 1, 'a'}

print(f"s: {s}")

s: {1, 2, 'a'}

s = {1, 2, 1, 'a', 'c'}

for i in s:
    print(i)

c
1
2
a

s = set('hello')

print(f"s: {s}")

s: {'h', 'e', 'o', 'l'}

3.常用操作+内置方法：常用操作和内置方法分为优先掌握（今天必须得记住）、需要掌握（一周内记住）两个部分。

1.1优先掌握(*************)

1. 长度len
2. 成员运算in和not in
3. |并集、union
4. &交集、intersection
5. -差集、difference
6. ^对称差集、symmetric_difference
7. ==
8. 父集：>、>= 、issuperset
9. 子集：<、<= 、issubset

1.长度len

s = {1, 2, 'a'}

print(f"len(s): {len(s)}")

len(s): 3

2.成员运算in和not in

# set之成员运算in和not in
s = {1, 2, 'a'}

print(f"1 in s: {1 in s}")

1 in s: True

3.|并集

# str之|并集
pythoners = {'lh', 'nick', 'hello', 'handsome'}
linuxers = {'nick', 'word', 'hi'}

print(f"pythoners|linuxers:{pythoners | linuxers}")
print(f"pythoners.union(linuxers): {pythoners.union(linuxers)}")

pythoners|linuxers:{'lh', 'hello', 'hi', 'nick', 'word', 'handsome'}
pythoners.union(linuxers): {'lh', 'hello', 'hi', 'nick', 'word', 'handsome'}

4.&交集

# str之&交集
pythoners = {'lh', 'nick', 'hello', 'handsome'}
linuxers = {'nick', 'word', 'hi'}

print(f"pythoners-linuxers:{pythoners & linuxers}")
print(f"pythoners.intersection(linuxers): {pythoners.intersection(linuxers)}")

pythoners-linuxers:{'nick'}
pythoners.intersection(linuxers): {'nick'}

5.-差集

# str之-差集
pythoners = {'lh', 'nick', 'hello', 'handsome'}
linuxers = {'nick', 'word', 'hi'}

print(f"pythoners-linuxers:{pythoners - linuxers}")
print(f"pythoners.difference(linuxers): {pythoners.difference(linuxers)}")

pythoners-linuxers:{'handsome', 'lh', 'hello'}
pythoners.difference(linuxers): {'handsome', 'lh', 'hello'}

6.^对称差集

# str之^对称差集
pythoners = {'lh', 'nick', 'hello', 'handsome'}
linuxers = {'nick', 'word', 'hi'}

print(f"pythoners^linuxers:{pythoners ^ linuxers}")
print(f"pythoners.symmetric_difference(linuxers): {pythoners.symmetric_difference(linuxers)}")

pythoners^linuxers:{'word', 'hi', 'handsome', 'lh', 'hello'}
pythoners.symmetric_difference(linuxers): {'word', 'hi', 'handsome', 'lh', 'hello'}

7.==

# str之==
pythoners = {'lh', 'nick', 'hello', 'handsome'}
linuxers = {'nick', 'word', 'hi'}
javers = {'nick', 'word', 'hi'}

print(f"pythoners==linuxers:{pythoners == linuxers}")
print(f"javers==linuxers: {javers == linuxers}")

pythonerslinuxers:False
javerslinuxers: True

8.父集：>、>=

# str之父集：>、>=
pythoners = {'lh', 'nick', 'hello', 'handsome'}
linuxers = {'nick', 'word', 'hi'}
javaers = {'nick', 'lh'}

print(f"pythoners>linuxers: {pythoners > linuxers}")
print(f"pythoners>=linuxers: {pythoners >= linuxers}")
print(f"pythoners>=javaers: {pythoners >= javaers}")
print(f"pythoners.issuperset(javaers): {pythoners.issuperset(javaers)}")

pythoners>linuxers: False
pythoners>=linuxers: False
pythoners>=javaers: True
pythoners.issuperset(javaers): True

9.子集：<、<=

# str之子集：<、<=
pythoners = {'lh', 'nick', 'hello', 'handsome'}
linuxers = {'nick', 'word', 'hi'}
javaers = {'nick', 'lh'}

print(f"pythoners<linuxers: {pythoners < linuxers}")
print(f"pythoners<=linuxers: {pythoners <= linuxers}")
print(f"javaers.issubset(javaers): {javaers.issubset(javaers)}")

pythoners<linuxers: False
pythoners<=linuxers: False
javaers.issubset(javaers): True

1.2需要掌握(************)

1. add
2. remove
3. difference_update
4. discard
5. isdisjoint

1.add

# set之add()
s = {1, 2, 'a'}
s.add(3)

print(s)

{1, 2, 3, 'a'}

2.remove()

# set之remove()
s = {1, 2, 'a'}
s.remove(1)

print(s)

{2, 'a'}

3.difference_update()

# str之difference_update()
pythoners = {'lh', 'nick', 'hello', 'handsome'}
linuxers = {'nick', 'word', 'hi'}

pythoners.difference_update(linuxers)

print(f"pythoners.difference_update(linuxers): {pythoners}")

pythoners.difference_update(linuxers): {'lh', 'hello', 'handsome'}

4.discard()

# set之discard()
s = {1, 2, 'a'}
# remove(3)  # 报错
s.discard(3)

print(s)

{1, 2, 'a'}

5.isdisjoint()

# set之isdisjoint()，集合没有共同的部分返回True，否则返回False
pythoners = {'lh', 'nick', 'hello', 'handsome'}
linuxers = {'nick', 'word', 'hi'}

pythoners.isdisjoint(linuxers)

print(f"pythoners.isdisjoint(linuxers): {pythoners.isdisjoint(linuxers)}")

pythoners.isdisjoint(linuxers): False