目录
- 一、集合类型内置方法(set)
- 1.1 优先掌握(*****)
- 1.2 需要掌握(****)
- 二、练习
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 = ['jason', 'nick', 'tank', 'sean']
linuxers = ['nick', 'egon', 'kevin']
# 即报名pythoners又报名linux的学生
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'} # s = set({1,2,'a'})
print(f"s: {s}")
s: {1, 2, 'a'}
s = {1, 2, 1, 'a', 'c'}
for i in s:
print(i)
1 2 c a
s = set('hello')
print(f"s: {s}")
s: {'e', 'o', 'h', '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(*****)
长度len
成员运算in和not in
|并集、union
&交集、intersection
-差集、difference
^对称差集、symmetric_difference
==
父集：>、>= 、issuperset
子集：<、<= 、issubset
1.长度len
# set之长度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 = {'jason', 'nick', 'tank', 'sean'}
linuxers = {'nick', 'egon', 'kevin'}
print(f"pythoners|linuxers: {pythoners|linuxers}")
print(f"pythoners.union(linuxers): {pythoners.union(linuxers)}")
pythoners|linuxers: {'egon', 'tank', 'kevin', 'jason', 'nick', 'sean'}
pythoners.union(linuxers): {'egon', 'tank', 'kevin', 'jason', 'nick', 'sean'}
4.&交集
# str之&交集
pythoners = {'jason', 'nick', 'tank', 'sean'}
linuxers = {'nick', 'egon', 'kevin'}
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 = {'jason', 'nick', 'tank', 'sean'}
linuxers = {'nick', 'egon', 'kevin'}
print(f"pythoners-linuxers: {pythoners-linuxers}")
print(f"pythoners.difference(linuxers): {pythoners.difference(linuxers)}")
pythoners-linuxers: {'tank', 'jason', 'sean'}
pythoners.difference(linuxers): {'tank', 'jason', 'sean'}
6.^对称差集
# str之^对称差集
pythoners = {'jason', 'nick', 'tank', 'sean'}
linuxers = {'nick', 'egon', 'kevin'}
print(f"pythoners^linuxers: {pythoners^linuxers}")
print(
f"pythoners.symmetric_difference(linuxers): {pythoners.symmetric_difference(linuxers)}")
pythoners^linuxers: {'egon', 'tank', 'kevin', 'jason', 'sean'}
pythoners.symmetric_difference(linuxers): {'egon', 'tank', 'kevin', 'jason', 'sean'}
7.==
# str之==
pythoners = {'jason', 'nick', 'tank', 'sean'}
linuxers = {'nick', 'egon', 'kevin'}
javers = {'nick', 'egon', 'kevin'}
print(f"pythoners==linuxers: {pythoners==linuxers}")
print(f"javers==linuxers: {javers==linuxers}")
pythoners==linuxers: False javers==linuxers: True
8.父集：>、>=
# str之父集：>、>=
pythoners = {'jason', 'nick', 'tank', 'sean'}
linuxers = {'nick', 'egon', 'kevin'}
javaers = {'jason', 'nick'}
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 = {'jason', 'nick', 'tank', 'sean'}
linuxers = {'nick', 'egon', 'kevin'}
javaers = {'jason', 'nick'}
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(****)
add
remove
difference_update
discard
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 = {'jason', 'nick', 'tank', 'sean'}
linuxers = {'nick', 'egon', 'kevin'}
pythoners.difference_update(linuxers)
print(f"pythoners.difference_update(linuxers): {pythoners}")
pythoners.difference_update(linuxers): {'tank', 'jason', 'sean'}
4.discard()
# set之discard()
s = {1, 2, 'a'}
# s.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 = {'jason', 'nick', 'tank', 'sean'}
linuxers = {'nick', 'egon', 'kevin'}
pythoners.isdisjoint(linuxers)
print(f"pythoners.isdisjoint(linuxers): {pythoners.isdisjoint(linuxers)}")
pythoners.isdisjoint(linuxers): False
二、练习
有如下列表，列表元素为不可hash类型，去重，得到新列表，且新列表一定要保持列表原来的顺序
stu_info_list = [
{'name':'nick','age':19,'sex':'male'},
{'name':'egon','age':18,'sex':'male'},
{'name':'tank','age':20,'sex':'female'},
{'name':'tank','age':20,'sex':'female'},
{'name':'egon','age':18,'sex':'male'},
]
stu_info_list = [
{'name': 'nick', 'age': 19, 'sex': 'male'},
{'name': 'egon', 'age': 18, 'sex': 'male'},
{'name': 'tank', 'age': 20, 'sex': 'female'},
{'name': 'tank', 'age': 20, 'sex': 'female'},
{'name': 'egon', 'age': 18, 'sex': 'male'},
]
new_stu_info_list = []
for stu_info in stu_info_list:
if stu_info not in new_stu_info_list:
new_stu_info_list.append(stu_info)
for new_stu_info in new_stu_info_list:
print(new_stu_info)
{'name': 'nick', 'age': 19, 'sex': 'male'}
{'name': 'egon', 'age': 18, 'sex': 'male'}
{'name': 'tank', 'age': 20, 'sex': 'female'}
4.存一个值or多个值：多个值，且值为不可变类型。
5.有序or无序：无序
s = {1, 2, 'a'}
print(f'first:{id(s)}')
s.add(3)
print(f'second:{id(s)}')
first:4480523848 second:4480523848
6.可变or不可变：可变数据类型