一、集合類型內置方法(set)
集合可以理解成一個集合體,學習Python的學生可以是一個集合體;學習Linux的學生可以是一個集合體。
pythoners = ['jason', 'nick', 'tank', 'sean']
linuxers = ['nick', 'egon', 'kevin']
# 即報名pythoners又報名linux的學生
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']
上述的列表方式求兩個集合體的關系運算非常復雜,因此有了我們的集合數據類型。
1.用途:用於關系運算的集合體,由於集合內的元素無序且集合元素不可重復,因此集合可以去重,但是去重后的集合會打亂原來元素的順序。
2.定義:{}內用逗號分隔開多個元素,每個元素必須是不可變類型。
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+內置方法:常用操作和內置方法分為優先掌握(今天必須得記住)、需要掌握(一周內記住)兩個部分。
1.1 優先掌握(*****)
-
長度len
-
成員運算in和not in
-
|並集、union
-
&交集、intersection
-
-差集、difference
-
^對稱差集、symmetric_difference
-
==
-
父集:>、>= 、issuperset
-
子集:<、<= 、issubset
1.長度len
# set之長度len
s = {1, 2, 'a'}
print(f"len(s): {len(s)}")
len(s): 3
2.成員運算in和not in
# set之成員運算in和not in
s = {1, 2, 'a'}
print(f"1 in s: {1 in s}")
1 in s: True
3.|並集
# str之|並集
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.^對稱差集
# str之^對稱差集
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) # 報錯
s.discard(3)
print(s)
{1, 2, 'a'}
5.isdisjoint()
# set之isdisjoint(),集合沒有共同的部分返回True,否則返回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、練習
有如下列表,列表元素為不可hash類型,去重,得到新列表,且新列表一定要保持列表原來的順序
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.存一個值or多個值:多個值,且值為不可變類型。
5.有序or無序:無序
s = {1, 2, 'a'}
print(f'first:{id(s)}')
s.add(3)
print(f'second:{id(s)}')
first:4480523848
second:4480523848
6.可變or不可變:可變數據類型