A collection set type
1 role
Relational operation, Deduplication
Use for relational operators can do, it is too much trouble.
l1=[1,2,3,4,5]
l2=[3,3,4,5,6]
l=[]
for x in l1:
if x in l2:
l.append(x)
print(x)2 is defined: {} in the plurality of elements separated by commas, the collection element has three characteristics:
1. the collection of elements are immutable type
2. unordered collection of elements within
3. the set of elements is not repeated
Definition of the empty set: s = set (), braces default empty dictionary.
3 type conversion
set({1,2,3})
res=set('helllllllo')
dict=({'k1':1,'k2':2})
s=set(dict)
print(res,s)Built Method 4 - Relational Operators
4.1 intersected
friend1={1,2,3,4,5}
friend2={3,4,5,6,7,8}
print(friend1 & friend2)4.2 Take union
print(friend1 | friend2)4.3 taking the difference current
print(friend1 - friend2)4.4 Take symmetric difference
print(friend1 ^ friend2)4.5 father and son set
When set S1 is a subset of said S2, S1 <S2
Built Method 5 - deduplication
Features: only for immutable types, can not guarantee that the original order.
General to re-write their own programs according to the situation, python can not provide all the solutions to problems
l=[
{'name':'lili','age':18,'sex':'male'},
{'name':'jack','age':73,'sex':'male'},
{'name':'tom','age':20,'sex':'female'},
{'name':'lili','age':18,'sex':'male'},
{'name':'lili','age':18,'sex':'male'},
]
new_l=[]
for dic in l:
if dic not in new_l:
new_l.append(dic)6 Other built-in method
6.1 .discard in parentheses is the value you want to delete (), delete the element does not exist if no error
s={1,2,3}
# s.discard(4) # 删除元素不存在do nothing
# print(s)
# s.remove(4) # 删除元素不存在则报错6.2 .update () to add a new collection, .add () to add a new element
# s.update({1,3,5})
# print(s)
# s.add(4)
# print(s)
6.3 .pop () removes an element of random
# s.update({1,3,5})
# print(s)