set

It is the basic data type of python to combine different elements to form a set

Properties of the collection:

1. De duplication, a list programming set, will automatically de duplication
2. Relation test: test the intersection, difference, union and other relations between two sets of data
3. It doesn't need to mean there's no index, so you can't get a fixed value
4. So this irrelevant data type is the way we read the contents: for loop or iterator

s = set('yuyang is DBA')
s1 = ['yu','yang']
print(s)
print(set(s1))

Execution result:

{'u', 'g', 'y', ' ', 'i', 'a', 'D', 'B', 'A', 'n', 's'}
{'yang', 'yu'}

Process finished with exit code 0

As can be seen from the above execution results, if there are duplicate characters in the set, the variable will only appear once when printing

 

• A collection object is a set of hashable values that do not need to be arranged: collection members can be keys to a dictionary.

• Hashable value, that is, an immutable type

a = ['yuyang is DBA',[1,2]]
s = set(a)
print(s)

ERROR:

Traceback (most recent call last):
  File "H:/python_scripts/study_scripts/daily/day14/set.py", line 2, in <module>
    s = set(a)
TypeError: unhashable type: 'list'         #This means that the list is a non hashable data type

Process finished with exit code 1

If variable objects such as lists and dictionaries appear when assigning values to sets like the above, an error will be reported

 

 

Set classification:

1. Variable set: elements can be added or deleted. Non hashable elements cannot be dictionary keys or other set elements
2. Immutable set: opposite to variable set

Collection related operations

1. Create collection

s = set('yuyang is DBA')
s1 = ['yu','yang']
print(s)
print(set(s1))

Execution result:

{'u', 'g', 'y', ' ', 'i', 'a', 'D', 'B', 'A', 'n', 's'}
{'yang', 'yu'}

Process finished with exit code 0

2. Access collection

a = ['yuyang','is','DBA',1]
s = set(a)
print(s)

for i in s:
    print(i)

print('end'.center(30,'-'))
print('yuyang' in s)
print('jiayanping' in s)

Execution result:

{1, 'DBA', 'is', 'yuyang'}
1
DBA
is
yuyang
-------------end--------------
True
False

Process finished with exit code 0

The collection is unordered, so if you need to access the value, you need to use for loop traversal, or iterator to implement

 

 

3. Add elements to collection
set.add()

a = ['yuyang','is','DBA',1]
s = set(a)
print(s)

for i in s:
    print(i)

print('end'.center(30,'-'))
print('yuyang' in s)
print('jiayanping' in s)

s.add('jiayanping')
print(s)

Execution result:

{1, 'is', 'yuyang', 'DBA'}
1
is
yuyang
DBA
-------------end--------------
True
False
{1, 'is', 'yuyang', 'DBA', 'jiayanping'}

Process finished with exit code 0

set.update()

a = ['yuyang','is','DBA',1]
s = set(a)
print(s)

s.update('jiayanping')
print(s)

Execution result:

{1, 'DBA', 'is', 'yuyang'}
{1, 'DBA', 'yuyang', 'y', 'p', 'is', 'i', 'j', 'a', 'g', 'n'}

Process finished with exit code 0

update only updates, and does not empty the previous content

The difference between add and update:

• add is to append the object as a whole
• update is to decompose objects into multiple elements in a sequence and append them to the collection one by one

Example:

a = ['yuyang','is','DBA',1]
s = set(a)
print(s)

s.update([1,2,'jiayanping'])
print(s)

Execution result:

{'yuyang', 'is', 1, 'DBA'}
{1, 2, 'is', 'jiayanping', 'yuyang', 'DBA'}

Process finished with exit code 0

4. Delete elements from collection

set.remove()

a = ['yuyang','is','DBA',1]
s = set(a)
print(s)

s.remove('is')
print(s)

Execution result:

{'DBA', 1, 'yuyang', 'is'}
{'DBA', 1, 'yuyang'}

Process finished with exit code 0

set.pop()

a = ['yuyang','is','DBA',1]
s = set(a)
print(s)

s.pop()          #Randomly delete an element
print(s)

Execution result:

{'yuyang', 'DBA', 1, 'is'}
{'DBA', 1, 'is'}

Process finished with exit code 0

set.clear()

a = ['yuyang','is','DBA',1]
s = set(a)
print(s)

s.clear()
print(s)

Execution result:

{1, 'DBA', 'yuyang', 'is'}
set()                              #This is an empty set

Process finished with exit code 0

Set type operator
1.in and not in

2. Set is equivalent to nonequivalence (= =,! =)

print(set('yuyang') == set('yuyangyuyangyuyang'))

result:

True

Since the set will be de duplicated, the two quantities are equal here

3. Subset, superset

print(set('yuyang') < set('yuyang_jiayanping'))

result:

True

4. Joint

Relationship test

 

• intersection
• Difference set
• subset

a = set([1,2,3,4,5])
b = set([4,5,6])
print(a.union(b))            #Union
print(a | b)                 #Union
print(a.intersection(b))     #intersection
print(a.difference(b))       #Difference set
print(b.difference(a))
print(b.symmetric_difference(a))   #Symmetric difference set reverse intersection
print(a-b)                   #Difference set
print(b-a)                  #Difference set
print(a^b)                  #Symmetric difference set reverse intersection
print(b^a)                  #Symmetric difference set reverse intersection
print(a&b)                  #intersection
print(a.issuperset(b))      #Superset (parent set) a > b
print(a.issubset(b))        #Subset a < B

result:

{1, 2, 3, 4, 5, 6}
{1, 2, 3, 4, 5, 6}
{4, 5}
{1, 2, 3}
{6}
{1, 2, 3, 6}
{1, 2, 3}
{6}
{1, 2, 3, 6}
{1, 2, 3, 6}
{4, 5}
False
False

Process finished with exit code 0

The easiest way to de duplicate a list is to convert it to a collection