Python basics: collections

Set (set): Different elements are grouped together to form a set, which is the basic data type of python. Collection classification: mutable set (set), immutable set (frozenset), the creation method is the same. Collection characteristics: disordered, unique, fast1. Create a set>>> s =
set('ian')>>> s{'a', 'n', 'i'}>>> len(s)3>>> li =
[' apple','pear','peach']>>> s =
set(li)>>> s{'peach', 'pear',
'apple'}> ;>> len(s)3 2. Accessing the collectionSince the collection itself is unordered, you cannot create an index or slice operation for the collection, you can only loop through or use in, not
in to access or determine collection elements. >>> 'apple' in
sTrue>>> 'banana'
in sFalse>>> for i in
s:... print(i)... peachpearapple 3. Update set and contentadd method: add the element to be passed in as a whole to the collection>>> a{'peach', 'pear',
'apple'}>>>
a.add('banana')>>> a{'peach', 'banana',
'pear', 'apple'} update method : Split the elements to be passed in and pass them into the set as individuals>>> a =
set(['apple'])>>> a{'apple'} >>>
a.update('peach')>>> a{'c', 'p', 'apple',
'h', 'e', ​​'a '} remove method: delete the specified element>>> a{'c', 'p', 'apple',
'h', 'e', ​​'a'}>>>
a.remove('apple')>>> a{'c', 'p', 'h', 'e',
'a'} pop method: delete an element and return> ;>> a{'h', 'e', ​​'a'}>>> c =
a.pop()>>> a{'e', 'a' }>>> c'h' #What is the difference between remove and pop? #remove just deletes the element, requires parameters, and has no return value. #pop takes out the element and assigns it, does not require parameters, and has a return value. 4. Set operator Intersection>>> ; a{'e', 'a'}>>> b{'c', 'a', 'b'}>>> a & b{'a'}>>> ;
a.intersection(b){'a'} Union (set)>>> a | b{'e', 'a', 'c', 'b'} Difference set> >>
a.difference(b)
#Who.difference, is the one in the loop{'e'}>>>
b.difference(a){'c', 'b'}>>>
a-b #Same difference{'e'}>>> b-a{'c', 'b'} Symmetry difference#You are not in me or I am not in either The values ​​in you, the two sets are cycled once each #It can also be understood as the union of the differences between the two sets>>>
a.symmetric_difference(b) #Which set is the same in the front{' c', 'e', ​​'b'}>>>
b.symmetric_difference(a){'e', 'c', 'b'}>>>
(a-b )|(b-a)
#The union of two sets of differences {'e', 'c', 'b'}

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