How to find out who bought the same time IPhone7 and 8?
iphone7 = ['alex','rain','jack','old_driver'] iphone8 = ['alex','shanshan','jack','old_boy'] both_list = [] for name in iphone7: if name in iphone8: both_list.append(name) print(both_list)
Is a set of unordered not repeated combinations of data, its main role is as follows:
- De-emphasis, the list becomes a collection, automatically go heavy
- Relationship test, the intersection of the two groups before the test array, poor, union relations, etc.
S = {1,2,3,2,4,5 >>> } # create collections
>>> S
{ . 1, 2,. 3,. 4,. 5 } only a # 2
>>>
>>> s.add ( 2 ) add duplicate # 2 can not be put into
>>> s.add (44 is )
>>> S
{ . 1, 2,. 3,. 4,. 5, 44 is }
>>> s.update ([2,3,4 , 5,5,99 ]) # were added to the collection of a plurality of values
>>> S
{ . 1, 2,. 3,. 4,. 5, 99, 44 is }
>>> s.discard (. 1 ) # remove elements, also not no error
>>> s.pop () # delete a random element is empty, the collection being given
2
>>> s.pop ()
. 3
>>> S
{ . 4,. 5, 99, 44 is }
>>>
>>> s.clear () # Empty
>>> S
SET ()
>>> s.add (. 1)
>>> s.copy ()
{}. 1
>>> s.add ([l, 2,3 ]) # immutable data can add
Traceback (MOST Recent Last Call):
File "<stdin>", Line. 1, in <Module1>
TypeError: unhashable type: 'List'
Set relationship test
- Intersection
- Difference set
- Union
>>> iphone7 = {'alex','rain','jack','old_driver'}
>>> iphone8 = {'alex','jack','shanshan','old_boy'}
Union
>>> iphone7.intersection(iphone8)
{'jack', 'alex'}
>>> iphone7 & iphone8
{'jack', 'alex'}
Difference set
>>> iphone7.difference(iphone8)
{'rain', 'old_driver'}
>>> iphone7 - iphone8
{'rain', 'old_driver'}
Union
>>> iphone8.union(iphone7)
{'old_boy', 'jack', 'shanshan', 'rain', 'old_driver', 'alex'}
>>> iphone8 | iphone7
{'old_boy', 'jack', 'shanshan', 'rain', 'old_driver', 'alex'}
Symmetric difference: only buy or only bought iphone8 iphone7 person, and taking the difference counter set
>>> iphone8.symmetric_difference(iphone7)
{'old_boy', 'rain', 'shanshan', 'old_driver'}
>>> iphone8 ^ iphone7
{'old_boy', 'rain', 'shanshan', 'old_driver'}
>>>
Inclusion relation
in, not in: == determine whether the elements in the collection! Analyzing two sets are equal =
Usually between two sets of three relationships, intersection, comprising disjoint.
set.isdisjoint (s): Analyzing two sets are not disjoint
set.issuperset (s): determining a set of other sets is not contained, it is equivalent to a> = b
set.issubset (s): determined set is not part of the other set, and so on for a <= b
>>> s
{1, 2, 3, 4}
>>> s2
{2, 3, 5, 6}
>>> s2.add(1)
>>> s2.add(4)
>>> s2
{1, 2, 3, 4, 5, 6}
>>> s.issubset(s2)
True
>>> s2.issuperset(s)
True
>>>