Python组合不按约束重复

Question

我有一个元组(Name, val 1, val 2, Class)

tuple = (("Jackson",10,12,"A"),
         ("Ryan",10,20,"A"),
         ("Michael",10,12,"B"),
         ("Andrew",10,20,"B"),
         ("McKensie",10,12,"C"),
         ("Alex",10,20,"D"))

我需要使用itertools combinations返回所有不重复类的组合。如何返回不重复类的组合。例如，第一个返回的语句是：tuple0, tuple2, tuple4, tuple5，依此类推。你知道吗

Answer 1

分组（按班级）：

>>> ts = (("Jackson",10,12,"A"),
...       ("Ryan",10,20,"A"),
...       ("Michael",10,12,"B"),
...       ("Andrew",10,20,"B"),
...       ("McKensie",10,12,"C"),
...       ("Alex",10,20,"D"))
>>> import itertools
>>> import operator
>>>
>>> by_class = operator.itemgetter(3)
>>>
>>> tuple_grps = [list(grp) for key, grp in itertools.groupby(sorted(ts, key=by_class), key=by_class)]
>>> tuple_grps
[[('Jackson', 10, 12, 'A'), ('Ryan', 10, 20, 'A')],
 [('Michael', 10, 12, 'B'), ('Andrew', 10, 20, 'B')],
 [('McKensie', 10, 12, 'C')],
 [('Alex', 10, 20, 'D')]]

然后，使用^{}获得所需的结果：

>>> for xs in itertools.product(*tuple_grps):
...     print(xs)
...
(('Jackson', 10, 12, 'A'), ('Michael', 10, 12, 'B'), ('McKensie', 10, 12, 'C'), ('Alex', 10, 20, 'D'))
(('Jackson', 10, 12, 'A'), ('Andrew', 10, 20, 'B'), ('McKensie', 10, 12, 'C'), ('Alex', 10, 20, 'D'))
(('Ryan', 10, 20, 'A'), ('Michael', 10, 12, 'B'), ('McKensie', 10, 12, 'C'), ('Alex', 10, 20, 'D'))
(('Ryan', 10, 20, 'A'), ('Andrew', 10, 20, 'B'), ('McKensie', 10, 12, 'C'), ('Alex', 10, 20, 'D'))

要获得任意长度的组合：

>>> for i in range(1, len(tuple_grps)+1):
...     for xs in itertools.combinations(tuple_grps, i):
...         for ys in itertools.product(*xs):
...             print(ys)
...
(('Jackson', 10, 12, 'A'),)
(('Ryan', 10, 20, 'A'),)
(('Michael', 10, 12, 'B'),)
(('Andrew', 10, 20, 'B'),)
(('McKensie', 10, 12, 'C'),)
(('Alex', 10, 20, 'D'),)
(('Jackson', 10, 12, 'A'), ('Michael', 10, 12, 'B'))
(('Jackson', 10, 12, 'A'), ('Andrew', 10, 20, 'B'))
(('Ryan', 10, 20, 'A'), ('Michael', 10, 12, 'B'))
(('Ryan', 10, 20, 'A'), ('Andrew', 10, 20, 'B'))
...
(('Ryan', 10, 20, 'A'), ('Andrew', 10, 20, 'B'), ('McKensie', 10, 12, 'C'), ('Alex', 10, 20, 'D'))