Python的組合沒有重複的約束

Question

我的元組(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")) 

我需要返回使用itertools combinations不復讀班的所有組合的元組。我怎樣才能返回不重複類的組合。例如，第一個返回的語句將是：tuple0, tuple2, tuple4, tuple5依此類推。

Answer 1

製作的基團（由類）：

>>> 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')]] 

然後，使用itertools.product以獲得所期望的結果：

>>> 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')) 

要獲得的組合的任何長度：

>>> 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'))