Question

我有三维列表。我需要删除重复的元素。是否有可能在没有遍历或更多pythonic方式的情况下过滤重复元素。在下面的例子中，890和930是重复的。

coords = [[(335, 270), (365, 270), (365, 410), (335, 410), (335, 270)],
          [(755, 270), (785, 270), (785, 410), (755, 410), (755, 270)], 
          [(615,250), (645, 250), (645, 390), (615, 390), (615, 250)],
          [(195, 250),(225, 250), (225, 390), (195, 390), (195, 250)], 
          [(890, -5), (930, -5), (930, 705), (890, 705), (890, -5)], 
          [(610, -5), (650, -5), (650, 705), (610, 705), (610, -5)],
          [(750, -5), (790, -5), (790, 705), (750, 705), (750, -5)],
          [(470, -5), (510, -5), (510, 705), (470, 705), (470, -5)], 
          [(890, -5), (930, -5), (930, 705), (890, 705), (890, -5)],
          [(610, -5), (650, -5), (650, 705), (610, 705), (610, -5)], 
          [(750, -5), (790, -5), (790, 705), (750, 705), (750, -5)],
          [(470, -5), (510, -5), (510, 705), (470, 705), (470, -5)],
          [(470, -5), (510, -5), (510, 705), (470, 705), (470, -5)],
          [(190, -5), (230, -5), (230, 705), (190, 705), (190, -5)], 
          [(330, -5), (370, -5), (370, 705), (330, 705), (330, -5)],
          [(50, -5), (90, -5), (90, 705), (50, 705), (50, -5)]]

Answer 1

您可以为list(set(sublist))中的每个列表致电coords。保证结果列表仅包含唯一元素，但不保留原始订单。

>>> [list(set(x)) for x in coords]
[[(365, 270), (335, 270), (365, 410), (335, 410)],
 [(785, 410), (755, 410), (785, 270), (755, 270)],
 [(645, 390), (615, 390), (645, 250), (615, 250)],
 [(225, 250), (195, 250), (225, 390), (195, 390)],
 [(930, -5), (890, -5), (890, 705), (930, 705)],
 [(650, -5), (610, -5), (610, 705), (650, 705)],
 [(750, -5), (790, -5), (790, 705), (750, 705)],
 [(470, -5), (510, -5), (510, 705), (470, 705)],
 [(930, -5), (890, -5), (890, 705), (930, 705)],
 [(650, -5), (610, -5), (610, 705), (650, 705)],
 [(750, -5), (790, -5), (790, 705), (750, 705)],
 [(470, -5), (510, -5), (510, 705), (470, 705)],
 [(470, -5), (510, -5), (510, 705), (470, 705)],
 [(190, -5), (230, -5), (230, 705), (190, 705)],
 [(370, -5), (330, -5), (330, 705), (370, 705)],
 [(90, -5), (50, -5), (50, 705), (90, 705)]]

如果您必须维护订单，则需要维护一个集合，将所有看到的元组添加到：

>>> result = []
>>> for subl in coords:
...    seen = set()
...    result.append([x for x in subl if x not in seen and not seen.add(x)])
...
>>> result
[[(335, 270), (365, 270), (365, 410), (335, 410)],
 [(755, 270), (785, 270), (785, 410), (755, 410)],
 [(615, 250), (645, 250), (645, 390), (615, 390)],
 [(195, 250), (225, 250), (225, 390), (195, 390)],
 [(890, -5), (930, -5), (930, 705), (890, 705)],
 [(610, -5), (650, -5), (650, 705), (610, 705)],
 [(750, -5), (790, -5), (790, 705), (750, 705)],
 [(470, -5), (510, -5), (510, 705), (470, 705)],
 [(890, -5), (930, -5), (930, 705), (890, 705)],
 [(610, -5), (650, -5), (650, 705), (610, 705)],
 [(750, -5), (790, -5), (790, 705), (750, 705)],
 [(470, -5), (510, -5), (510, 705), (470, 705)],
 [(470, -5), (510, -5), (510, 705), (470, 705)],
 [(190, -5), (230, -5), (230, 705), (190, 705)],
 [(330, -5), (370, -5), (370, 705), (330, 705)],
 [(50, -5), (90, -5), (90, 705), (50, 705)]]

在列表中设置操作列表并过滤重复的坐标

1 个答案: