如何根据特定元素(在本例中为第2个元素)从2d列表中删除连续重复项.

我尝试了几种与itertools的组合,但没有运气.

任何人都可以建议我如何解决这个问题？

INPUT

192.168.1.232  >>>>>   173.194.36.64 , 14 , 15 , 16
192.168.1.232  >>>>>   173.194.36.64 , 14 , 15 , 17
192.168.1.232  >>>>>   173.194.36.119 , 23 , 30 , 31
192.168.1.232  >>>>>   173.194.36.98 , 24 , 40 , 41
192.168.1.232  >>>>>   173.194.36.98 , 24 , 40 , 62
192.168.1.232  >>>>>   173.194.36.74 , 25 , 42 , 43
192.168.1.232  >>>>>   173.194.36.74 , 25 , 42 , 65
192.168.1.232  >>>>>   173.194.36.74 , 26 , 44 , 45
192.168.1.232  >>>>>   173.194.36.74 , 26 , 44 , 66
192.168.1.232  >>>>>   173.194.36.78 , 27 , 46 , 47

OUTPUT

192.168.1.232  >>>>>   173.194.36.64 , 14 , 15 , 16
192.168.1.232  >>>>>   173.194.36.119 , 23 , 30 , 31
192.168.1.232  >>>>>   173.194.36.98 , 24 , 40 , 41
192.168.1.232  >>>>>   173.194.36.74 , 25 , 42 , 43
192.168.1.232  >>>>>   173.194.36.78 , 27 , 46 , 47

这是预期的产出.

UPDATE

上面给出的是列表的精美打印形式.

实际列表看起来像这样.

>>> for x  in connection_frame:
    print x


['192.168.1.232', '173.194.36.64', 14, 15, 16]
['192.168.1.232', '173.194.36.64', 14, 15, 17]
['192.168.1.232', '173.194.36.119', 23, 30, 31]
['192.168.1.232', '173.194.36.98', 24, 40, 41]
['192.168.1.232', '173.194.36.98', 24, 40, 62]
['192.168.1.232', '173.194.36.74', 25, 42, 43]
['192.168.1.232', '173.194.36.74', 25, 42, 65]
['192.168.1.232', '173.194.36.74', 26, 44, 45]
['192.168.1.232', '173.194.36.74', 26, 44, 66]
['192.168.1.232', '173.194.36.78', 27, 46, 47]
['192.168.1.232', '173.194.36.78', 27, 46, 67]
['192.168.1.232', '173.194.36.78', 28, 48, 49]
['192.168.1.232', '173.194.36.78', 28, 48, 68]
['192.168.1.232', '173.194.36.79', 29, 50, 51]
['192.168.1.232', '173.194.36.79', 29, 50, 69]
['192.168.1.232', '173.194.36.119', 32, 52, 53]
['192.168.1.232', '173.194.36.119', 32, 52, 74]

解决方法:

因此,您希望保留顺序并仅保留弹出连续条目,我不知道您可以使用任何花哨的内置.所以这是“蛮力”方法：

>>> remList = []
>>> for i in range(len(connection_frame)):
...     if (i != len(connection_frame)-)1 and (connection_frame[i][1] == connection_frame[i+1][1]):
...         remList.append(i)
...
for i in remList:
    connection_frame.pop(i)
['192.168.1.232', '173.194.36.119', 32, 52, 53]
['192.168.1.232', '173.194.36.79', 29, 50, 51]
['192.168.1.232', '173.194.36.78', 28, 48, 49]
['192.168.1.232', '173.194.36.78', 27, 46, 67]
['192.168.1.232', '173.194.36.78', 27, 46, 47]
['192.168.1.232', '173.194.36.74', 26, 44, 45]
['192.168.1.232', '173.194.36.74', 25, 42, 65]
['192.168.1.232', '173.194.36.74', 25, 42, 43]
['192.168.1.232', '173.194.36.98', 24, 40, 41]
['192.168.1.232', '173.194.36.64', 14, 15, 16]
>>>
>>> for conn in connection_frame:
...     print conn
...
['192.168.1.232', '173.194.36.64', 14, 15, 17]
['192.168.1.232', '173.194.36.119', 23, 30, 31]
['192.168.1.232', '173.194.36.98', 24, 40, 62]
['192.168.1.232', '173.194.36.74', 26, 44, 66]
['192.168.1.232', '173.194.36.78', 28, 48, 68]
['192.168.1.232', '173.194.36.79', 29, 50, 69]
['192.168.1.232', '173.194.36.119', 32, 52, 74]
>>>

或者,如果你想一次性完成所有这一切与列表理解：

>>> new_frame = [conn for conn in connection_frame if not connection_frame.index(conn) in [i for i in range(len(connection_frame)) if (i != len(connection_frame)-1) and (connection_frame[i][1] == connection_frame[i+1][1])]]
>>>
>>> for conn in new_frame:
...     print conn
...
['192.168.1.232', '173.194.36.64', 14, 15, 17]
['192.168.1.232', '173.194.36.119', 23, 30, 31]
['192.168.1.232', '173.194.36.98', 24, 40, 62]
['192.168.1.232', '173.194.36.74', 26, 44, 66]
['192.168.1.232', '173.194.36.78', 28, 48, 68]
['192.168.1.232', '173.194.36.79', 29, 50, 69]
['192.168.1.232', '173.194.36.119', 32, 52, 74]