从 2D 列表中删除连续重复项，python？
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【中文标题】从 2D 列表中删除连续重复项，python？【英文标题】：Remove consecutive duplicates from a 2D list , python? 【发布时间】：2014-04-19 21:15:43 【问题描述】：如何根据特定元素（在本例中为第二个元素）从 2d 列表中删除连续重复项。
我尝试了一些与 itertools 的组合，但没有运气。
谁能建议我如何解决这个问题？
输入
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
输出
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
这是预期的输出。
更新
上面给出的是一个打印得很好的列表形式。
实际列表如下所示。
>>> 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]
【问题讨论】：
看看itertools.groupby。
您只需将它们打成一组即可删除重复项
您使用的实际数据类型是什么。这些行是字符串、元组等吗？
我也相信 OP 只希望删除与某些其他元素重复的元素。不仅仅是没有重复。
@thecreator232，如果顺序无关紧要，我们怎么能有有意义的连续条目？
【参考方案1】：
所以因为你想保持顺序并且只弹出连续的条目，我不知道你可以使用任何花哨的内置。所以这里是“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，如果您想通过列表理解一次性完成所有操作：
>>> 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]
【讨论】：
@thecreator232，你需要改变什么？让我知道，以便我可以在这里更新它 'if (connection_frame[i][1] == connection_frame[i+1][1]) and (connection_frame[i][2] == connection_frame[i+1][2] ) : connection_frame.remove(connection_frame[i+1])' @thecreator232 好的，我明白你在做什么。在您迭代列表时，我会小心更改列表，您可能会得到一些奇怪的结果，而且它通常被认为是糟糕的形式，所以我不会在这里更改它 感谢您的提醒。【参考方案2】：使用itertools.groupby():
import itertools
data = """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""".split("\n")
for k, g in itertools.groupby(data, lambda l:l.split()[2]):
print next(g)
打印出来
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
（这使用了一个字符串列表，但很容易适应列表列表。）
【讨论】：
如果 OP 的数据结构确实是字符串列表，则此方法有效，但他只是表示它可能是字符串列表（可能是[['192.168.1.1', '>>>>', ...], ...]，这会使答案稍微复杂一些。是的，它绝对会删除不连续的重复项。
@aruisdante：答案末尾有一条关于此的评论（您可能需要重新加载才能看到它）。
是的，看到它，但这将删除不连续的重复，这不是 OP 想要的。
@aruisdante：不，这不会删除不连续的重复项。我在这里错过了什么吗？
@aruisdante：如果data 是列表列表，那么它是相同的：result = (next(g) for _, g in groupby(data, key=lambda x: x[1]))【参考方案3】：
Pandas.groupby 是itertools.groupby 的替代品，它还允许您跟踪原始列表的连续/非连续元素 --- 通过提供行号而不是迭代器。像这样的：
df = pandas.DataFrame(connection_frame)
print df
Out:
0 1 2 3 4
0 '192.168.1.232' '173.194.36.64' 14 15 16
1 '192.168.1.232' '173.194.36.64' 14 15 17
2 '192.168.1.232' '173.194.36.119' 23 30 31
3 '192.168.1.232' '173.194.36.98' 24 40 41
4 '192.168.1.232' '173.194.36.98' 24 40 62
5 '192.168.1.232' '173.194.36.74' 25 42 43
6 '192.168.1.232' '173.194.36.74' 25 42 65
7 '192.168.1.232' '173.194.36.74' 26 44 45
8 '192.168.1.232' '173.194.36.74' 26 44 66
9 '192.168.1.232' '173.194.36.78' 27 46 47
10 '192.168.1.232' '173.194.36.78' 27 46 67
11 '192.168.1.232' '173.194.36.78' 28 48 49
12 '192.168.1.232' '173.194.36.78' 28 48 68
13 '192.168.1.232' '173.194.36.79' 29 50 51
14 '192.168.1.232' '173.194.36.79' 29 50 69
15 '192.168.1.232' '173.194.36.119' 32 52 53
16 '192.168.1.232' '173.194.36.119' 32 52 74
然后，您可以按第二列对它们进行分组并将组打印为
gps = df.groupby(2).groups
print gps
Out:
' 14': [0, 1],
' 23': [2],
' 24': [3, 4],
' 25': [5, 6],
' 26': [7, 8],
' 27': [9, 10],
' 28': [11, 12],
' 29': [13, 14],
' 32': [15, 16]
查看各个行号？有很多方法可以删除gps 的每个列表中的连续重复项。这是一个：
valid_rows = list()
for g in gps.values():
old_row = g[0]
valid_rows.append(old_row)
for row_id in range(1, len(g)):
new_row = g[row_id]
if new_row - old_row != 1:
valid_rows.append(new_row)
old_row = new_row
print valid_rows
Out: [5, 3, 9, 7, 0, 2, 15, 13, 11]
最后，通过valid_rows 索引pandas DataFrame。
print df.ix[sorted(valid_rows)]
Out:
0 '192.168.1.232' '173.194.36.64' 14 15 16
2 '192.168.1.232' '173.194.36.119' 23 30 31
3 '192.168.1.232' '173.194.36.98' 24 40 41
5 '192.168.1.232' '173.194.36.74' 25 42 43
7 '192.168.1.232' '173.194.36.74' 26 44 45
9 '192.168.1.232' '173.194.36.78' 27 46 47
11 '192.168.1.232' '173.194.36.78' 28 48 49
13 '192.168.1.232' '173.194.36.79' 29 50 51
15 '192.168.1.232' '173.194.36.119' 32 52 53
【讨论】：
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