通过列对(5-tuple)索引Pandas

Question

我试图为网络5元组设置流id，原始数据帧看起来像:

tup = [['192.168.0.1', '1032', '192.168.0.2', '443'],
['192.168.0.1', '1032', '192.168.0.2', '443'],
['192.168.0.1', '1034', '192.168.0.2', '443'],
['192.168.0.2', '443', '192.168.0.1', '1034'],
['192.168.0.1', '1034', '192.168.0.2', '443'],
['192.168.0.1', '1034', '192.168.0.2', '443'],
['192.168.0.2', '443', '192.168.0.1', '1034'],
['192.168.0.2', '443', '192.168.0.1', '1034'],
['192.168.0.1', '1032', '192.168.0.2', '443'],
['192.168.0.2', '443', '192.168.0.1', '1032']]
df = pd.DataFrame(tup,columns=['src','src_port','dst','dst_port'])

对于来自同一流(入/出)的流量，流id应该设置为:

src src_port    dst dst_port    flow_id
0   192.168.0.1 1032    192.168.0.2 443 1
1   192.168.0.1 1032    192.168.0.2 443 1
2   192.168.0.1 1034    192.168.0.2 443 2
3   192.168.0.2 443 192.168.0.1 1034    2
4   192.168.0.1 1034    192.168.0.2 443 2
5   192.168.0.1 1034    192.168.0.2 443 2
6   192.168.0.2 443 192.168.0.1 1034    2
7   192.168.0.2 443 192.168.0.1 1034    2
8   192.168.0.1 1032    192.168.0.2 443 1
9   192.168.0.2 443 192.168.0.1 1032    1

我将数据帧转换为值并将它们排序在一起，但在设置正确的流量索引时卡住了。

有没有更快/更优雅的方法?

Answer 1

一个想法是成对排序-嵌套元组，然后调用factorize:

a = df[['src','src_port','dst','dst_port']].to_numpy()
s = [tuple(sorted(((x[0], x[1]), (x[2], x[3])))) for x in a]
df['flow_id'] = pd.factorize(s)[0] + 1
print (df)
src src_port          dst dst_port  flow_id
0  192.168.0.1     1032  192.168.0.2      443        1
1  192.168.0.1     1032  192.168.0.2      443        1
2  192.168.0.1     1034  192.168.0.2      443        2
3  192.168.0.2      443  192.168.0.1     1034        2
4  192.168.0.1     1034  192.168.0.2      443        2
5  192.168.0.1     1034  192.168.0.2      443        2
6  192.168.0.2      443  192.168.0.1     1034        2
7  192.168.0.2      443  192.168.0.1     1034        2
8  192.168.0.1     1032  192.168.0.2      443        1
9  192.168.0.2      443  192.168.0.1     1032        1