在熊猫和数据处理中匹配元组

Question

以下是我的数据框的简化块。我想处理

first.csv

,No.,Time,Source,Destination,Protocol,Length,Info,src_dst_pair 
325778,112.305107,02:e0,Broadcast,ARP,64,Who has 253.244.230.77? Tell 253.244.230.67,"('02:e0', 'Broadcast')" 
801130,261.868118,02:e0,Broadcast,ARP,64,Who has 253.244.230.156? Tell 253.244.230.67,"('02:e0', 'Broadcast')" 
700094,222.055094,02:e0,Broadcast,ARP,60,Who has 253.244.230.77? Tell 253.244.230.156,"('02:e0', 'Broadcast')" 
766542,766543,247.796156,100.118.138.150,41.177.26.176,TCP,66,32222 > http [SYN] Seq=0,"('100.118.138.150', '41.177.26.176')" 
767405,248.073313,100.118.138.150,41.177.26.176,TCP,64,32222 > http [ACK] Seq=1,"('100.118.138.150', '41.177.26.176')" 
767466,248.083268,100.118.138.150,41.177.26.176,HTTP,380,Continuation [Packet capture],"('100.118.138.150', '41.177.26.176')" 

我有（最后一个元素）的src_dst_pair所有独特元素

uniq_src_dst_pair = numpy.unique(data.src_dst_pair.ravel()) 
[('02:e0', 'Broadcast') ('100.118.138.150', '41.177.26.176')] 

我怎样才能做到在大熊猫以下

每个元素位于uniq_src_dst_pair中，请针对df.src_dst_pair进行检查。如果匹配，增加df.Length并将其存储在单独的列

我的预期结果为：

('02:e0', 'Broadcast') : 188 
('100.118.138.150', '41.177.26.176') : 510 

我怎样才能做到这一点？

下面是我尝试

import pandas 
import numpy 


data = pandas.read_csv('first.csv') 
print data 

uniq_src_dst_pair = numpy.unique(data.src_dst_pair.ravel()) 
print uniq_src_dst_pair 
print len(uniq_src_dst_pair) 

# following is hardcoded, but need to be more general for the above list 
match1 = data[data.src_dst_pair == "('02:e0:ed:0a:fb:5f', 'Broadcast')"] # doesn't work 

Answer 1

您的CSV文件被搞砸了。您不应该在标题中留下第一个逗号，并且在第4个非标题行中有一个额外的字段。修复这一点，你可以使用：

In [6]: data.groupby('src_dst_pair').Length.sum() 
Out[6]: 
src_dst_pair 
('02:e0', 'Broadcast')     188 
('100.118.138.150', '41.177.26.176') 510 
Name: Length, dtype: int64 

然而，你的决赛场上，“src_dst_pair”是多余的，如果这是你想要完成的，因为你可以简单地做类似如下的内容：

In [8]: data.groupby(['Source','Destination']).Length.sum() 
Out[8]: 
Source   Destination 
02:e0   Broadcast  188 
100.118.138.150 41.177.26.176 510 
Name: Length, dtype: int64