depark
depark

Reputation: 3

pandas merge rows inside of single dataframe

New to Pandas and have a question that I cannot answer on my own. For context, this is output from a firewall. it generates millions of packets and I am trying to aggregate that data into a firewall ruleset. The best way I've come up with is to identify traffic based on the destination IP.

The source/dest ports will change if they are ephemeral so it's important that I aggregate them into the same row. That way I can determine the range of ports for the ruleset.

RAW CSV:

dvc,"src_interface",transport,"src_ip","src_port","dest_ip","dest_port",direction,action,cause,count "Firewall-1",outside,tcp,"4.4.4.4",53,"1.1.1.1",1025,outbound,allowed,"",2 "Firewall-1",outside,tcp,"4.4.4.4",53,"1.1.1.1",1026,outbound,allowed,"",2 "Firewall-1",outside,tcp,"4.4.4.4",22,"1.1.1.1",1028,outbound,allowed,"",2 "Firewall-1",outside,tcp,"3.3.3.3",22,"2.2.2.2",2200,outbound,allowed,"",2

Dataframe:

dvc src_interface transport   src_ip  src_port        dest_ip  dest_port direction   action  cause  count
0  Firewall-1       outside       tcp  4.4.4.4       53  1.1.1.1       1025  outbound  allowed    NaN      2
1  Firewall-1       outside       tcp  4.4.4.4       53  1.1.1.1       1026  outbound  allowed    NaN      2
2  Firewall-1       outside       tcp  4.4.4.4       53  1.1.1.1       1028  outbound  allowed    NaN      2
3  Firewall-1       outside       tcp  3.3.3.3       22  2.2.2.2       2200  outbound  allowed    NaN      2

How would I go about merging rows with the same dest_ip?

CODE:

df = pd.concat([pd.read_csv(f) for f in glob.glob('*.csv')], ignore_index = True)
index_cols = df.columns.tolist()
index_cols.remove('dest_ip')
df = df.groupby(index_cols, as_index=False)['dest_ip'].apply(list)
print(df)

Expected Output:

Firewall-1 outside tcp 4.4.4.4 53 1.1.1.1 1025-1026,1028 outbound allowed nan 2
Firewall-1 outside tcp 3.3.3.3 22 2.2.2.2 2200 outbound allowed nan 2

Most examples I've found online involve joining two dataframes whereas I only have the one. Any help would be appreciated. Thanks in advance!

Upvotes: 0

Views: 11005

Answers (2)

robertwest
robertwest

Reputation: 942

Try this. Group all the columns where you expected information to be duplicated and then aggregate the different “dest_port” values into a list:

df = pd.DataFrame([
            ["Firewall-1","outside","tcp","4.4.4.4",53,"1.1.1.1",1025,"outbound","allowed","",2], 
            ["Firewall-1","outside","tcp","4.4.4.4",53,"1.1.1.1",1026,"outbound","allowed","",2], 
            ["Firewall-1","outside","tcp","4.4.4.4",22,"1.1.1.1",1028,"outbound","allowed","",2], 
            ["Firewall-1","outside","tcp","3.3.3.3",22,"2.2.2.2",2200,"outbound", "allowed","",2]
        ], 
        columns=["dvc","src_interface","transport","src_ip","src_port","dest_ip","dest_port","direction", "action", "cause", "count"])

index_cols = df.columns.tolist()
index_cols.remove("dest_port") 
df = df.groupby(index_cols)["dest_port"].apply(list)
df = df.reset_index()

this results in 3 remaining rows and not 2 as in your desired output:

   dvc              src_interface transport   src_ip         src_port  dest_ip direction   action cause  count     dest_port
0  Firewall-1       outside       tcp         3.3.3.3        22  2.2.2.2  outbound  allowed            2        [2200]
1  Firewall-1       outside       tcp         4.4.4.4        22  1.1.1.1  outbound  allowed            2        [1028]
2  Firewall-1       outside       tcp         4.4.4.4        53  1.1.1.1  outbound  allowed            2  [1025, 1026]

Upvotes: 2

DeathbyGreen
DeathbyGreen

Reputation: 347

I think the following might do what you're looking for:

    import pandas as pd
    #create practice dataframe. will remove rows if values in 'key' are duplicate
    df = pd.DataFrame({'key':[1,1,3,4],'color':[1,2,3,2],'house':[1,2,3,7]})
    print(df.drop_duplicates(['key']))

Original dataframe:

    key  color  house
    1      1      1
    1      2      2
    3      3      3
    4      2      7

Output dataframe:

    key  color  house
    1      1      1
    3      3      3
    4      2      7

Upvotes: 0

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