Session generation from log file analysis with pandas

Question

I'm analysing a Apache log file and I have imported it in to a pandas dataframe.

'65.55.52.118 - - [30/May/2013:06:58:52 -0600] "GET /detailedAddVen.php?refId=7954&uId=2802 HTTP/1.1" 200 4514 "-" "Mozilla/5.0 (compatible; bingbot/2.0; +http://www.bing.com/bingbot.htm)"'

My dataframe:

I want to group this in to sessions based on IP, Agent and Time difference (If the duration of time is greater than 30 mins it should be a new session).

It is easy to group the dataframe by IP and Agent but how to check this time difference?Hope the problem is clear.

sessions = df.groupby(['IP', 'Agent']).size()

UPDATE : df.index is like follows:

<class 'pandas.tseries.index.DatetimeIndex'>
[2013-05-30 06:00:41, ..., 2013-05-30 22:29:14]
Length: 31975, Freq: None, Timezone: None

Answer 1

I would do this using a shift and a cumsum (here's a simple example, with numbers instead of times - but they would work exactly the same):

In [11]: s = pd.Series([1., 1.1, 1.2, 2.7, 3.2, 3.8, 3.9])

In [12]: (s - s.shift(1) > 0.5).fillna(0).cumsum(skipna=False)  # *
Out[12]:
0    0
1    0
2    0
3    1
4    1
5    2
6    2
dtype: int64

* the need for skipna=False appears to be a bug.

Then you can use this in a groupby apply:

In [21]: df = pd.DataFrame([[1.1, 1.7, 2.5, 2.6, 2.7, 3.4], list('AAABBB')]).T

In [22]: df.columns = ['time', 'ip']

In [23]: df
Out[23]:
  time ip
0  1.1  A
1  1.7  A
2  2.5  A
3  2.6  B
4  2.7  B
5  3.4  B

In [24]: g = df.groupby('ip')

In [25]: df['session_number'] = g['time'].apply(lambda s: (s - s.shift(1) > 0.5).fillna(0).cumsum(skipna=False))

In [26]: df
Out[26]:
  time ip  session_number
0  1.1  A               0
1  1.7  A               1
2  2.5  A               2
3  2.6  B               0
4  2.7  B               0
5  3.4  B               1

Now you can groupby 'ip' and 'session_number' (and analyse each session).

Answer 2

Andy Hayden's answer is lovely and concise, but it gets very slow if you have a large number of users/IP addresses to group over. Here's another method that's much uglier but also much faster.

import pandas as pd
import numpy as np

sample = lambda x: np.random.choice(x, size=10000)
df = pd.DataFrame({'ip': sample(range(500)), 
                   'time': sample([1., 1.1, 1.2, 2.7, 3.2, 3.8, 3.9])})
max_diff = 0.5 # Max time difference

def method_1(df):
    df = df.sort_values('time')
    g = df.groupby('ip')
    df['session'] = g['time'].apply(
        lambda s: (s - s.shift(1) > max_diff).fillna(0).cumsum(skipna=False)
        )
    return df['session']


def method_2(df):
    # Sort by ip then time 
    df = df.sort_values(['ip', 'time'])

    # Get locations where the ip changes 
    ip_change = df.ip != df.ip.shift()
    time_or_ip_change = (df.time - df.time.shift() > max_diff) | ip_change
    df['session'] = time_or_ip_change.cumsum()

    # The cumsum operated over the whole series, so subtract out the first 
    # value for each IP
    df['tmp'] = 0
    df.loc[ip_change, 'tmp'] = df.loc[ip_change, 'session']
    df['tmp'] = np.maximum.accumulate(df.tmp)
    df['session'] = df.session - df.tmp

    # Delete the temporary column
    del df['tmp']
    return df['session']

r1 = method_1(df)
r2 = method_2(df)

assert (r1.sort_index() == r2.sort_index()).all()

%timeit method_1(df)
%timeit method_2(df)

400 ms ± 195 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
11.6 ms ± 2.04 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)