Split events into sessions

The problem

Group each user's events into sessions: a new session starts when more than gap_minutes have passed since that user's previous event. Return user, sessions and events, sorted by user.

Function: def sessionize(events,gap_minutes):

Example input (a JSON list of arguments): [[{"user": "u1", "at": "2026-01-01T11:00:00"}, {"user": "u2", "at": "2026-01-01T09:00:00"}, {"user": "u1", "at": "2026-01-01T10:00:00"}, {"user": "u1", "at": "2026-01-01T11:30:00"}, {"user": "u1", "at": "2026-01-01T10:20:00"}], 30]
Expected output: [{"user": "u1", "sessions": 2, "events": 4}, {"user": "u2", "sessions": 1, "events": 1}]

Inputs satisfy the problem statement. Sequences contain at most 200 elements unless a smaller bound is stated. Use JSON values and the supplied function signature.

Hints

1. Sort by user and time, then take diff() of the timestamps within each user.
2. Check the supplied examples, empty/boundary cases and the stated answer policy before optimizing.

Worked answer

Try it yourself first. When you are ready, open the answer below.

def sessionize(events,gap_minutes):
    import pandas as pd
    if not events:return []
    df=pd.DataFrame(events);df['at']=pd.to_datetime(df['at']);df=df.sort_values(['user','at'])
    gap=df.groupby('user')['at'].diff()
    df['start']=gap.isna()|(gap>pd.Timedelta(minutes=gap_minutes))
    out=df.groupby('user').agg(sessions=('start','sum'),events=('at','size')).reset_index()
    return [{"user":r.user,"sessions":int(r.sessions),"events":int(r.events)} for r in out.itertuples()]

Why it works

Sort by user and time, then take diff() of the timestamps within each user. A session starts at each user's first event (the gap is missing) or when the gap is longer than the limit. Summing those start flags per user counts sessions. A gap exactly equal to the limit stays in the same session.

Complexity: O(n log n) time / O(n) space.

Time: O(n log n) time · Space: O(n) space