Subtracting values across grouped data frames in Pandas

Question

I have a set of IDs and Timestamps, and want to calculate the "total time elapsed per ID" by getting the difference of the oldest / earliest timestamps, grouped by ID.

Data

id   timestamp
1    2018-02-01 03:00:00
1    2018-02-01 03:01:00
2    2018-02-02 10:03:00
2    2018-02-02 10:04:00
2    2018-02-02 11:05:00

Expected Result

(I want the delta converted to minutes)

id   delta
1    1
2    62

I have a for loop, but it's very slow (10+ min for 1M+ rows). I was wondering if this was achievable via pandas functions?

# gb returns a DataFrameGroupedBy object, grouped by ID
gb = df.groupby(['id'])

# Create the resulting df
cycletime = pd.DataFrame(columns=['id','timeDeltaMin'])

def calculate_delta():
    for id, groupdf in gb:
        time = groupdf.timestamp
        # returns timestamp rows for the current id

        time_delta = time.max() - time.min()

        # convert Timedelta object to minutes
        time_delta = time_delta / pd.Timedelta(minutes=1) 

        # insert result to cycletime df
        cycletime.loc[-1] = [id,time_delta]
        cycletime.index += 1

Thinking of trying next:
- Multiprocessing

score 2 · Accepted Answer · answered May 11 '18 at 17:50

First ensure datetimes are OK:

df.timestamp = pd.to_datetime(df.timestamp)

Now find the number of minutes in the difference between the maximum and minimum for each id:

import numpy as np

>>> (df.timestamp.groupby(df.id).max() - df.timestamp.groupby(df.id).min()) / np.timedelta64(1, 'm')
id
1     1.0
2    62.0
Name: timestamp, dtype: float64

harvpan · Answer 2 · 2018-05-11T17:59:27.343

You can sort by id and tiemstamp, then groupby id and then find the difference between min and max timestamp per group.

df['timestamp'] = pd.to_datetime(df['timestamp'])
result = df.sort_values(['id']).groupby('id')['timestamp'].agg(['min', 'max'])
result['diff'] = (result['max']-result['min']) / np.timedelta64(1, 'm')
result.reset_index()[['id', 'diff']]

Output:

    id  diff
0   1   1.0
1   2   62.0

score 1 · Answer 3 · answered May 11 '18 at 18:16

Another one:

import pandas as pd
import numpy as np
import datetime
ids = [1,1,2,2,2]
times = ['2018-02-01 03:00:00','2018-02-01 03:01:00','2018-02-02 
10:03:00','2018-02-02 10:04:00','2018-02-02 11:05:00']
df = pd.DataFrame({'id':ids,'timestamp':pd.to_datetime(pd.Series(times))})
df.set_index('id', inplace=True)
print(df.groupby(level=0).diff().sum(level=0)['timestamp'].dt.seconds/60)

Subtracting values across grouped data frames in Pandas

3 Answers3