Subtract last timestamp from first timestamp for each Id in Pandas Dataframe

Question

I have a dataframe (df) with the following structure:

retweet_datetime	tweet_id	tweet_datetime
2020-04-24 03:33:15	85053699	2020-04-24 02:28:22
2020-04-24 02:43:35	85053699	2020-04-24 02:28:22
2020-04-18 04:24:03	86095361	2020-04-18 00:06:01
2020-04-18 00:19:08	86095361	2020-04-18 00:06:01
2020-04-18 00:18:38	86095361	2020-04-18 00:06:01
2020-04-18 00:07:08	86095361	2020-04-18 00:06:01

The retweet_datetime is sorted from latest to newest retweets.

I’d like to create two new columns as follows:

tweet_lifetime1: the difference between the last retweet time and the first retweet time, i.e., for each tweet_id: last retweet_datetime – first retweet_datetime
tweet_lifetime2: the difference between the last retweet time and tweet creation time (tweet_datetime)

Update

For example, for the tweet id: "86095361":

tweet_lifetime1 = 2020-04-18 04:24:03 - 2020-04-18 00:07:08 (04:16:55)
tweet_lifetime2 = 2020-04-18 04:24:03 - 2020-04-18 00:06:01 (04:18:02)

The expected output df:

retweet_datetime	tweet_id	tweet_datetime	lifetime1	lifetime2
2020-04-24 03:33:15	85053699	2020-04-24 02:28:22	00:49:40	01:04:53
2020-04-18 04:24:03	86095361	2020-04-18 00:06:01	04:16:55	04:18:02

I’ve seen several similar posts, but they mostly subtract consecutive rows. For example, I can subtract the time difference between each retweet_datetimes for each tweet id as follows:

df2 = df.assign(delta = df.groupby('tweet_id')['retweet_datetime'].diff())

Asked By: mOna

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Source

Answer 1

Groupby and agg with first to select the first row per group, then groupby and agg retweet_datetime with last to select last value per group. Then join aggregated dataframes and calculate lifetime1 and lifetime2 by subtracting relevant columns

g = df.groupby('tweet_id')
s = g.first().join(g['retweet_datetime'].agg(['last']))

s['lifetime1'] = s['retweet_datetime'] - s.pop('last')
s['lifetime2'] = s['retweet_datetime'] - s['tweet_datetime']

            retweet_datetime      tweet_datetime       lifetime1       lifetime2
tweet_id                                                                        
85053699 2020-04-24 03:33:15 2020-04-24 02:28:22 0 days 00:49:40 0 days 01:04:53
86095361 2020-04-18 04:24:03 2020-04-18 00:06:01 0 days 04:16:55 0 days 04:18:02

Answered By: Shubham Sharma

Answer 2

Use named aggregation with subtract column with Series.sub, DataFrame.pop is used for drop column tmp after processing:

df1 = (df.groupby('tweet_id', as_index=False)
         .agg(retweet_datetime=('retweet_datetime','first'),
              tmp = ('retweet_datetime','last'),
              tweet_datetime = ('tweet_datetime','last')))

df1['lifetime1'] = df1['retweet_datetime'].sub(df1.pop('tmp'))
df1['lifetime2'] = df1['retweet_datetime'].sub(df1['tweet_datetime'])
print (df1)
   tweet_id    retweet_datetime      tweet_datetime       lifetime1  
0  85053699 2020-04-24 03:33:15 2020-04-24 02:28:22 0 days 00:49:40   
1  86095361 2020-04-18 04:24:03 2020-04-18 00:06:01 0 days 04:16:55   

        lifetime2  
0 0 days 01:04:53  
1 0 days 04:18:02

If need format HH:MM:SS use:

def f(x):
    ts = x.total_seconds()
    hours, remainder = divmod(ts, 3600)
    minutes, seconds = divmod(remainder, 60)
    return ('{:02d}:{:02d}:{:02d}').format(int(hours), int(minutes), int(seconds))

df1['lifetime1'] = df1['retweet_datetime'].sub(df1.pop('tmp')).apply(f)
df1['lifetime2'] = df1['retweet_datetime'].sub(df1['tweet_datetime']).apply(f)
print (df1)
   tweet_id    retweet_datetime      tweet_datetime lifetime1 lifetime2
0  85053699 2020-04-24 03:33:15 2020-04-24 02:28:22  00:49:40  01:04:53
1  86095361 2020-04-18 04:24:03 2020-04-18 00:06:01  04:16:55  04:18:02

Answered By: jezrael

Subtract last timestamp from first timestamp for each Id in Pandas Dataframe

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