Python 从 Pandas Dataframe 中每个 Id 的第一个时间戳中减go 最后一个时间戳

使用named aggregation和Series.sub的减法列，DataFrame.pop用于处理后的下降列tmp:

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  

如果需要格式HH:MM:SS，请使用:

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