如何添加一列timedelta并将其与另一列关联?

时间:2019-06-11 15:42:27

标签: python python-3.x pandas

我有2个数据框,一个是按user_id分组的,并计数显示user_id的次数。另一个数据框是用户使用服务的时间和日期。我想做的是按照最早的日期-最晚的日期从第二个数据帧中计算出时间增量,然后将时间增量添加到第一个数据帧中,甚至还有一列可以从时间增量中提取日期。我认为可能需要进行循环以迭代user_id。我已经尝试了很多次,但无法获得想要的结果。

df1 = pd.DataFrame({'user_id': ['8', '2','5', '1', '10', '4'], 'usage_times':[466,423,401,350,352,333]})
df2 = pd.DataFrame({'user_id': ['1', '5','5', '8', '8', '1'], 'Date':['2010-11-16 16:44:52','2010-06-01 00:34:38','2010-05-31 05:01:24','2010-06-01 00:29:30','2010-09-11 23:55:00','2010-08-10 13:00:00']})
df1:
user_id   usage_times
8         466
2         423
5         401
1         350
10        352
4         333
df2:
user_id                 Date
1        2010-11-16 16:44:52
5        2010-06-01 00:34:38
5        2010-05-31 05:01:24
8        2010-06-01 00:29:30
8        2010-09-11 23:55:00
1        2010-08-10 13:00:00

我尝试过的代码是:

for users in top_users.user_id:
    latest_trip = df_final[(df_final['user_id'] == users)]['start_at'].max()
    earliest_trip = df_final[(df_final['user_id'] == users)]['start_at'].min()
    usage_period = earliest_trip - latest_trip
    times = days_hours_minutes(usage_period)
    top_users['period'] = top_users.apply(lambda x: list(x) for x in times)

我想要的数据框变成这样:

df1:
user_id   usage_times   period                days
8         466           100 days, 00:23:45    100
2         423           15 days, 00:05:45     15
5         401           104 days, 00:23:45    104
1         350           72 days, 00:15:45     72
10        352           40 days, 00:23:45     40
4         333           28 days, 00:43:45     28

2 个答案:

答案 0 :(得分:0)

IIUC,您可以merge df1和df2,并使用groupby

创建期间
df = df1.merge(df2, on='user_id')
df['period'] = df.groupby('user_id')['Date'].transform(lambda x: x.max() -  x.min() )
df['days'] = df['period'].dt.days
df.drop_duplicates('user_id', inplace=True)
df.drop(columns=['Date'], inplace = True)
df.head()


    user_id usage_times period              days
0   8       466         102 days 23:25:30   102
2   5       401         0 days 19:33:14     0
4   1       350         98 days 03:44:52    98

答案 1 :(得分:0)

必须完成两个不同的步骤。

首先,您需要获取期间。为此,您可以在df2上使用groupby,然后在日期差处使用aggregate

df2 = df2.groupby(["user_id"]).agg(lambda x: x.max() - x.min())

然后,您可以merge df1和df2:

df_res = df1.merge(df2, on='user_id')