将函数应用于熊猫中分组数据的单个列

Question

Theres a pandas dataframe like shown below

      Bank       date               creationdate
0     JP Morgan  2010-07-22 2010-07-22 12:17:38.187000
1     JP Morgan  2010-07-31 2010-07-31 12:41:57.083000
2     JP Morgan  2010-11-18 2010-11-18 19:24:15.503000
3     JP Morgan  2011-03-08 2011-03-08 18:57:31.477000
4     JP Morgan  2011-04-27 2011-04-27 13:13:01.357000
5     JP Morgan  2011-05-01 2011-05-01 17:19:28.773000
6     JP Morgan  2011-05-06 2011-05-06 19:40:51.757000
7     JP Morgan  2011-05-10 2011-05-10 01:14:52.503000
8     JP Morgan  2011-05-23 2011-05-23 20:36:18.490000
9     JP Morgan  2011-05-25 2011-05-25 15:51:08.650000
10    JP Morgan  2011-05-28 2011-05-28 21:08:30.270000
11    JP Morgan  2011-05-29 2011-05-29 04:18:26.693000
12    JP Morgan  2011-06-03 2011-06-03 16:54:13.770000
13    JP Morgan  2011-06-08 2011-06-08 18:35:50.450000
14    JP Morgan  2011-06-08 2011-06-08 18:37:12.493000
15    JP Morgan  2011-06-08 2011-06-08 18:37:45.593000

I want to find out the mean of the differences of creationdates of each date. For this I am doing a groupby and call diff and then mean on the grouped data

df_grouped = date_df.groupby(['bank', 'date'], as_index = False)
mean = df_grouped['creationdate'].diff().mean()

but this gives me a mean of all differences instead of giving mean for difference corresponding to each date.

Please suggest how can I achieve the mean of differences for each date

Answer 1

I think you can do this with .aggregate in one step, rather than trying to get things done in two steps:

In [30]:

print df_grouped['creationdate'].aggregate(lambda x: (np.diff(x)).mean())
         Bank       date    creationdate
0   JP Morgan 2010-07-22             NaT
1   JP Morgan 2010-07-31             NaT
2   JP Morgan 2010-11-18             NaT
3   JP Morgan 2011-03-08             NaT
4   JP Morgan 2011-04-27             NaT
5   JP Morgan 2011-05-01             NaT
6   JP Morgan 2011-05-06             NaT
7   JP Morgan 2011-05-10             NaT
8   JP Morgan 2011-05-23             NaT
9   JP Morgan 2011-05-25             NaT
10  JP Morgan 2011-05-28             NaT
11  JP Morgan 2011-05-29             NaT
12  JP Morgan 2011-06-03             NaT
13  JP Morgan 2011-06-08 00:00:57.571500

In the sample data that you showed, only 2011-06-08 has more than one value and will result in a number other than NaT