[英]Applying function to column in grouped pandas dataframe and returning output as a new column
[英]Applying function to a single column of a grouped data in pandas
有一個熊貓數據框,如下所示
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
我想找出每個日期的創建日期差異的平均值。 為此,我正在進行groupby並調用diff,然后對分組數據進行平均
df_grouped = date_df.groupby(['bank', 'date'], as_index = False)
mean = df_grouped['creationdate'].diff().mean()
但這給了我所有差異的平均值,而不是給出每個日期對應的差異平均值。
請提出如何獲得每個日期的差異平均值
我認為您可以使用.aggregate
一步完成此.aggregate
,而不是嘗試分兩步完成:
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
在您顯示的樣本數據中,只有2011-06-08
具有多個值,並且將得出NaT
以外的數字
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