Get values from current till last column values in pandas groupby

Question

Image following pandas dataframe:

+----+------+-------+
| ID | Name | Value |
+----+------+-------+
| 1  | John | 1     |
+----+------+-------+
| 1  | John | 4     |
+----+------+-------+
| 1  | John | 10    |
+----+------+-------+
| 1  | John | 50    |
+----+------+-------+
| 1  | Adam | 6     |
+----+------+-------+
| 1  | Adam | 3     |
+----+------+-------+
| 2  | Jen  | 9     |
+----+------+-------+
| 2  | Jen  | 6     |
+----+------+-------+

I want to apply groupby function and create a new column which stores the Value values as a list from the current till the last groupby value.

Like that:

+----+------+-------+----------------+
| ID | Name | Value | NewCol         |
+----+------+-------+----------------+
| 1  | John | 1     | [1, 4, 10, 50] |
+----+------+-------+----------------+
| 1  | John | 4     | [4, 10, 50]    |
+----+------+-------+----------------+
| 1  | John | 10    | [10, 50]       |
+----+------+-------+----------------+
| 1  | John | 50    | [50]           |
+----+------+-------+----------------+
| 1  | Adam | 6     | [6, 3]         |
+----+------+-------+----------------+
| 1  | Adam | 3     | [3]            |
+----+------+-------+----------------+
| 2  | Jen  | 9     | [9, 6]         |
+----+------+-------+----------------+
| 2  | Jen  | 6     | [9]            |
+----+------+-------+----------------+

Is this anyhow possible using pandas groupby function?

Answer 1

Use GroupBy.transform with custom lambda functions:

f = lambda x: [x.iloc[i:len(x)].tolist() for i, y in enumerate(x)]
df['new'] = df.groupby(['Name', 'ID'])['Value'].transform(f)

Or:

f = lambda x: [y[::-1].tolist() for y in x.expanding()]
df['new'] = df.iloc[::-1].groupby(['Name', 'ID'])['Value'].transform(f)
print (df)
   ID  Name  Value             new
0   1  John      1  [1, 4, 10, 50]
1   1  John      4     [4, 10, 50]
2   1  John     10        [10, 50]
3   1  John     50            [50]
4   1  Adam      6          [6, 3]
5   1  Adam      3             [3]
6   2   Jen      9          [9, 6]
7   2   Jen      6             [6]