Pandas dataframe rows to dict of lists, using first value of each row as key

Question

import pandas as pd

---

SETUP

I have a dataframe:

df = pd.DataFrame({'A': ['A0', 'A1', 'A2', 'A3'],
                   'B': ['B0', 'B1', 'B2', 'B3'],
                   'C': ['C0', 'C1', 'C2', 'C3'],
                   'D': ['D0', 'D1', 'D2', 'D3']},
                   index=[0, 1, 2, 3])

ie:

+----+-----+-----+-----+-----+
|    | A   | B   | C   | D   |
|----+-----+-----+-----+-----|
|  0 | A0  | B0  | C0  | D0  |
|  1 | A1  | B1  | C1  | D1  |
|  2 | A2  | B2  | C2  | D2  |
|  3 | A3  | B3  | C3  | D3  |
+----+-----+-----+-----+-----+

_{(using print(tabulate(df, headers='keys', tablefmt='psql')), related Q)}

---

PROBLEM

I would like to convert the above dataframe into this dict:

{'A0': ['A0', 'B0', 'C0', 'D0'],
 'A1': ['A1', 'B1', 'C1', 'D1'],
 'A2': ['A2', 'B2', 'C2', 'D2'],
 'A3': ['A3', 'B3', 'C3', 'D3']}

The first element from each row are the keys, the rows of the dataframes are the values for the dict as lists.

---

SOLUTIONS

A Using .iterrows(), which seems bad practice:

`{row[1][0]: list(row[1]) for row in df.iterrows() for alias in row[1]}`

B Using .groupby() (and this):

gbdict=df.groupby('A').apply(lambda row: row.to_dict(orient='row')).to_dict()
{key: list(gbdict[key][0].values()) for key in gbdict.keys()}

They both produce the required output.

---

QUESTION

Is there a more efficient way to achieve the above goal?

If there could be a way without the for loop, ie the dict comprehension, that'd be great.

Answer 1

Something like this , I think this will faster than agg ~

yourdict = dict(zip(df.A,df.values.tolist()))
Out[123]: 
{'A0': ['A0', 'B0', 'C0', 'D0'],
 'A1': ['A1', 'B1', 'C1', 'D1'],
 'A2': ['A2', 'B2', 'C2', 'D2'],
 'A3': ['A3', 'B3', 'C3', 'D3']}

Answer 2

Use set_index then agg using list on axis=1 and use Series.to_dict:

dct = df.set_index(df['A']).agg(list, axis=1).to_dict()

Result:

{'A0': ['A0', 'B0', 'C0', 'D0'],
 'A1': ['A1', 'B1', 'C1', 'D1'],
 'A2': ['A2', 'B2', 'C2', 'D2'],
 'A3': ['A3', 'B3', 'C3', 'D3']}