split pandas column into two columns-CodePudding

In this data.frame, I want to split the id_date into two columns id and date. I checked previous answers and tried to use str.split but none of my trials worked.

DATA

dict = {'id_date': {0: ('A', datetime.date(1972, 2, 28)), 1: ('A', datetime.date(1972, 2, 28)), 2: ('A', datetime.date(1972, 2, 28)), 3: ('A', datetime.date(1972, 12, 18)), 4: ('A', datetime.date(1973, 1, 22)), 5: ('A', datetime.date(1973, 1, 22)), 6: ('A', datetime.date(1973, 1, 22)), 7: ('A', datetime.date(1973, 1, 22)), 8: ('A', datetime.date(1973, 1, 23)), 9: ('A', datetime.date(1973, 1, 23)), 10: ('A', datetime.date(1973, 1, 23))}}

import pandas as pd
import datetime

df = pd.DataFrame.from_dict(dict)

CodePudding user response：

Use the DataFrame constructor:

out = pd.DataFrame(df['id_date'].to_list(), columns=['id', 'date'])

output:

   id        date
0   A  1972-02-28
1   A  1972-02-28
2   A  1972-02-28
3   A  1972-12-18
4   A  1973-01-22
5   A  1973-01-22
6   A  1973-01-22
7   A  1973-01-22
8   A  1973-01-23
9   A  1973-01-23
10  A  1973-01-23

As new columns:

df[['id', 'date']] = pd.DataFrame(df['id_date'].to_list())

output:

            id_date id        date
0   (A, 1972-02-28)  A  1972-02-28
1   (A, 1972-02-28)  A  1972-02-28
2   (A, 1972-02-28)  A  1972-02-28
3   (A, 1972-12-18)  A  1972-12-18
4   (A, 1973-01-22)  A  1973-01-22
5   (A, 1973-01-22)  A  1973-01-22
6   (A, 1973-01-22)  A  1973-01-22
7   (A, 1973-01-22)  A  1973-01-22
8   (A, 1973-01-23)  A  1973-01-23
9   (A, 1973-01-23)  A  1973-01-23
10  (A, 1973-01-23)  A  1973-01-23

CodePudding user response：

Use:

df['id'] = df['id_date'].str[0]
df['date'] = df['id_date'].str[1]

OUTPUT

            id_date id        date
0   (A, 1972-02-28)  A  1972-02-28
1   (A, 1972-02-28)  A  1972-02-28
2   (A, 1972-02-28)  A  1972-02-28
3   (A, 1972-12-18)  A  1972-12-18
4   (A, 1973-01-22)  A  1973-01-22
5   (A, 1973-01-22)  A  1973-01-22
6   (A, 1973-01-22)  A  1973-01-22
7   (A, 1973-01-22)  A  1973-01-22
8   (A, 1973-01-23)  A  1973-01-23
9   (A, 1973-01-23)  A  1973-01-23
10  (A, 1973-01-23)  A  1973-01-23

CodePudding user response：

You can do this :

import datetime
import pandas as pd

dict = {'id_date': {0: ('A', datetime.date(1972, 2, 28)), 1: ('A', datetime.date(1972, 2, 28)), 2: ('A', datetime.date(1972, 2, 28)), 3: ('A', datetime.date(1972, 12, 18)), 4: ('A', datetime.date(1973, 1, 22)), 5: ('A', datetime.date(1973, 1, 22)), 6: ('A', datetime.date(1973, 1, 22)), 7: ('A', datetime.date(1973, 1, 22)), 8: ('A', datetime.date(1973, 1, 23)), 9: ('A', datetime.date(1973, 1, 23)), 10: ('A', datetime.date(1973, 1, 23))}}

df = pd.DataFrame.from_dict(dict)

df['id'] = [el[0] for el in df['id_date']]
df['date'] = [el[1] for el in df['id_date']]

print(df)

output:

            id_date id        date
0   (A, 1972-02-28)  A  1972-02-28
1   (A, 1972-02-28)  A  1972-02-28
2   (A, 1972-02-28)  A  1972-02-28
3   (A, 1972-12-18)  A  1972-12-18
4   (A, 1973-01-22)  A  1973-01-22
5   (A, 1973-01-22)  A  1973-01-22
6   (A, 1973-01-22)  A  1973-01-22
7   (A, 1973-01-22)  A  1973-01-22
8   (A, 1973-01-23)  A  1973-01-23
9   (A, 1973-01-23)  A  1973-01-23
10  (A, 1973-01-23)  A  1973-01-23

CodePudding user response：

Or use assign, to keep it cleaner.

df = (pd.DataFrame.from_dict(dict) .assign(id=lambda x:x['id_date'].str[0],date =lambda x:x['id_date'].str[1]))