pandas extract start and end date from date range colum

Question

I have a column that contains a date range and I'd just like to have it extracted to the start and end dates respectively. Not sure if this is doable with datetime.strptime directly

df_have = pd.DataFrame([[1, '01 Jan 2019-04 Jan 2019'], [2, '07 Jan 2019-11 Jan 2019']], columns=['Index', 'Range'])

Index   Range
0   1   01 Jan 2019-04 Jan 2019
1   2   07 Jan 2019-11 Jan 2019

df_want = pd.DataFrame([[1, '01 Jan 2019', '04 Jan 2019'], [2, '07 Jan 2019', '11 Jan 2019']], columns=['Index', 'Start', 'End'])

Index   Start   End
0   1   01 Jan 2019 04 Jan 2019
1   2   07 Jan 2019 11 Jan 2019

Thanks

Rakesh · Accepted Answer · 2019-02-07T11:21:02.473

3

Use str.split

Ex:

import pandas as pd

df_have = pd.DataFrame([[1, '01 Jan 2019-04 Jan 2019'], [2, '07 Jan 2019-11 Jan 2019']], columns=['Index', 'Range'])
df_have[["start", "end"]] = df_have.pop("Range").str.split("-", expand=True)    #Thanks @ jezrael
print(df_have)

Output:

   Index        start          end
0      1  01 Jan 2019  04 Jan 2019
1      2  07 Jan 2019  11 Jan 2019

edited Feb 07 '19 at 11:21

answered Feb 07 '19 at 10:48

Rakesh

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1

I see now that I've thought about the q from a `datetime` perspective whereas it could easily be solved using `str.split`. Thanks for the answer though – Grizzly2501 Feb 07 '19 at 12:00

score 0 · Answer 2 · answered Feb 07 '19 at 11:03

import pandas as pd
df_have['Start']=df_have['Range'].map(lambda x:x.split('-')[0])
df_have['End']=df_have['Range'].map(lambda x:x.split('-')[1])
df_have.drop('Range', axis=1)
print(df_have) 
   Index        Start          End
0      1  01 Jan 2019  04 Jan 2019
1      2  07 Jan 2019  11 Jan 2019

pandas extract start and end date from date range colum

2 Answers2