Replace the day in a date in to fit the pd.to_datetime format

Question

I have a data frame with multiple columns and multiple rows. In one of these columns there are dates that take the form of mm/dd/yyyy.

I am trying to convert this using df['col'] = pd.to_datetime(df['col']) but am getting the following error because there are multiple records that have 00 in the place of a missing month or day:

ValueError: day is out of range for month

I don't want to do df['col'] = pd.to_datetime(df['col'], errors = 'coerce') because I want to keep whatever data is there.

I would like all the dates that are missing days or months or both (e.g 11/00/2018, 00/13/2018, or 00/00/2018) to have the value 01 where the value is missing (e.g 11/01/2018, 01/13/2018, 01/01/2018).

Answer 1

You could use the following regex to replace 00:

import pandas as pd
data = ['11/00/2018', '00/13/2018', '00/00/2018']

df = pd.DataFrame(data=data, columns=['col'])
replace = df['col'].replace('00/', '01/', regex=True)
result = pd.to_datetime(replace)
print(result)

Output

0   2018-11-01
1   2018-01-13
2   2018-01-01
Name: col, dtype: datetime64[ns]