Is it possible to split a pandas column after last integer?

I am trying to split a pandas column into two separate, where the first should contain just the date and the second string. But I don't want to split it after a certain character, like counting where the last integer instead I want to make a code that is applicable in general.

My col looks like this :

And I want it to look like this:

df_t['date'] = df_t['date_time'].str[0:19]
df_t["name"] = df_t["date_time"].str[19: ]
    
    
tid = df_t.drop(["date_time"], axis = 1)

This is the way I did it but I need a general way as mentioned above

You can use str.extract together with regular expressions:

import pandas as pd

# Sample data
data = {'Column A': ['01.01.2000John Doe', '01.01.2002Jane Doe']}
df = pd.DataFrame(data)

# Regular expression pattern
pattern = r'(?P<Date>\d{2}\.\d{2}\.\d{4})(?P<Name>.*)'

# Extracting the date and name into separate columns
df[['Column A', 'Column B']] = df['Column A'].str.extract(pattern)

print(df)

Explanation:

• The pattern variable contains the regular expression pattern. The expression (?P<Date>\d{2}.\d{2}.\d{4}) captures the date, and (?P<Name>.*) captures the name.
• The ?P<> syntax is used to name the captured groups, which makes it easier to create the new columns in the DataFrame.

EDIT

import pandas as pd

# Sample data
data = {
    '1Column A': ['2000-01-01 00:00:00John Doe', '2002-01-01 00:00:00Jane Doe'],
    '2Column B': ['2000-01-01 00:00:00Alice', '2002-01-01 00:00:00Bob'],
    '3Column C': ['Some other data', 'Not a date and name'],
}

df = pd.DataFrame(data)

# Regular expression pattern
pattern = r'(?P<Date>\d{4}-\d{2}-\d{2} \d{2}:\d{2}:\d{2})(?P<Name>.*)'

# Iterate through columns and apply the pattern conditionally
for col in df.columns:
    if col.startswith("1") or col.startswith("2"):
        # Extract date and name into separate columns with suffixes
        df[[f"{col}_date", f"{col}_name"]] = df[col].str.extract(pattern)
        # Drop the original column
        df.drop(col, axis=1, inplace=True)

print(df)

You can simply use indexing:

df['Column A'], df['Column B'] = df['Column A'].str[:10], df['Column A'].str[10:]
print(df)

# Output
     Column A  Column B
0  01.01.2000  John Doe
1  01.01.2002  Jane Doe

If you want to convert as datetime64:

df['Column A'], df['Column B'] = \
    pd.to_datetime(df['Column A'].str[:10], dayfirst=True), df['Column A'].str[10:]
print(df)

# Output
    Column A  Column B
0 2000-01-01  John Doe
1 2002-01-01  Jane Doe