I have a dataframe with a column composed by date object and a column composed by time object. I have to merge the two columns.

Personally, I think it is so ugly the following solution. Why I have to cast to str? I crafted my solution based on this answer

#importing all the necessary libraries 
import pandas as pd 
import datetime

#I have only to create a Minimal Reproducible Example
time1 = datetime.time(3,45,12)
time2 = datetime.time(3,49,12)
date1 = datetime.datetime(2020, 5, 17)
date2 = datetime.datetime(2021, 5, 17)
date_dict= {"time1":[time1,time2],"date1":[date1,date2]}


df["TimeMerge"] = pd.to_datetime(df.date1.astype(str)+' '+df.time1.astype(str))
  • 1
    \$\begingroup\$ i agree it would be nice if they overloaded + for date and time, but your current str/to_datetime() code is the fastest way to do it (even if it looks uglier) \$\endgroup\$
    – tdy
    May 12, 2021 at 3:32
  • 1
    \$\begingroup\$ if your real data is coming from a csv, pd.read_csv(..., parse_dates=[['date1','time1']]) would probably be the "prettiest" and fastest option \$\endgroup\$
    – tdy
    May 12, 2021 at 3:38

1 Answer 1


We can let pandas handle this for us and use DataFrame.apply and datetime.datetime.combine like this:

df["TimeMerge"] = df.apply(lambda row: datetime.datetime.combine(row.date1, row.time1), axis=1)

Although the following approach is more explicit and might therefore be more readable if you're not familiar with DataFrame.apply, I would strongly recommend the first approach.

You could also manually map datetime.datetime.combine over a zip object of date1 and time1:

def combine_date_time(d_t: tuple) -> datetime.datetime:
    return datetime.datetime.combine(*d_t)

df["TimeMerge"] = pd.Series(map(combine_date_time, zip(df.date1, df.time1)))

You can also inline it as an anonymous lambda function:

df["TimeMerge"] = pd.Series(map(lambda d_t: datetime.datetime.combine(*d_t), zip(df.date1, df.time1)))

This is handy for simple operations, but I would advise against this one-liner in this case.

By the way, the answer your were looking for can also be found under the question you linked.

  • 1
    \$\begingroup\$ "can and should" seems too conclusive though. apply(axis=1) is usually the slowest option. in this case, it's 2x slower than pd.to_datetime() at 200 rows, 3x slower at 2K rows, 3.5x slower at 2M rows. \$\endgroup\$
    – tdy
    May 12, 2021 at 3:28
  • \$\begingroup\$ Thank you for the input, I was expecting df.apply to be faster than that. I reduced it to "can". \$\endgroup\$ May 12, 2021 at 10:21
  • \$\begingroup\$ Thank you @riskypenguin I missed the specific answer :) \$\endgroup\$ May 18, 2021 at 6:40

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