Python – Add months to a datetime column in pandas

ipythonpandaspythonpython-2.7python-3.x

I have a dataframe df with 2 columns as below –

               START_DATE             MONTHS
0              2015-03-21                240
1              2015-03-21                240
2              2015-03-21                240
3              2015-03-21                240
4              2015-03-21                240
5              2015-01-01                120
6              2017-01-01                240
7                     NaN                NaN
8                     NaN                NaN
9                     NaN                NaN

The datatypes of the 2 columns are objects.

>>> df.dtypes
START_DATE    object
MONTHS        object
dtype: object

Now, I want to create a new column "Result" by adding df['START_DATE'] & df['MONTHS']. So, I have done the below –

from dateutil.relativedelta import relativedelta  

df['START_DATE'] = pd.to_datetime(df['START_DATE'])
df['MONTHS'] = df['MONTHS'].astype(float)

df['offset'] = df['MONTHS'].apply(lambda x: relativedelta(months=x))

df['Result'] = df['START_DATE'] + df['offset'] 

Here, I get the below error –

TypeError: incompatible type [object] for a datetime/timedelta operation

Note: Wanted to convert df['Months'] to int but wouldn't work as the field had Nulls.

Can you please give me some directions.Thanks.

Best Solution

This is a vectorized way to do this, so should be quite performant. Note that it doesn't handle month crossings / endings (and doesn't deal well with DST changes. I believe that's why you get the times).

In [32]: df['START_DATE'] + df['MONTHS'].values.astype("timedelta64[M]")
Out[32]: 
0   2035-03-20 20:24:00
1   2035-03-20 20:24:00
2   2035-03-20 20:24:00
3   2035-03-20 20:24:00
4   2035-03-20 20:24:00
5   2024-12-31 10:12:00
6   2036-12-31 20:24:00
7                   NaT
8                   NaT
9                   NaT
Name: START_DATE, dtype: datetime64[ns]

If you need exact MonthEnd/Begin handling, this is an appropriate method. (Use MonthsOffset to get the same day)

In [33]: df.dropna().apply(lambda x: x['START_DATE'] + pd.offsets.MonthEnd(x['MONTHS']), axis=1)
Out[33]: 
0   2035-02-28
1   2035-02-28
2   2035-02-28
3   2035-02-28
4   2035-02-28
5   2024-12-31
6   2036-12-31
dtype: datetime64[ns]