Extracting the first day of month of a datetime type column in pandas

We can use date offset in conjunction with Series.dt.normalize:

In [60]: df['month'] = df['purchase_date'].dt.normalize() - pd.offsets.MonthBegin(1)

In [61]: df
Out[61]:
   user_id       purchase_date      month
0        1 2015-01-23 14:05:21 2015-01-01
1        2 2015-02-05 05:07:30 2015-02-01
2        3 2015-02-18 17:08:51 2015-02-01
3        4 2015-03-21 17:07:30 2015-03-01
4        5 2015-03-11 18:32:56 2015-03-01
5        6 2015-03-03 11:02:30 2015-03-01

Or much nicer solution from @BradSolomon

In [95]: df['month'] = df['purchase_date'] - pd.offsets.MonthBegin(1, normalize=True)

In [96]: df
Out[96]:
   user_id       purchase_date      month
0        1 2015-01-23 14:05:21 2015-01-01
1        2 2015-02-05 05:07:30 2015-02-01
2        3 2015-02-18 17:08:51 2015-02-01
3        4 2015-03-21 17:07:30 2015-03-01
4        5 2015-03-11 18:32:56 2015-03-01
5        6 2015-03-03 11:02:30 2015-03-01

Try this ..

df['month']=pd.to_datetime(df.purchase_date.astype(str).str[0:7]+'-01')

Out[187]: 
   user_id        purchase_date       month
0        1  2015-01-23 14:05:21  2015-01-01
1        2  2015-02-05 05:07:30  2015-02-01
2        3  2015-02-18 17:08:51  2015-02-01
3        4  2015-03-21 17:07:30  2015-03-01
4        5  2015-03-11 18:32:56  2015-03-01
5        6  2015-03-03 11:02:30  2015-03-01

For me df['purchase_date'] - pd.offsets.MonthBegin(1) didn’t work (it fails for the first day of the month), so I’m subtracting the days of the month like this:

df['purchase_date'] - pd.to_timedelta(df['purchase_date'].dt.day - 1, unit='d')

@Eyal: This is what I did to get the first day of the month using pd.offsets.MonthBegin and handle the scenario where day is already first day of month.

import datetime

from_date= pd.to_datetime('2018-12-01')

from_date = from_date - pd.offsets.MonthBegin(1, normalize=True) if not from_date.is_month_start else from_date

from_date

result: Timestamp('2018-12-01 00:00:00')

from_date= pd.to_datetime('2018-12-05')

from_date = from_date - pd.offsets.MonthBegin(1, normalize=True) if not rom_date.is_month_start else from_date

from_date

result: Timestamp('2018-12-01 00:00:00')

Most proposed solutions don’t work for the first day of the month.

Following solution works for any day of the month:

df['month'] = df['purchase_date'] + pd.offsets.MonthEnd(0) - pd.offsets.MonthBegin(normalize=True)

[EDIT]

Another, more readable, solution is:

from pandas.tseries.offsets import MonthBegin
df['month'] = df['purchase_date'].dt.normalize().map(MonthBegin().rollback)

Be aware not to use:

df['month'] = df['purchase_date'].map(MonthBegin(normalize=True).rollback)

because that gives incorrect results for the first day due to a bug: https://github.com/pandas-dev/pandas/issues/32616

To extract the first day of every month, you could write a little helper function that will also work if the provided date is already the first of month. The function looks like this:

def first_of_month(date):
    return date + pd.offsets.MonthEnd(-1) + pd.offsets.Day(1)

You can apply this function on pd.Series:

df['month'] = df['purchase_date'].apply(first_of_month)

With that you will get the month column as a Timestamp. If you need a specific format, you might convert it with the strftime() method.

df['month_str'] = df['month'].dt.strftime('%Y-%m-%d')

How about this easy solution?
As purchase_date is already in datetime64[ns] format, you can use strftime to format the date to always have the first day of month.

df['date'] = df['purchase_date'].apply(lambda x: x.strftime('%Y-%m-01'))

print(df)
 user_id   purchase_date       date
0   1   2015-01-23 14:05:21 2015-01-01
1   2   2015-02-05 05:07:30 2015-02-01
2   3   2015-02-18 17:08:51 2015-02-01
3   4   2015-03-21 17:07:30 2015-03-01
4   5   2015-03-11 18:32:56 2015-03-01
5   6   2015-03-03 11:02:30 2015-03-01

Because we used strftime, now the date column is in object (string) type:

print(df.dtypes)
user_id                   int64
purchase_date    datetime64[ns]
date                     object
dtype: object

Now if you want it to be in datetime64[ns], just use pd.to_datetime():

df['date'] = pd.to_datetime(df['date'])

print(df.dtypes)
user_id                   int64
purchase_date    datetime64[ns]
date             datetime64[ns]
dtype: object

try this Pandas libraries, where ‘purchase_date’ is date parameter placed into the module.

date['month_start'] = pd.to_datetime(sched_slim.purchase_date)
.dt.to_period('M')
.dt.to_timestamp()

Just adding my 2 cents, for the sake of completeness:

1 – transform purchase_date to date, instead of datetime. This will remove hour, minute, second, etc…

df['purchase_date'] = df['purchase_date'].dt.date

2 – apply the datetime replace, to use day 1 instead of the original:

df['purchase_date_begin'] = df['purchase_date'].apply(lambda x: x.replace(day=1))

This replace method is available on the datetime library:

from datetime import date

today = date.today()
month_start = today.replace(day=1)

and you can replace day, month, year, etc…