Pandas: Change day

Question

I have a series in datetime format, and need to change the day to 1 for each entry. I have thought of numerous simple solutions, but none of them w

Answer 1

You can use .apply and datetime.replace, eg:

import pandas as pd
from datetime import datetime

ps = pd.Series([datetime(2014, 1, 7), datetime(2014, 3, 13), datetime(2014, 6, 12)])
new = ps.apply(lambda dt: dt.replace(day=1))

Gives:

0   2014-01-01
1   2014-03-01
2   2014-06-01
dtype: datetime64[ns]

Answer 2

The other answer works, but any time you use apply, you slow your code down a lot. I was able to get an 8.5x speedup by writing a quick vectorized Datetime replace for a series.

def vec_dt_replace(series, year=None, month=None, day=None):
    return pd.to_datetime(
        {'year': series.dt.year if year is None else year,
         'month': series.dt.month if month is None else month,
         'day': series.dt.day if day is None else day})

Apply:

%timeit dtseries.apply(lambda dt: dt.replace(day=1))
# 4.17 s ± 38.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Vectorized:

%timeit vec_dt_replace(dtseries, day=1)
# 491 ms ± 6.48 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Note that you could face errors by trying to change dates to ones that don't exist, like trying to change 2012-02-29 to 2013-02-29. Use the errors argument of pd.to_datetime to ignore or coerce them.

Data generation: Generate series with 1 million random dates:

import pandas as pd
import numpy as np

# Generate random dates. Modified from: https://stackoverflow.com/a/50668285
def pp(start, end, n):
    start_u = start.value // 10 ** 9
    end_u = end.value // 10 ** 9

    return pd.Series(
        (10 ** 9 * np.random.randint(start_u, end_u, n)).view('M8[ns]'))

start = pd.to_datetime('2015-01-01')
end = pd.to_datetime('2018-01-01')
dtseries = pp(start, end, 1000000)
# Remove time component
dtseries = dtseries.dt.normalize()