Numbers Of Day In Month

I have a data frame with a date time index, and I would like to multiply some columns with the number of days in that month. TUFNWGTP TELFS t070101 t070102 t

Solution 1:

There is now a Series.dt.days_in_month attribute for datetime series. Here is an example based on Jeff's answer.

In [3]: df = pd.DataFrame({'date': pd.date_range('20120101', periods=15, freq='M')})

In [4]: df['year'] = df['date'].dt.year

In [5]: df['month'] = df['date'].dt.month

In [6]: df['days_in_month'] = df['date'].dt.days_in_month

In [7]: df
Out[7]:
         date  year  month  days_in_month
0  2012-01-31  2012      1             31
1  2012-02-29  2012      2             29
2  2012-03-31  2012      3             31
3  2012-04-30  2012      4             30
4  2012-05-31  2012      5             31
5  2012-06-30  2012      6             30
6  2012-07-31  2012      7             31
7  2012-08-31  2012      8             31
8  2012-09-30  2012      9             30
9  2012-10-31  2012     10             31
10 2012-11-30  2012     11             30
11 2012-12-31  2012     12             31
12 2013-01-31  2013      1             31
13 2013-02-28  2013      2             28
14 2013-03-31  2013      3             31

Solution 2:

pd.tslib.monthrange is an unadvertised / undocumented function that handles the days_in_month calculation (adjusting for leap years). This could/should prob be added as a property to Timestamp/DatetimeIndex.

In [34]: df = DataFrame({'date' : pd.date_range('20120101',periods=15,freq='M') })

In [35]: df['year'] = df['date'].dt.year

In [36]: df['month'] = df['date'].dt.month

In [37]: df['days_in_month'] = df.apply(lambda x: pd.tslib.monthrange(x['year'],x['month'])[1], axis=1)

In [38]: df
Out[38]: 
         date  year  month  days_in_month
0  2012-01-31  2012      1             31
1  2012-02-29  2012      2             29
2  2012-03-31  2012      3             31
3  2012-04-30  2012      4             30
4  2012-05-31  2012      5             31
5  2012-06-30  2012      6             30
6  2012-07-31  2012      7             31
7  2012-08-31  2012      8             31
8  2012-09-30  2012      9             30
9  2012-10-31  2012     10             31
10 2012-11-30  2012     11             30
11 2012-12-31  2012     12             31
12 2013-01-31  2013      1             31
13 2013-02-28  2013      2             28
14 2013-03-31  2013      3             31

Solution 3:

Here is a little clunky hand-made method to get the number of days in a month

import datetime

def days_in_month(dt):
  next_month = datetime.datetime(
      dt.year + dt.month / 12, dt.month % 12 + 1, 1)
  start_month = datetime.datetime(dt.year, dt.month, 1)
  td = next_month - start_month
  return td.days

For example:

>>> days_in_month(datetime.datetime.strptime('2013-12-12', '%Y-%m-%d'))
31
>>> days_in_month(datetime.datetime.strptime('2013-02-12', '%Y-%m-%d'))
28
>>> days_in_month(datetime.datetime.strptime('2012-02-12', '%Y-%m-%d'))
29
>>> days_in_month(datetime.datetime.strptime('2012-01-12', '%Y-%m-%d'))
31
>>> days_in_month(datetime.datetime.strptime('2013-11-12', '%Y-%m-%d'))
30

I let you figure out how to read your table and do the multiplication yourself :)

Solution 4:

import pandas as pd
from pandas.tseries.offsets import MonthEnd

df['dim'] = (pd.to_datetime(df.index) + MonthEnd(0)).dt.day

You can omit pd.to_datetime(), if your index is already DatetimeIndex.