Storing day and month without year in Python

Question

I simply need to store a month and a date (a birthday) WITHOUT a year value.

When looking at the documentation for pandas (mostly because it has useful time manipulation features), specifically Periods, it states that you need to pass in a timespan and a frequency. I'm not quite sure how to use the frequency part regarding to my problem. What frequency would be most appropriate? Is using pd even the best way to do this?

Thanks!

Answer 1

That isn't how Period's work. They represent a specific period in time. Similar to how Timestamp's represent specific points in time.

You seem to want a generic Month-Day label. You can use strftime to get the string but you'll lose any date manipulation.

Consider the series of timestamps with timestamp indices.

s = pd.date_range('2011-01-31', periods=12, freq='M').to_series()
s

2011-01-31   2011-01-31
2011-02-28   2011-02-28
2011-03-31   2011-03-31
2011-04-30   2011-04-30
2011-05-31   2011-05-31
2011-06-30   2011-06-30
2011-07-31   2011-07-31
2011-08-31   2011-08-31
2011-09-30   2011-09-30
2011-10-31   2011-10-31
2011-11-30   2011-11-30
2011-12-31   2011-12-31
Freq: M, dtype: datetime64[ns]

You can get just the month and day like this (see this site for a summary of strftime)

s.dt.strftime('%m-%d')

2011-01-31    01-31
2011-02-28    02-28
2011-03-31    03-31
2011-04-30    04-30
2011-05-31    05-31
2011-06-30    06-30
2011-07-31    07-31
2011-08-31    08-31
2011-09-30    09-30
2011-10-31    10-31
2011-11-30    11-30
2011-12-31    12-31
Freq: M, dtype: object

Answer 2

I believe you are asking if there is an object like datetime whithout a specified year. There doesn't seem to be. Period object has always a year specified: utilizing them won't save memory.

t=pd.Period('03-04',freq='D')
t.year==1
t.month==3
t.day==4

You should use dateTime objects and simply ignore the year they are in. The only obstacle are leap years. For starters you can choose a year 2 years away from any leap year or refer to the Timestamps.isocalender value choosing the starting year with a similar strategy, making your year 12 years away from the closest with 53 weeks (isocalendar has a fixed exact 52 weeks, with a 53 weeks year every 24 years (7 days/week * 4 years/addLeapDay).

TDate15.Date=[d.replace(year=1904) for d in TDate15.Date]#   
TDate15=TDate15.groupby('Date').agg({'Data_Value':['min','max']})
#vs
TDate15=TDate15.groupby([d2015.Date.dt.month,d2015.Date.dt.day]).agg({'Data_Value':['min','max']})
#if you use a reference year as in the first exemple, you get an index of Datetime.
#vs
#using directly groupby day & month, you might need to convert to datetime further, wich would looked at least like:
#TDate15.index=[pd.to_datetime(d,m) for m,d in TDate15.index]
#However I had to do: 
i=list(zip(TDate15.index.labels[0],TDate15.index.labels[1]))
x=[pd.to_datetime( str(TDate05.index.levels[0][m]) + '-' + str(TDate05.index.levels[1][d]) +'-' + '1904',
                format='%m-%d-%Y') \
  for m,d in i]

The complete solution I see is to tweek a bit the datetime class by making it always come back to the same non-leap year. For merging or grouping, strftime does the job for a one time shot. The outputs are strings so for multiple manip you'll need to transform back and forth string-datetime. I hope you don't have a real issue with memory space used because of the 'year'.