The following sequence is an extract of the pandas DataFrame that I've got:
>>> df_t
value
2011-01-31 -5.575000
2011-03-31 7.700000
2011-05-31 15.966667
2011-07-31 10.683333
2011-08-31 10.454167
2011-10-31 9.320833
2011-12-31 -0.358333
2012-01-31 -11.550000
2012-03-31 1.700000
2012-05-31 12.333333
2012-07-31 12.816667
2012-08-31 11.837500
2012-10-31 2.733333
2012-12-31 4.075000
2013-01-31 2.450000
2013-03-31 -4.262500
2013-05-31 11.491667
2013-07-31 14.812500
2013-08-31 13.920833
2013-10-31 4.125000
2013-12-31 0.075000
How can I delete March 31st in every leap year? I tried something like:
def isleap(year):
return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)
if isleap(df_t.index.year):
df_t=df_t[df_t.index.dayofyear!=91]
...but obviously, this was too straightforward in my head. Is the only solution to loop through the whole dataframe and check at every step if the year is a leap year and the date is 91st day of year or is there any easier solution available?
EDIT: The issue is not how to determine whether a year is a leap year, but, if so, to delete March 31st in the above dataframe.