delete leap days in pandas

Use dt accessor:

df = df[~((df.Date.dt.month == 2) & (df.Date.dt.day == 29))]

Add dt accessor because working with Series, not with DatetimeIndex:

df = df[~((df.Date.dt.month == 2) & (df.Date.dt.day == 29))]

Or invert condition with chaining | for bitwise OR and != for not equal:

df = df[(df.Date.dt.month != 2) | (df.Date.dt.day != 29)]

Or use strftime for convert to MM-DD format:

df = df[df.Date.dt.strftime('%m-%m') != '02-29']

Another way you can try below in incase your Date column is not proper datetime rather a str.

df[~df.Date.str.endswith('02-29')]

OR , if it's in datetime format even you can try converting to str.

df[~df.Date.astype(str).str.endswith('02-29')]

OR, Even use contains:

df[~df.Date.str.contains('02-29')]

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