pd.Period off by hundred of years for dates past ~2263

[eyurtsev]

In [1]: import pandas as pd

In [2]: pd.Period(year=2300, month=1, day=1, freq='M').end_time
Out[2]: Timestamp('1715-07-14 00:25:26.290448383')

Input year is '2300', output shows '1715'

output of pd.show_versions()

In [3]: pd.show_versions()

INSTALLED VERSIONS

commit: None
python: 2.7.11.final.0
python-bits: 64
OS: Linux
OS-release: 3.19.0-25-generic
machine: x86_64
processor: x86_64
byteorder: little
LC_ALL: None
LANG: en_US.UTF-8

pandas: 0.18.1
nose: None
pip: 7.1.2
setuptools: 21.0.0
Cython: None
numpy: 1.11.0
scipy: 0.17.0
statsmodels: 0.6.1
xarray: None
IPython: 4.2.0
sphinx: None
patsy: 0.4.1
dateutil: 2.5.3
pytz: 2016.4
blosc: None
bottleneck: None
tables: None
numexpr: None
matplotlib: 1.5.1
openpyxl: None
xlrd: None
xlwt: None
xlsxwriter: None
lxml: None
bs4: 4.4.1
html5lib: 0.9999999
httplib2: None
apiclient: None
sqlalchemy: None
pymysql: None
psycopg2: None
jinja2: 2.8
boto: 2.39.0
pandas_datareader: 0.2.1

[jreback]

http://pandas.pydata.org/pandas-docs/stable/timeseries.html#timestamp-limitations

there are limitations on the repr of Timestamps.

[jreback]

you can represent as Periods that is in fact the point. But conversions are out-of-bounds.

[jreback]

I suppose this should actually raise.

[jreback]

cc @MaximilianR
@sinhrks

[eyurtsev]

+1 for raising an Exception

Is another way to get the year-month-day corresponding to end_time/start_time of the given period?