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DOC : Update the pandas.Period.hour docstring #20312

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Mar 12, 2018
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DOC : Update the pandas.Period.hour docstring #20312

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Update Period.hour docstring
IHackPy Mar 12, 2018
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update required changes
IHackPy Mar 12, 2018
c968c6d
Update period.pyx
TomAugspurger Mar 12, 2018
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18 changes: 18 additions & 0 deletions pandas/_libs/tslibs/period.pyx
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Original file line number Diff line number Diff line change
Expand Up @@ -1241,6 +1241,24 @@ cdef class _Period(object):

@property
def hour(self):
"""
Get hours of a day that a Period falls on.
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Maybe "Get the hour of the day component of a Period."? Since a Period can span multiple days.

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Yup Period can span multiple days .Its really hard to define accurate sentance ... day conponent of the period sounds more clear.


Returns
-------
int
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For the explanation note that it's betwene 0 and 23.

int
    The hour as an integer, between 0 and 23.

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sure


See Also
--------
Period.minute : Get the minute of hour
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"minute of hour" -> "minute of the Period." (end with a .).

Period.second : Get the second of hour
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"second of hour" -> "second of the Period." (end with a .).


Examples
--------
>>> p = pd.Period("2018-03-11 13:03:12.050000")
>>> p.hour
13
"""
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Can you add an example with a period longer than a day, to show that it uses 0?

In [4]: pd.Period("2017-01-01", freq="M").hour
Out[4]: 0

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can I add this too

pd.Period("2017-01-01", freq="M").hour
0

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Yes, please do (with the >>> like you've done for the other).

base, mult = get_freq_code(self.freq)
return phour(self.ordinal, base)

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