DOC : Update the pandas.Period.hour docstring #20312
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IHackPy
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Mar 12, 2018
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IHackPy
commented
TomAugspurger
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pandas/_libs/tslibs/period.pyx
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| @@ -1241,6 +1241,24 @@ cdef class _Period(object): | |||
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| @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.
pandas/_libs/tslibs/period.pyx
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| 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.
pandas/_libs/tslibs/period.pyx
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| See Also | ||
| -------- | ||
| Period.minute : Get the minute of hour |
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"minute of hour" -> "minute of the Period." (end with a .).
pandas/_libs/tslibs/period.pyx
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| See Also | ||
| -------- | ||
| Period.minute : Get the minute of hour | ||
| Period.second : Get the second of hour |
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"second of hour" -> "second of the Period." (end with a .).
pandas/_libs/tslibs/period.pyx
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| >>> 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).
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Thanks @himanshuawasthi95 ! |
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thanks @TomAugspurger |
