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DOC: Solution for issue #60044 #60058

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DOC: Solution for issue #60044 #60058

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DOC: Solution for issue #60044
ZKaoChi Oct 16, 2024
039af7e
reduce characters for each line
ZKaoChi Oct 16, 2024
53559a4
DOC: Make the value parameter of pandas.Timedelta can accept float
ZKaoChi Oct 17, 2024
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6 changes: 4 additions & 2 deletions pandas/_libs/tslibs/timedeltas.pyx
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Expand Up @@ -1864,10 +1864,12 @@ class Timedelta(_Timedelta):

Parameters
----------
value : Timedelta, timedelta, np.timedelta64, str, or int
value : Timedelta, timedelta, np.timedelta64, str, int or float
Input value.
unit : str, default 'ns'
Denote the unit of the input, if input is an integer.
If input is an integer, denote the unit of the input.
If input is a float, denote the unit of integer parts.
(If unit = None, ignore decimal parts. Otherwise proportionally convert them.)
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@5j9 5j9 Oct 17, 2024 •
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I just noticed that: pd.Timedelta(0.123456789, 'microseconds') returns 0 days 00:00:00.000000123.

So it's not the unit that determines the rounding, but the lowest timedelta resolution which is currently 1ns.
I'd suggest:
"If value is a float, the decimal parts of the value that require a precision higher than a nanosecond will be ignored."
(This could be added to value's description instead.)

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Thanks a lot for your accommodating and suggestion! I will make some changes to the content and my commit title.


Possible values:

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