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DOC: Solution for issue #60044 #60058

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DOC: Solution for issue #60044 #60058

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ZKaoChi
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@ZKaoChi ZKaoChi commented Oct 16, 2024

@5j9
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5j9 commented Oct 16, 2024

@ZKaoChi , See the error message for the failed test:

cython-lint.............................................................................................Failed
- hook id: cython-lint
- exit code: 1

/code/pandas/_libs/tslibs/timedeltas.pyx:1871:89: E501 line too long (149 > 88 characters)

(Keep lines below 89 characters.)

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ZKaoChi commented Oct 16, 2024

@5j9 I see. Thanks a lot!!!

5j9
5j9 reviewed Oct 17, 2024
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pandas/_libs/tslibs/timedeltas.pyx Outdated
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.

5j9
5j9 reviewed Oct 17, 2024
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Another suggestion: I'd use a more descriptive commit title. Something that gives more information about the issue at hand rather than just referring to issue number.

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@5j9 5j9 5j9 left review comments

@MarcoGorelli MarcoGorelli Awaiting requested review from MarcoGorelli MarcoGorelli is a code owner

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DOC: The value parameter of pandas.Timedelta can also accept float
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@ZKaoChi @5j9