Fix pandas.Timedelta range #12728
Fix pandas.Timedelta range #12728
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| # Resolution is in nanoseconds | ||
| # (2**63)/(1*24*60*60*(10**9)) = 106751.991167 | ||
| Timedelta.min = Timedelta(-106751.991167, 'D') |
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Can we specify this in ns instead of fractional days, which might have rounding errors? e.g., Timedelta(np.iinfo(np.int64).min, 'ns')
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Actually, looks like you need min + 1 since min corresponds to NaT:
In [6]: Timedelta(np.iinfo(np.int64).min, 'ns')
Out[6]: NaT
In [7]: Timedelta(np.iinfo(np.int64).min + 1, 'ns')
Out[7]: Timedelta('-106752 days +00:12:43.145224')
In [8]: Timedelta(np.iinfo(np.int64).max, 'ns')
Out[8]: Timedelta('106751 days 23:47:16.854775')
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This concerns the 2nd commit, in this PR @jreback, can you confirm that some of the tests at rely on buggy output. Specifically, that rng = timedelta_range('1 day 10:11:12', freq='h', periods=500)
s = Series(np.arange(len(rng)), index=rng)
result = s['5 day':'6 day']
expected = s.iloc[86:110]
assert_series_equal(result, expected)If this is a second bug fix, then I will address the changes documentation. Maybe there is even a better way other than peaking at the indices and creating the |
| @@ -1680,7 +1680,7 @@ def test_partial_slice(self): | |||
| s = Series(np.arange(len(rng)), index=rng) | |||
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| result = s['5 day':'6 day'] | |||
| expected = s.iloc[86:134] | |||
| expected = s.iloc[86:110] | |||
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why are you changing this?
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| @@ -109,6 +109,31 @@ The ``unit`` keyword argument specifies the unit of the Timedelta: | |||
| to_timedelta(np.arange(5), unit='s') | |||
| to_timedelta(np.arange(5), unit='d') | |||
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| timdelta limits | |||
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pls add a whatsnew note, point to the new docs (with a thanks. |
*Problem* Pandas Timedelta derives from `datetime.timedelta` and increases the resolution of the timedeltas to nanoseconds. As such Pandas.Timedelta has a smaller range of values. *Solution* This change modifies the advertised `min` and `max` timedeltas.
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thanks @has2k1 |
git diff upstream/master | flake8 --diffProblem
Pandas Timedelta derives from
datetime.timedeltaand increasethe resolution of the timedeltas. As such the Pandas.Timedelta
object can only have a smaller range of values.
Solution
This change modifies the properties that report
the range and resolution to reflect Pandas capabilities.
Reference
https://github.com/python/cpython/blob/8d1d7e6816753248768e4cc1c0370221814e9cf1/Lib/datetime.py#L651-L654