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REF: dont use np.abs in NDFrame.__abs__ #43847

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143 changes: 71 additions & 72 deletions pandas/core/generic.py
Original file line number Diff line number Diff line change
Expand Up @@ -1589,6 +1589,77 @@ def bool(self):

self.__nonzero__()

@final
def abs(self: NDFrameT) -> NDFrameT:
"""
Return a Series/DataFrame with absolute numeric value of each element.

This function only applies to elements that are all numeric.

Returns
-------
abs
Series/DataFrame containing the absolute value of each element.

See Also
--------
numpy.absolute : Calculate the absolute value element-wise.

Notes
-----
For ``complex`` inputs, ``1.2 + 1j``, the absolute value is
:math:`\\sqrt{ a^2 + b^2 }`.

Examples
--------
Absolute numeric values in a Series.

>>> s = pd.Series([-1.10, 2, -3.33, 4])
>>> s.abs()
0 1.10
1 2.00
2 3.33
3 4.00
dtype: float64

Absolute numeric values in a Series with complex numbers.

>>> s = pd.Series([1.2 + 1j])
>>> s.abs()
0 1.56205
dtype: float64

Absolute numeric values in a Series with a Timedelta element.

>>> s = pd.Series([pd.Timedelta('1 days')])
>>> s.abs()
0 1 days
dtype: timedelta64[ns]

Select rows with data closest to certain value using argsort (from
`StackOverflow <https://stackoverflow.com/a/17758115>`__).

>>> df = pd.DataFrame({
... 'a': [4, 5, 6, 7],
... 'b': [10, 20, 30, 40],
... 'c': [100, 50, -30, -50]
... })
>>> df
a b c
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
>>> df.loc[(df.c - 43).abs().argsort()]
a b c
1 5 20 50
0 4 10 100
2 6 30 -30
3 7 40 -50
"""
res_mgr = self._mgr.apply(np.abs)
return self._constructor(res_mgr).__finalize__(self, name="abs")

@final
def __abs__(self: NDFrameT) -> NDFrameT:
return self.abs()
Expand Down Expand Up @@ -9822,78 +9893,6 @@ def _tz_localize(ax, tz, ambiguous, nonexistent):
# ----------------------------------------------------------------------
# Numeric Methods

@final
def abs(self: NDFrameT) -> NDFrameT:
"""
Return a Series/DataFrame with absolute numeric value of each element.

This function only applies to elements that are all numeric.

Returns
-------
abs
Series/DataFrame containing the absolute value of each element.

See Also
--------
numpy.absolute : Calculate the absolute value element-wise.

Notes
-----
For ``complex`` inputs, ``1.2 + 1j``, the absolute value is
:math:`\\sqrt{ a^2 + b^2 }`.

Examples
--------
Absolute numeric values in a Series.

>>> s = pd.Series([-1.10, 2, -3.33, 4])
>>> s.abs()
0 1.10
1 2.00
2 3.33
3 4.00
dtype: float64

Absolute numeric values in a Series with complex numbers.

>>> s = pd.Series([1.2 + 1j])
>>> s.abs()
0 1.56205
dtype: float64

Absolute numeric values in a Series with a Timedelta element.

>>> s = pd.Series([pd.Timedelta('1 days')])
>>> s.abs()
0 1 days
dtype: timedelta64[ns]

Select rows with data closest to certain value using argsort (from
`StackOverflow <https://stackoverflow.com/a/17758115>`__).

>>> df = pd.DataFrame({
... 'a': [4, 5, 6, 7],
... 'b': [10, 20, 30, 40],
... 'c': [100, 50, -30, -50]
... })
>>> df
a b c
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
>>> df.loc[(df.c - 43).abs().argsort()]
a b c
1 5 20 50
0 4 10 100
2 6 30 -30
3 7 40 -50
"""
# error: Incompatible return value type (got "ndarray[Any, dtype[Any]]",
# expected "NDFrameT")
return np.abs(self) # type: ignore[return-value]

@final
def describe(
self: NDFrameT,
Expand Down