pandas.Series.abs#

Series.abs()[source]#

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 \(\sqrt{ a^2 + b^2 }\).

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).

>>> 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