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