abs(self: 'NDFrameT') -> 'NDFrameT'
This function only applies to elements that are all numeric.
For complex
inputs, 1.2 + 1j
, the absolute value is $\sqrt{ a^2 + b^2 }$
.
Series/DataFrame containing the absolute value of each element.
Return a Series/DataFrame with absolute numeric value of each element.
numpy.absolute
Calculate the absolute value element-wise.
Absolute numeric values in a Series.
This example is valid syntax, but we were not able to check execution>>> 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.
This example is valid syntax, but we were not able to check execution>>> s = pd.Series([1.2 + 1j])
... s.abs() 0 1.56205 dtype: float64
Absolute numeric values in a Series with a Timedelta element.
This example is valid syntax, but we were not able to check execution>>> 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).
This example is valid syntax, but we were not able to check execution>>> df = pd.DataFrame({This example is valid syntax, but we were not able to check execution
... '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 -50See :
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