SciPy

numpy.spacing

numpy.spacing(x, /, out=None, *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature, extobj]) = <ufunc 'spacing'>

Return the distance between x and the nearest adjacent number.

Parameters:
x : array_like

Values to find the spacing of.

out : ndarray, None, or tuple of ndarray and None, optional

A location into which the result is stored. If provided, it must have a shape that the inputs broadcast to. If not provided or None, a freshly-allocated array is returned. A tuple (possible only as a keyword argument) must have length equal to the number of outputs.

where : array_like, optional

Values of True indicate to calculate the ufunc at that position, values of False indicate to leave the value in the output alone.

**kwargs

For other keyword-only arguments, see the ufunc docs.

Returns:
out : ndarray or scalar

The spacing of values of x. This is a scalar if x is a scalar.

Notes

It can be considered as a generalization of EPS: spacing(np.float64(1)) == np.finfo(np.float64).eps, and there should not be any representable number between x + spacing(x) and x for any finite x.

Spacing of +- inf and NaN is NaN.

Examples

>>> np.spacing(1) == np.finfo(np.float64).eps
True

Previous topic

numpy.nextafter

Next topic

numpy.lcm