SciPy

numpy.floor

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

Return the floor of the input, element-wise.

The floor of the scalar x is the largest integer i, such that i <= x. It is often denoted as \lfloor x \rfloor.

Parameters:

x : array_like

Input data.

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:

y : ndarray or scalar

The floor of each element in x.

See also

ceil, trunc, rint

Notes

Some spreadsheet programs calculate the “floor-towards-zero”, in other words floor(-2.5) == -2. NumPy instead uses the definition of floor where floor(-2.5) == -3.

Examples

>>> a = np.array([-1.7, -1.5, -0.2, 0.2, 1.5, 1.7, 2.0])
>>> np.floor(a)
array([-2., -2., -1.,  0.,  1.,  1.,  2.])

Previous topic

numpy.fix

Next topic

numpy.ceil