numpy.divide(x1, x2, /, out=None, *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature, extobj]) = <ufunc 'divide'>

Divide arguments element-wise.


x1 : array_like

Dividend array.

x2 : array_like

Divisor array.

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.


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


y : ndarray or scalar

The quotient x1/x2, element-wise. Returns a scalar if both x1 and x2 are scalars.

See also

Set whether to raise or warn on overflow, underflow and division by zero.


Equivalent to x1 / x2 in terms of array-broadcasting.

Behavior on division by zero can be changed using seterr.

In Python 2, when both x1 and x2 are of an integer type, divide will behave like floor_divide. In Python 3, it behaves like true_divide.


>>> np.divide(2.0, 4.0)
>>> x1 = np.arange(9.0).reshape((3, 3))
>>> x2 = np.arange(3.0)
>>> np.divide(x1, x2)
array([[ NaN,  1. ,  1. ],
       [ Inf,  4. ,  2.5],
       [ Inf,  7. ,  4. ]])

Note the behavior with integer types (Python 2 only):

>>> np.divide(2, 4)
>>> np.divide(2, 4.)

Division by zero always yields zero in integer arithmetic (again, Python 2 only), and does not raise an exception or a warning:

>>> np.divide(np.array([0, 1], dtype=int), np.array([0, 0], dtype=int))
array([0, 0])

Division by zero can, however, be caught using seterr:

>>> old_err_state = np.seterr(divide='raise')
>>> np.divide(1, 0)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
FloatingPointError: divide by zero encountered in divide
>>> ignored_states = np.seterr(**old_err_state)
>>> np.divide(1, 0)

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