numpy.divide — NumPy v1.13 Manual (original) (raw)
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.
| Parameters: | 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. **kwargs For other keyword-only arguments, see theufunc docs. |
|---|---|
| Returns: | 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.
Notes
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.
Examples
np.divide(2.0, 4.0) 0.5 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) 0 np.divide(2, 4.) 0.5
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 "", line 1, in FloatingPointError: divide by zero encountered in divide
ignored_states = np.seterr(**old_err_state) np.divide(1, 0) 0