/NumPy 1.13

numpy.where

`numpy.where(condition[, x, y])`

Return elements, either from `x` or `y`, depending on `condition`.

If only `condition` is given, return `condition.nonzero()`.

Parameters: condition : array_like, bool When True, yield `x`, otherwise yield `y`. x, y : array_like, optional Values from which to choose. `x`, `y` and `condition` need to be broadcastable to some shape. out : ndarray or tuple of ndarrays If both `x` and `y` are specified, the output array contains elements of `x` where `condition` is True, and elements from `y` elsewhere. If only `condition` is given, return the tuple `condition.nonzero()`, the indices where `condition` is True.

Notes

If `x` and `y` are given and input arrays are 1-D, `where` is equivalent to:

```[xv if c else yv for (c,xv,yv) in zip(condition,x,y)]
```

Examples

```>>> np.where([[True, False], [True, True]],
...          [[1, 2], [3, 4]],
...          [[9, 8], [7, 6]])
array([[1, 8],
[3, 4]])
```
```>>> np.where([[0, 1], [1, 0]])
(array([0, 1]), array([1, 0]))
```
```>>> x = np.arange(9.).reshape(3, 3)
>>> np.where( x > 5 )
(array([2, 2, 2]), array([0, 1, 2]))
>>> x[np.where( x > 3.0 )]               # Note: result is 1D.
array([ 4.,  5.,  6.,  7.,  8.])
>>> np.where(x < 5, x, -1)               # Note: broadcasting.
array([[ 0.,  1.,  2.],
[ 3.,  4., -1.],
[-1., -1., -1.]])
```

Find the indices of elements of `x` that are in `goodvalues`.

```>>> goodvalues = [3, 4, 7]
>>> ix = np.isin(x, goodvalues)
>>> ix
array([[False, False, False],
[ True,  True, False],
[False,  True, False]], dtype=bool)
>>> np.where(ix)
(array([1, 1, 2]), array([0, 1, 1]))
```