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 and y need to have the same shape as condition.

Returns :

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.

See also

nonzero, choose

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.in1d(x.ravel(), goodvalues).reshape(x.shape)
>>> ix
array([[False, False, False],
       [ True,  True, False],
       [False,  True, False]], dtype=bool)
>>> np.where(ix)
(array([1, 1, 2]), array([0, 1, 1]))