np.where

2019-11-06 08:48:04

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有時(shí)間過來整理numpy.where(condition[,x,y])

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

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

Parameters:

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 sameshape as condition.
Returns:	out : ndarray or tuple of ndarrays If both x and y are specified, the output array containselements ofx wherecondition is True, and elements fromy elsewhere. If only condition is given, return the tuple`condition.nonzero()`, the indices wherecondition is True.

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 sameshape as condition.

Returns:

out : ndarray or tuple of ndarrays

If both x and y are specified, the output array containselements ofx wherecondition is True, and elements fromy elsewhere.
If only condition is given, return the tuplecondition.nonzero(), the indices wherecondition is True.

[xv if c else yv for (c,xv,yv) in zip(condition,x,y)]>>> 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)>>> ixarray([[False, False, False],       [ True,  True, False],       [False,  True, False]], dtype=bool)>>> np.where(ix)(array([1, 1, 2]), array([0, 1, 1]))