Indexing a numpy array with a list of tuples

Why can't I index an ndarray using a list of tuple indices like so?

idx = [(x1, y1), ... (xn, yn)]
X[idx]

Instead I have to do something unwieldy like

idx2 = numpy.array(idx)
X[idx2[:, 0], idx2[:, 1]] # or more generally:
X[tuple(numpy.vsplit(idx2.T, 1)[0])]

Is there a simpler, more pythonic way?

回答1:

You can use a list of tuples, but the convention is different from what you want. numpy expects a list of row indices, followed by a list of column values. You, apparently, want to specify a list of (x,y) pairs.

http://docs.scipy.org/doc/numpy/reference/arrays.indexing.html#integer-array-indexing The relevant section in the documentation is 'integer array indexing'.

Here's an example, seeking 3 points in a 2d array. (2 points in 2d can be confusing):

In [223]: idx
Out[223]: [(0, 1, 1), (2, 3, 0)]
In [224]: X[idx]
Out[224]: array([2, 7, 4])

Using your style of xy pairs of indices:

In [230]: idx1 = [(0,2),(1,3),(1,0)]
In [231]: [X[i] for i in idx1]
Out[231]: [2, 7, 4]

In [240]: X[tuple(np.array(idx1).T)]
Out[240]: array([2, 7, 4])

X[tuple(zip(*idx1))] is another way of doing the conversion. The tuple() is optional in Python2. zip(*...) is a Python idiom that reverses the nesting of a list of lists.

You are on the right track with:

In [242]: idx2=np.array(idx1)
In [243]: X[idx2[:,0], idx2[:,1]]
Out[243]: array([2, 7, 4])

My tuple() is just a bit more compact (and not necessarily more 'pythonic'). Given the numpy convention, some sort of conversion is necessary.

(Should we check what works with n-dimensions and m-points?)

回答2:

Use tuple of numpy arrays which can be directly passed to matrix to get the elements

Index = tuple(np.array(list(zip(*index_tuple))))
new_array = list(prev_array[index])