numpy: how to get a max from an argmax result

问题

I have a numpy array of arbitrary shape, e.g.:

a = array([[[ 1,  2],
            [ 3,  4],
            [ 8,  6]],

          [[ 7,  8],
           [ 9,  8],
           [ 3, 12]]])
a.shape = (2, 3, 2)

and a result of argmax over the last axis:

np.argmax(a, axis=-1) = array([[1, 1, 0],
                               [1, 0, 1]])

I'd like to get max:

np.max(a, axis=-1) = array([[ 2,  4,  8],
                            [ 8,  9, 12]])

But without recalculating everything. I've tried:

a[np.arange(len(a)), np.argmax(a, axis=-1)]

But got:

IndexError: shape mismatch: indexing arrays could not be broadcast together with shapes (2,) (2,3) 

How to do it? Similar question for 2-d: numpy 2d array max/argmax

回答1:

You can use advanced indexing -

In [17]: a
Out[17]: 
array([[[ 1,  2],
        [ 3,  4],
        [ 8,  6]],

       [[ 7,  8],
        [ 9,  8],
        [ 3, 12]]])

In [18]: idx = a.argmax(axis=-1)

In [19]: m,n = a.shape[:2]

In [20]: a[np.arange(m)[:,None],np.arange(n),idx]
Out[20]: 
array([[ 2,  4,  8],
       [ 8,  9, 12]])

For a generic ndarray case of any number of dimensions, as stated in the comments by @hpaulj, we could use np.ix_, like so -

shp = np.array(a.shape)
dim_idx = list(np.ix_(*[np.arange(i) for i in shp[:-1]]))
dim_idx.append(idx)
out = a[dim_idx]

回答2:

For ndarray with arbitrary shape, you can flatten the argmax indices, then recover the correct shape, as so:

idx = np.argmax(a, axis=-1)
flat_idx = np.arange(a.size, step=a.shape[-1]) + idx.ravel()
maximum = a.ravel()[flat_idx].reshape(*a.shape[:-1])

来源：https://stackoverflow.com/questions/40357335/numpy-how-to-get-a-max-from-an-argmax-result