numpy: how to get a max from an argmax result

Question

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

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

          [[ 7,  8],
           [ 9,  8],
         


        

Answer 1

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])

Answer 2

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]

Answer 3

For arbitrary-shape arrays, the following should work :)

a = np.arange(5 * 4 * 3).reshape((5,4,3))

# for last axis
argmax = a.argmax(axis=-1)
a[tuple(np.indices(a.shape[:-1])) + (argmax,)]

# for other axis (eg. axis=1)
argmax = a.argmax(axis=1)
idx = list(np.indices(a.shape[:1]+a.shape[2:]))
idx[1:1] = [argmax]
a[tuple(idx)]

or

a = np.arange(5 * 4 * 3).reshape((5,4,3))

argmax = a.argmax(axis=0)
np.choose(argmax, np.moveaxis(a, 0, 0))

argmax = a.argmax(axis=1)
np.choose(argmax, np.moveaxis(a, 1, 0))

argmax = a.argmax(axis=2)
np.choose(argmax, np.moveaxis(a, 2, 0))

argmax = a.argmax(axis=-1)
np.choose(argmax, np.moveaxis(a, -1, 0))