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pythonnumpyargmax

numpy: how to get a max from an argmax result


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


Solution

  • 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]