sygi sygi - 1 month ago 6
Python Question

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

Answer

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]
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