N largest values in each row of ndarray

You can use np.partition in the same way as the question you linked: the sorting is already along the last axis:

In [2]: a = np.array([[ 5,  4,  3,  2,  1],
               [10,  9,  8,  7,  6]])
In [3]: b = np.partition(a, -3)    # top 3 values from each row
In [4]: b[:,-3:]
Out[4]: 
array([[ 3,  4,  5],
       [ 8,  9, 10]])

You can use np.argsort along the rows with axis = 1 like so -

import numpy as np

# Find sorted indices for each row
sorted_row_idx = np.argsort(A, axis=1)[:,A.shape[1]-N::]

# Setup column indexing array
col_idx = np.arange(A.shape[0])[:,None]

# Use the column-row indices to get specific elements from input array. 
# Please note that since the column indexing array isn't of the same shape 
# as the sorted row indices, it will be broadcasted
out = A[col_idx,sorted_row_idx]

Sample run -

In [417]: A
Out[417]: 
array([[0, 3, 3, 2, 5],
       [4, 2, 6, 3, 1],
       [2, 1, 1, 8, 8],
       [6, 6, 3, 2, 6]])

In [418]: N
Out[418]: 3

In [419]: sorted_row_idx = np.argsort(A, axis=1)[:,A.shape[1]-N::]

In [420]: sorted_row_idx
Out[420]: 
array([[1, 2, 4],
       [3, 0, 2],
       [0, 3, 4],
       [0, 1, 4]], dtype=int64)

In [421]: col_idx = np.arange(A.shape[0])[:,None]

In [422]: col_idx
Out[422]: 
array([[0],
       [1],
       [2],
       [3]])

In [423]: out = A[col_idx,sorted_row_idx]

In [424]: out
Out[424]: 
array([[3, 3, 5],
       [3, 4, 6],
       [2, 8, 8],
       [6, 6, 6]])

If you would like to have the elements in descending order, you can use this additional step -

In [425]: out[:,::-1]
Out[425]: 
array([[5, 3, 3],
       [6, 4, 3],
       [8, 8, 2],
       [6, 6, 6]])