Numpy get index of row with second-largest value

Question

I am aware of how to get the index of the row containing the largest element with this snippet:

np.argmax(np.max(x, axis=0))

However, how could I get the index of the second largest row? I figure that I can cut off a row of x then try again, but is there a more elegant solution to this which can be scaled for the 3rd largest, 4th, etc?

Answer 1

You can use np.argsort(np.max(x, axis=0))[-2].

This scales to any index you want by changing the slicing index from -2 to -index.

Answer 2

Use argsort on flattened version and then use np.unravel_index to get row, col indices -

row,col = np.unravel_index(np.argsort(x.ravel()),x.shape)

Then, the largest row index would be row[-1], second largest in row[-2] and so on. Similarly, for columns, use col.

So, for convenience, you can flip the elements and use :

row,col = row[::-1],col[::-1]

Then, the largest ones would be at [0], second largest at [1] and so on.