PYTHON - Calculate Min and Index of Min in an NdArray

Question

I have an n*n array, and I want to find the min in the array, and get the index of the min in [x,y] format

Of course, this can be done using for loops and using temporary variables, but I am looking for a more sophisticated process to do this.

Example -

    [[1,2,8],
     [7,4,2],
     [9,1,7],
     [0,1,5],
     [6,-4,3]]

I should get the following output -

    Output-
    Min = -4
    Index = [4,1]

Can I implement something similar?

TIA.

Answer 1

Global minimum value and index

Flatten the array, get the argmin index. Get the corresponding row-col indices from it with np.unravel_index. Also, index into the flattened array with the earlier obtained flattened argmin index for the minimum value.

def smallest_val_index(a):
    idx = a.ravel().argmin()
    return a.ravel()[idx], np.unravel_index(idx, a.shape)

Sample run -

In [182]: a
Out[182]: 
array([[ 1,  2,  8],
       [ 7,  4,  2],
       [ 9,  1,  7],
       [ 0,  1,  5],
       [ 6, -4,  3]])

In [183]: val, indx = smallest_val_index(a)

In [184]: val
Out[184]: -4

In [185]: indx
Out[185]: (4, 1)

---

Global maximum value and index

Similarly, to get the global maximum value, use argmax -

def largest_val_index(a):
    idx = a.ravel().argmax()
    return a.ravel()[idx], np.unravel_index(idx, a.shape)

Sample run -

In [187]: a
Out[187]: 
array([[ 1,  2,  8],
       [ 7,  4,  2],
       [ 9,  1,  7],
       [ 0,  1,  5],
       [ 6, -4,  3]])

In [188]: largest_val_index(a)
Out[188]: (9, (2, 0))