如何找到除一个特定索引以外的最小数字

Question

edit) Sorry my question wasn't clear. So, I wanted to find the min of a certain row and col without taking their intersection point, as @ParthSindhu said :)

I would like to find the min number from 2d array except the one number. (I'm using numpy array)

array([[30, 15, 41, 26, 12],
       [ 4, 19, 22, 40,  1],
       [41, 21,  0, 43, 22],
       [ 9, 40,  6, 10, 30],
       [24, 49, 22,  8, 41]])

For example, in row 2 and col 2, I would like to find the smallest number in each row and col except 0.

So, the answer want is 21 in row 2, and 6 in col 2.

I've tried to implement this code with 1d array,

a = np.arange(9, -1, -1)     # a = array([9, 8, 7, 6, 5, 4, 3, 2, 1, 0])
b = a[np.arange(len(a))!=3]  # b = array([9, 8, 7, 5, 4, 3, 2, 1, 0])

But, I could only find the one in the row but not in col.

a[np.arange(len(a))!=1].min()

The code just above returns 6

How could I do the same thing with col?

Sorry if the question is not so clear.

Answer 1

You Could this in case you are ignoring the intersection point:

import numpy as np
a = np.array([[30, 15, 41, 26, 12],
     [ 4, 19, 22, 40,  1],
     [41, 21,  0, 43, 22],
     [ 9, 40,  6, 10, 30],
     [24, 49, 22,  8, 41]]
row = 2
col = 2
col_indices = np.delete(np.arange(a.shape[0]), row)
row_indices = np.delete(np.arange(a.shape[1]), col)
col_min = a[col_indices, col].min()
row_min = a[row, row_indices].min()
print(col_min, row_min)

I'm sure there are better ways than this, this is just how i would do it.

Answer 2

You can use np.amin(a, axis = 1) to get an array with the smallest number in each row.

a = np.array([[30, 15, 41, 26, 12],
              [ 4, 19, 22, 40,  1],
              [41, 21,  0, 43, 22],
              [ 9, 40,  6, 10, 30],
              [24, 49, 22,  8, 41]])

print(np.amin(a, axis = 1))

This result in

>> [12  1  0  6  8]

Now you can run this again to find the smallest number in this array.

Dummy = np.amin(a, axis = 1)
print(np.amin(Dummy))

And you get the smallest number.

>> 0

You can change the axis if you set axis to 0 . So you can perform this operation on each axis of the array.

a = np.array([[30, 15, 41, 26, 12],
              [ 4, 19, 22, 40,  1],
              [41, 21,  0, 43, 22],
              [ 9, 40,  6, 10, 30],
              [24, 49, 22,  8, 41]])

print(np.amin(a, axis = 0))

>> [ 4 15  0  8  1]

Answer 3

You could use masked arrays:

a = np.array([[30, 15, 41, 26, 12],
              [ 4, 19, 22, 40,  1],
              [41, 21,  0, 43, 22],
              [ 9, 40,  6, 10, 30],
              [24, 49, 22,  8, 41]])

masked_a = np.ma.masked_array(a, mask=a == 0)
min_cols = masked_a.min(axis=0).data
min_rows = masked_a.min(axis=1).data

print(min_rows)
print(min_cols)

[12  1 21  6  8]
[ 4 15  6  8  1]

Answer 4

One possible way, replace 0 with inf :

a = np.array([[30, 15, 41, 26, 12],
       [ 4, 19, 22, 40,  1],
       [41, 21,  0, 43, 22],
       [ 9, 40,  6, 10, 30],
       [24, 49, 22,  8, 41]])

no_zero = np.where(a==0, np.inf, a)
no_zero.min(axis=0) # array([ 4., 15.,  6.,  8.,  1.])
no_zero.min(axis=1) # array([12.,  1., 21.,  6.,  8.])