The built-in max takes a key function, you can pass that as abs:

>>> max([3, 7, -10], key=abs)
-10

You can call abs again on the result to normalise the result:

>>> abs(max([3, 7, -10], key=abs))
10

Let, df the dataframe with two columns, apply conditional absolute minimum over rows using axis=1

Here, x[np.argmin(x.abs())] checks the row index with absolute minimum and return the the original value with sign.

Also, we would need pd.isnull(x.min()) check for 'All-NaN slice encountered i.e when both the columns have NaN values.

In [3]: df.apply(lambda x: x.min() if pd.isnull(x.min()) else
                           x[np.argmin(x.abs())], axis=1)
Out[3]:
0    NaN
1      2
2      1
3      0
4     -1
5     -2
6     -3
7     -4
8      4
9      3
10     2
dtype: float64

You can use the key named parameter of min:

>>> min(-5, -10, 10, -2, -1, key=abs)
-1

See also: documentation for the built-in function min.

You can use min with abs as a key function:

import pandas as pd

def get_min_absvalue(values):
    return min(values, key = abs)

lst = [-2.1, 2.2, -1.5]

ser = pd.Series(lst)
min_absvalue = get_min_absvalue(ser)

print(min_absvalue)

Output:

-1.5

You can replace the values smaller than 0.1 by for example 1, using np.where:

np.where(np.abs(matrix)< 0.1,1,np.abs(matrix))

Then apply np.amin on top :

np.amin(np.where(np.abs(matrix)< 0.1,1,np.abs(matrix)),axis=-1)

Result:

array([[0.10321885],
       [0.12167502],
       [0.18718385],
       [0.15494743],
       [0.21274028]])