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

A more idiomatic and readable solution would be to use itertools.combinations(list, 2) to take all pairs.

max(abs(a) - abs(b) for a, b in itertools.combinations(int_list, 2))

However, that is O(n²). A smarter O(n) solution would be to subtract the minimum absolute value from the maximum absolute value.

absolutes = [abs(elem) for elem in int_list]
print max(absolutes) - min(absolutes)

You should just code as you would write the steps:

outlist = []                          # create an empty output list
for x in data:                        # for each sublist in data
    val = max(abs(x[0]), abs(x[2]))   # find max of absolute of 1st and 3rd
    outlist.append(val)               # add above value to output list

print(outlist)                        # print the output list

Output:

[1.01201792, 2.68189991, 5.92202221, 10.19026759, 15.30091085]

If there may be more than 3 elements in sublists, you can use following:

outlist = []                        # create an empty output list
for x in data:                      # for each sublist in data
    x = [x[0]]+x[2:]                # create a new list without 2nd element
    val = max(x)                    # find max of new list
    outlist.append(val)             # add above value to output list

print(outlist)                      # print the output list