max(max(a),-min(a))

It's the fastest for now, since no intermediate list is created (for 100 000 values):

In [200]: %timeit max(max(a),-min(a))
100 loops, best of 3: 8.82 ms per loop

In [201]: %timeit abs(max(a,key=abs))
100 loops, best of 3: 13.8 ms per loop

In [202]: %timeit max(map(abs,a))
100 loops, best of 3: 13.2 ms per loop

In [203]: %timeit max(abs(n) for n in a)
10 loops, best of 3: 19.9 ms per loop

In [204]: %timeit np.abs(a).max()
100 loops, best of 3: 11.4 ms per loop

OK, after reading and understanding your question and not being able to find a vectorised approach, we can define a custom function and call apply and pass each row.

So this will check if either column is null if so return the min value, it then compares the abs value of either column and then returns the column that has smallest abs value but the original value including sign:

In [269]:

def func(x):
    if pd.isnull(x.a) or pd.isnull(x.b):
        return x.min()
    elif np.abs(x.a) < np.abs(x.b):
        return x.a
    else:
        return x.b

df.apply(lambda row: func(row), axis = 1)
Out[269]:
0    NaN
1      2
2      1
3      0
4     -1
5     -2
6     -3
7     -4
8      4
9      3
10     2
dtype: float64