>>> from operator import itemgetter
>>> a = range(100)
>>> itemgetter(5,13,25)(a)
(5, 13, 25)

The most direct way to do this with lists is to use a list comprehension:

c = [b[i] for i in index]

But, depending on exactly what your data looks like and what else you need to do with it, you could use numpy arrays - in which case:

c = b[index]

would do what you want, and would avoid the potential memory overhead for large slices - numpy arrays are stored more efficiently than lists, and slicing takes a view into the array rather than making a partial copy.

First, note that indexing in Python begins at 0. So the indices you need will be [0, 2].

You can then use a list comprehension:

temp = ['a', 'b', 'c', 'd']
idx = [0, 2]

res = [temp[i] for i in idx]            # ['a', 'c']

With built-ins, you may find map performs better:

res = map(temp.__getitem__, idx)        # ['a', 'c']

Since you are using Python 2.7, this returns a list. For Python 3.x, you would need to pass the map object to list.

If you are looking to avoid a Python-level loop altogether, you may wish to use a 3rd party library such as NumPy:

import numpy as np

temp = np.array(['a', 'b', 'c', 'd'])
res = temp[idx]

# array(['a', 'c'], 
#       dtype='<U1')

res2 = np.delete(temp, idx)

# array(['b', 'd'], 
#       dtype='<U1')

This returns a NumPy array, which you can then be converted to a list via res.tolist().