Python: Select subset from list based on index set

Use the built in function zip

property_asel = [a for (a, truth) in zip(property_a, good_objects) if truth]

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EDIT

Just looking at the new features of 2.7. There is now a function in the itertools module which is similar to the above code.

http://docs.python.org/library/itertools.html#itertools.compress

itertools.compress('ABCDEF', [1,0,1,0,1,1]) =>
  A, C, E, F

You could just use list comprehension:

property_asel = [val for is_good, val in zip(good_objects, property_a) if is_good]

or

property_asel = [property_a[i] for i in good_indices]

The latter one is faster because there are fewer good_indices than the length of property_a, assuming good_indices are precomputed instead of generated on-the-fly.

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Edit: The first option is equivalent to itertools.compress available since Python 2.7/3.1. See @Gary Kerr's answer.

property_asel = list(itertools.compress(property_a, good_objects))

I see 2 options.

1. Using numpy:

   property_a = numpy.array([545., 656., 5.4, 33.])
   property_b = numpy.array([ 1.2,  1.3, 2.3, 0.3])
   good_objects = [True, False, False, True]
   good_indices = [0, 3]
   property_asel = property_a[good_objects]
   property_bsel = property_b[good_indices]
   

2. Using a list comprehension and zip it:

   property_a = [545., 656., 5.4, 33.]
   property_b = [ 1.2,  1.3, 2.3, 0.3]
   good_objects = [True, False, False, True]
   good_indices = [0, 3]
   property_asel = [x for x, y in zip(property_a, good_objects) if y]
   property_bsel = [property_b[i] for i in good_indices]