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Example: from __future__ import division import numpy as np n = 8 """masking lists""" lst = range(n) print lst # the mask (filter) msk = [(el>3) and (el<=6) for el in lst] print msk # use of the mask print [lst[i] for i in xrange(len(lst)) if msk[i]] """masking arrays""" ary = np.arange(n) print ary # the mask (filter) msk = (ary>3)&(ary<=6) print msk # use of the mask print ary[msk] # very elegant and the results are: >>> [0, 1, 2, 3, 4, 5, 6, 7] [False, False, False, False, True, True, True, False] [4, 5, 6] [0 1 2 3 4 5 6 7] [False False False False True True True False] [4 5 6] As you see the operation of masking on array is more elegant compared to list. If you try to use the array masking scheme on list you'll get an error: >>> lst[msk] Traceback (most recent call last): File "<interactive input>", line 1, in <module> TypeError: only integer arrays with one element can be converted to an index The question is to find an elegant masking for list s. Updates: The answer by jamylak was accepted for introducing compress however the points mentioned by Joel Cornett made the solution complete to a desired form of my interest. >>> mlist = MaskableList >>> mlist(lst)[msk] >>> [4, 5, 6]

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