Python: Elegant and efficient ways to mask a list

Question

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 lists.

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] 

Answer 1

If you are using numpy:

>>> import numpy as np >>> a = np.arange(8) >>> mask = np.array([False, False, False, False, True, True, True, False], dtype=np.bool) >>> a[mask] array([4, 5, 6]) 

If you are not using numpy you are looking for itertools.compress

>>> from itertools import compress >>> a = range(8) >>> mask = [False, False, False, False, True, True, True, False] >>> list(compress(a, mask)) [4, 5, 6]