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]
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]