Slicing numpy array with another array

Question

I\'ve got a large one-dimensional array of integers I need to take slices off. That\'s trivial, I\'d just do a[start:end]. The problem is that I need more of th

Answer 1

This can (almost?) be done in pure numpy using masked arrays and stride tricks. First, we create our mask:

>>> indices = numpy.arange(a.size)
>>> mask = ~((indices >= start[:,None]) & (indices < end[:,None]))

Or more simply:

>>> mask = (indices < start[:,None]) | (indices >= end[:,None])

The mask is False (i.e. values not masked) for those indices that are >= to the start value and < the end value. (Slicing with None (aka numpy.newaxis) adds a new dimension, enabling broadcasting.) Now our mask looks like this:

>>> mask
array([[ True, False,  True,  True,  True,  True,  True,  True,  True,
         True,  True,  True],
       [ True,  True,  True,  True,  True, False, False, False, False,
        False,  True,  True],
       [ True,  True,  True,  True,  True,  True,  True, False, False,
         True,  True,  True]], dtype=bool)

Now we have to stretch the array to fit the mask using stride_tricks:

>>> as_strided = numpy.lib.stride_tricks.as_strided
>>> strided = as_strided(a, mask.shape, (0, a.strides[0]))
>>> strided
array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11],
       [ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11],
       [ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11]], dtype=int16)

This looks like a 3x12 array, but each row points at the same memory. Now we can combine them into a masked array:

>>> numpy.ma.array(strided, mask=mask)
masked_array(data =
 [[-- 1 -- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- -- 5 6 7 8 9 -- --]
 [-- -- -- -- -- -- -- 7 8 -- -- --]],
             mask =
 [[ True False  True  True  True  True  True  True  True  True  True  True]
 [ True  True  True  True  True False False False False False  True  True]
 [ True  True  True  True  True  True  True False False  True  True  True]],
       fill_value = 999999)

This isn't quite the same as what you asked for, but it should behave similarly.