Numpy Indexing: Return the rest

You can do this nicely with boolean masks:

a = numpy.arange(10)

mask = np.ones(len(a), dtype=bool) # all elements included/True.
mask[[7,2,8]] = False              # Set unwanted elements to False

print a[mask]
# Gives (removing entries 7, 2 and 8):
[0 1 3 4 5 6 9]

Addition (taken from @mgilson). The binary mask created can be used nicely to get back the original slices with a[~mask] however this is only the same if the original indices were sorted.

---

EDIT: Moved down, as I had to realize that I would consider np.delete buggy at this time (Sep. 2012).

You could also use np.delete, though masks are more powerful (and in the future I think that should be an OK option). At the moment however its slower then the above, and will create unexpected results with negative indices (or steps when given a slice).

print np.delete(a, [7,2,8])

For this simple 1D case, I'd actually use a boolean mask:

a = numpy.arange(10)
include_index = numpy.arange(4)
include_idx = set(include_index)  #Set is more efficient, but doesn't reorder your elements if that is desireable
mask = numpy.array([(i in include_idx) for i in xrange(len(a))])

Now you can get your values:

included = a[mask]  # array([0, 1, 2, 3])
excluded = a[~mask] # array([4, 5, 6, 7, 8, 9])

Note that a[mask] doesn't necessarily yield the same thing as a[include_index] since the order of include_index matters for the output in that scenario (it should be roughly equivalent to a[sorted(include_index)]). However, since the order of your excluded items isn't well defined, this should work Ok.

---

EDIT

A better way to create the mask is:

mask = np.zeros(a.shape,dtype=bool)
mask[include_idx] = True

(thanks to seberg).

It's more like:

a = numpy.array([1, 2, 3, 4, 5, 6, 7, 4])
exclude_index = numpy.arange(5)
include_index = numpy.setdiff1d(numpy.arange(len(a)), exclude_index)
a[include_index]
# array([6, 7, 4])

# Notice this is a little different from
numpy.setdiff1d(a, a[exclude_index])
# array([6, 7]