Change the values of a NumPy array that are NOT in a list of indices

Question

I have a NumPy array like:

a = np.arange(30)

I know that I can replace the values located at positions indices=[2,3,4] using f

Answer 1

I don't know of a clean way to do something like this:

mask = np.ones(a.shape,dtype=bool) #np.ones_like(a,dtype=bool)
mask[indices] = False
a[~mask] = 999
a[mask] = 888

Of course, if you prefer to use the numpy data-type, you could use dtype=np.bool_ -- There won't be any difference in the output. it's just a matter of preference really.

Answer 2

Only works for 1d arrays:

a = np.arange(30)
indices = [2, 3, 4]

ia = np.indices(a.shape)

not_indices = np.setxor1d(ia, indices)
a[not_indices] = 888

Answer 3

Just overcome similar situation, solved this way:

a = np.arange(30)
indices=[2,3,4]

a[indices] = 999

not_in_indices = [x for x in range(len(a)) if x not in indices]

a[not_in_indices] = 888

Answer 4

Obviously there is no general not operator for sets. Your choices are:

1. Subtracting your indices set from a universal set of indices (depends on the shape of a), but that will be a bit difficult to implement and read.
2. Some kind of iteration (probably the for-loop is your best bet since you definitely want to use the fact that your indices are sorted).

3. Creating a new array filled with new value, and selectively copying indices from the old one.

   b = np.repeat(888, a.shape)
   b[indices] = a[indices]