Use numpy.delete(), which returns a new array with sub-arrays along an axis deleted.

numpy.delete(a, index)

For your specific question:

import numpy as np

a = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9])
index = [2, 3, 6]

new_a = np.delete(a, index)

print(new_a)
# Output: [1, 2, 5, 6, 8, 9]

Note that numpy.delete() returns a new array since array scalars are immutable, similar to strings in Python, so each time a change is made to it, a new object is created. I.e., to quote the delete() docs:

"A copy of arr with the elements specified by obj removed. Note that delete does not occur in-place..."

If the code I post has output, it is the result of running the code.

You can find the index/indices of the object using np.argwhere, and then delete the object(s) using np.delete.

Example:

x = np.array([1,2,3,4,5])
index = np.argwhere(x==3)
y = np.delete(x, index)
print(x, y)

Python sets does a nice job of giving differences. It doesn't, though, maintain order

np.array(list(set(tuple(x) for x in a.tolist()).difference(set(tuple(x) for x in b.tolist()))))

Or to use boolean indexing, use broadcasting to create an outer equals, and sum with any and all

A = np.all((a[None,:,:]==b[:,None,:]),axis=-1)
A = np.any(A,axis=0)
a[~A,:]

Or make a and b complex:

ac = np.dot(a,[1,1j])
bc = np.dot(b,[1,1j])
A = np.any(ac==bc[:,None],axis=0)
a[~A,:]

or to use setxor1d

xx = np.setxor1d(ac,bc)
# array([ 2.+1.j,  3.+1.j,  3.+3.j])
np.array([xx.real,xx.imag],dtype=int).T

=================

In [222]: ac = np.dot(a,[1,1j])
     ...: bc = np.dot(b,[1,1j])
In [223]: ac
Out[223]: array([ 1.+1.j,  2.+1.j,  3.+1.j,  3.+2.j,  3.+3.j,  5.+5.j])
In [225]: bc
Out[225]: array([ 1.+1.j,  5.+5.j,  3.+2.j])
In [226]: ac == bc[:,None]
Out[226]: 
array([[ True, False, False, False, False, False],
       [False, False, False, False, False,  True],
       [False, False, False,  True, False, False]], dtype=bool)