Find the non-intersecting values of two arrays

np.setdiff1d(a,b)

This will return non intersecting value of first argument with second argument
Example:

a = [1,2,3]
b = [1,3]
np.setdiff1d(a,b)  -> returns [2]
np.setdiff1d(b,a)  -> returns []

You can use setxor1d. According to the documentation:

Find the set exclusive-or of two arrays.
Return the sorted, unique values that are in only one (not both) of the input arrays.

Usage is as follows:

import numpy

a = ['Brian', 'Steve', 'Andrew', 'Craig']
b = ['Andrew','Steve']

c = numpy.setxor1d(a, b)

Executing this will result in c having a value of array(['Brian', 'Craig']).

Given that none of the objects shown in your question are Numpy arrays, you don't need Numpy to achieve this:

c = list(set(a).symmetric_difference(b))

If you have to have a Numpy array as the output, it's trivial to create one:

c = np.array(set(a).symmetric_difference(b))

(This assumes that the order in which elements appear in c does not matter. If it does, you need to state what the expected order is.)

P.S. There is also a pure Numpy solution, but personally I find it hard to read:

c = np.setdiff1d(np.union1d(a, b), np.intersect1d(a, b))

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