How can I find the missing value more concisely?

The following code checks if x and y are distinct values (the variables x, y, z can only have values a, b, or c) and if so, sets z to the third character:

if x == 'a' and y == 'b' or x == 'b' and y == 'a':
    z = 'c'
elif x == 'b' and y == 'c' or x == 'c' and y == 'b':
    z = 'a'
elif x == 'a' and y == 'c' or x == 'c' and y == 'a':
    z = 'b'

Is is possible to do this in a more, concise, readable and efficient way?

z = (set(("a", "b", "c")) - set((x, y))).pop()

I am assuming that one of the three cases in your code holds. If this is the case, the set set(("a", "b", "c")) - set((x, y)) will consist of a single element, which is returned by pop().

Edit: As suggested by Raymond Hettinger in the comments, you could also use tuple unpacking to extract the single element from the set:

z, = set(("a", "b", "c")) - set((x, y))

The strip method is another option that runs quickly for me:

z = 'abc'.strip(x+y) if x!=y else None

Sven's excellent code did just a little too much work and chould have used tuple unpacking instead of pop(). Also, it could have added a guard if x != y to check for x and y being distinct. Here is what the improved answer looks like:

# create the set just once
choices = {'a', 'b', 'c'}

x = 'a'
y = 'b'

# the main code can be used in a loop
if x != y:
    z, = choices - {x, y}

Here are the comparative timings with a timing suite to show the relative performance:

import timeit, itertools

setup_template = '''
x = %r
y = %r
choices = {'a', 'b', 'c'}
'''

new_version = '''
if x != y:
    z, = choices - {x, y}
'''

original_version = '''
if x == 'a' and y == 'b' or x == 'b' and y == 'a':
    z = 'c'
elif x == 'b' and y == 'c' or x == 'c' and y == 'b':
    z = 'a'
elif x == 'a' and y == 'c' or x == 'c' and y == 'a':
    z = 'b'
'''

for x, y in itertools.product('abc', repeat=2):
    print '\nTesting with x=%r and y=%r' % (x, y)
    setup = setup_template % (x, y)
    for stmt, name in zip([original_version, new_version], ['if', 'set']):
        print min(timeit.Timer(stmt, setup).repeat(7, 100000)),
        print '\t%s_version' % name

Here are the results of the timings:

Testing with x='a' and y='a'
0.0410830974579     original_version
0.00535297393799    new_version

Testing with x='a' and y='b'
0.0112571716309     original_version
0.0524711608887     new_version

Testing with x='a' and y='c'
0.0383319854736     original_version
0.048309803009      new_version

Testing with x='b' and y='a'
0.0175108909607     original_version
0.0508949756622     new_version

Testing with x='b' and y='b'
0.0386209487915     original_version
0.00529098510742    new_version

Testing with x='b' and y='c'
0.0259420871735     original_version
0.0472128391266     new_version

Testing with x='c' and y='a'
0.0423510074615     original_version
0.0481910705566     new_version

Testing with x='c' and y='b'
0.0295209884644     original_version
0.0478219985962     new_version

Testing with x='c' and y='c'
0.0383579730988     original_version
0.00530385971069    new_version

These timings show that the original-version performance varies quite a bit depending on which if-statements are triggered by the various the input values.

z = (set('abc') - set(x + y)).pop()

Here are all of the scenarios to show that it works:

>>> (set('abc') - set('ab')).pop()   # x is a/b and y is b/a
'c'
>>> (set('abc') - set('bc')).pop()   # x is b/c and y is c/b
'a'
>>> (set('abc') - set('ac')).pop()   # x is a/c and y is c/a
'b'

If the three items in question weren't "a", "b" and "c", but rather 1, 2 and 3, you could also use a binary XOR:

z = x ^ y

More generally, if you want to set z to the remaining one of three numbers a, b and c given two numbers x and y from this set, you can use

z = x ^ y ^ a ^ b ^ c

Of course you can precompute a ^ b ^ c if the numbers are fixed.

This approach can also be used with the original letters:

z = chr(ord(x) ^ ord(y) ^ 96)

Example:

>>> chr(ord("a") ^ ord("c") ^ 96)
'b'

Don't expect anyone reading this code to immediately figure out what it means :)