# How can I find the missing value more concisely?

## Question:

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?

## Answers:

```
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))
```

```
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'
```

I think it should looks like that:

```
z = (set(("a", "b", "c")) - set((x, y))).pop() if x != y else None
```

I think the solution by Sven Marnach and F.J is beautiful, but it’s not faster in my little test. This is Raymond’s optimized version using a pre-computed `set`

:

```
$ python -m timeit -s "choices = set('abc')"
-s "x = 'c'"
-s "y = 'a'"
"z, = choices - set(x + y)"
1000000 loops, best of 3: 0.689 usec per loop
```

This is the original solution:

```
$ python -m timeit -s "x = 'c'"
-s "y = 'a'"
"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'"
10000000 loops, best of 3: 0.310 usec per loop
```

Note that this is the **worst possible input** for the `if`

-statements since all six comparisons will have to be tried out. Testing with all values for `x`

and `y`

gives:

```
x = 'a', y = 'b': 0.084 usec per loop
x = 'a', y = 'c': 0.254 usec per loop
x = 'b', y = 'a': 0.133 usec per loop
x = 'b', y = 'c': 0.186 usec per loop
x = 'c', y = 'a': 0.310 usec per loop
x = 'c', y = 'b': 0.204 usec per loop
```

The `set`

-based variant shows the same performance for different inputs, but it is consistently between **2 and 8 times slower**. The reason is that the `if`

-based variant runs much simpler code: equality tests compared to hashing.

I think both types of solutions are valuable: it’s important to know that creating “complicated” data structures like sets cost you something in performance — while they give you a lot in readability and *development speed*. The complex data types are also much better when the code change: it’s easy to extend the set-based solution to four, five, … variables whereas the if-statements quickly turn into a maintenance nightmare.

```
z = 'a'*('a' not in x+y) or 'b'*('b' not in x+y) or 'c'
```

or less hackish and using Conditional Assignment

```
z = 'a' if ('a' not in x+y) else 'b' if ('b' not in x+y) else 'c'
```

but probably the dict solution is faster… you’d have to time it.

Try this option, using dictionaries:

```
z = {'ab':'c', 'ba':'c', 'bc':'a', 'cb':'a', 'ac':'b', 'ca':'b'}[x+y]
```

Of course, if the `x+y`

key is not present in the map, it’ll produce a `KeyError`

which you’ll have to handle.

If the dictionary is precomputed a single time and stored for future use, the access will be much faster, since no new data structures will have to be created for each evaluation, only a string concatenation and a dictionary lookup are needed:

```
lookup_table = {'ab':'c', 'ba':'c', 'bc':'a', 'cb':'a', 'ac':'b', 'ca':'b'}
z = lookup_table[x+y]
```

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 🙂

Using list comprehension, assuming like others that one of the three cases in your code holds:

```
l = ['a', 'b', 'c']
z = [n for n in l if n not in [x,y]].pop()
```

Or, like in the accepted answer, taking advantage of the tuple to unpack it,

```
z, = [n for n in l if n not in [x,y]]
```

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.

The `strip`

method is another option that runs quickly for me:

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

See if this works

```
if a not in xy
z= 'a'
if b not in xy
z='b'
if c not in xy
z='c'
```