问题
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?
回答1:
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))
回答2:
The strip method is another option that runs quickly for me:
z = 'abc'.strip(x+y) if x!=y else None
回答3:
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.
回答4:
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'
回答5:
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 :)
回答6:
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.
回答7:
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]
回答8:
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.
回答9:
I think it should looks like that:
z = (set(("a", "b", "c")) - set((x, y))).pop() if x != y else None
回答10:
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]]
回答11:
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'
来源：https://stackoverflow.com/questions/8792440/how-can-i-find-the-missing-value-more-concisely