Average of two strings in alphabetical/lexicographical order

Question

Suppose you take the strings \'a\' and \'z\' and list all the strings that come between them in alphabetical order: [\'a\',\'b\',\'c\' ... \'x\',\'y\',\'z\']. Take the midpoint

Answer 1

Thanks for everyone who answered, but I ended up writing my own solution because the others weren't exactly what I needed. I am trying to average app engine key names, and after studying them a bit more I discovered they actually allow any 7-bit ASCII characters in the names. Additionally I couldn't really rely on the solutions that converted the key names first to floating point, because I suspected floating point accuracy just isn't enough.

To take an average, first you add two numbers together and then divide by two. These are both such simple operations that I decided to just make functions to add and divide base 128 numbers represented as lists. This solution hasn't been used in my system yet so I might still find some bugs in it. Also it could probably be a lot shorter, but this is just something I needed to get done instead of trying to make it perfect.

# Given two lists representing a number with one digit left to decimal point and the
# rest after it, for example 1.555 = [1,5,5,5] and 0.235 = [0,2,3,5], returns a similar
# list representing those two numbers added together.
#
def ladd(a, b, base=128):
        i = max(len(a), len(b))
        lsum = [0] * i  
        while i > 1:
                i -= 1
                av = bv = 0
                if i < len(a): av = a[i]
                if i < len(b): bv = b[i]
                lsum[i] += av + bv
                if lsum[i] >= base:
                        lsum[i] -= base
                        lsum[i-1] += 1
        return lsum

# Given a list of digits after the decimal point, returns a new list of digits
# representing that number divided by two.
#
def ldiv2(vals, base=128):
        vs = vals[:]
        vs.append(0)
        i = len(vs)
        while i > 0:
                i -= 1
                if (vs[i] % 2) == 1:
                        vs[i] -= 1
                        vs[i+1] += base / 2
                vs[i] = vs[i] / 2
        if vs[-1] == 0: vs = vs[0:-1]
        return vs

# Given two app engine key names, returns the key name that comes between them.
#
def average(a_kn, b_kn):
        m = lambda x:ord(x)
        a = [0] + map(m, a_kn)
        b = [0] + map(m, b_kn)
        avg = ldiv2(ladd(a, b))
        return "".join(map(lambda x:chr(x), avg[1:]))

print average('a', 'z') # m@
print average('aa', 'zz') # n-@
print average('aa', 'az') # am@
print average('cat', 'doggie') # d(mstr@
print average('google', 'microsoft') # jlim.,7s:
print average('microsoft', 'google') # jlim.,7s: