How can I get the next string, in alphanumeric ordering, in Python?

Question

I need a simple program that given a string, returns to me the next one in the alphanumeric ordering (or just the alphabetic ordering).

f(\"aaa\")=\"aab\"
f(         


        

Answer 1

Sucks that python doesn't have what ruby has: String#next So here's a shitty solution to deal with alpha-numerical strings:

def next_string(s):
  a1 = range(65, 91)  # capital letters
  a2 = range(97, 123) # letters
  a3 = range(48, 58)  # numbers
  char = ord(s[-1])
  for a in [a1, a2, a3]:
    if char in a:
      if char + 1 in a:
        return s[:-1] + chr(char + 1)
      else:
        ns = next_string(s[:-1]) if s[:-1] else chr(a[0])
        return ns + chr(a[0])

print next_string('abc')  # abd
print next_string('123')  # 124
print next_string('ABC')  # ABD

# all together now
print next_string('a0')   # a1
print next_string('1a')   # 1b
print next_string('9A')   # 9B

# with carry-over
print next_string('9')    # 00
print next_string('z')    # aa
print next_string('Z')    # AA

# cascading carry-over
print next_string('a9')   # b0
print next_string('0z')   # 1a
print next_string('Z9')   # AA0

print next_string('199')  # 200
print next_string('azz')  # baa
print next_string('Zz9')  # AAa0

print next_string('$a')   # $b
print next_string('$_')   # None... fix it yourself

Not great. Kinda works for me.