Incrementing string in Python

Question

I am writing a function to increment a 3-letter (a-z) string. For example:
Input: aaa
Output: baa

Input: zba
Output: aca

So the order is as following

aaa
baa
...
zaa
aba
bba
cba
...
zba
aca
bca
cca
...
zca
ada
...
zzz
aaa

I wrote the following function next_code() and it works, but I am wondering if there is a more elegant way to implement it rather than looping through individual letters in the string:

# 0 = a; 25 = z
def digit_to_char(digit):
    return chr(ord('a') + digit)

# a = 0; z = 25
def char_to_digit(char):
    return ord(char)-ord('a')

def next_code(code):
    # if used up all codes, loop from start
    if code == 'zzz':
        return next_code('aaa')
    else:
        code = list(code)
        # loop over letters and see which one we can increment
        for (i, letter) in enumerate(code):
            if letter == 'z':
                # go on to the next letter
                code[i] = 'a'
                continue
            else:
                # increment letter
                code[i] = digit_to_char(char_to_digit(letter) + 1)
                return ("".join(code))
                break



print (next_code('aab'))

Joran Beasley · Accepted Answer · 2016-06-21 23:41:38Z

6

just use itertools product

>>> import itertools
>>> from string import ascii_lowercase
>>> strings = itertools.product(*[ascii_lowercase]*3)
>>> "".join(next(strings,"No More Combos..."))
'aaa'
>>> "".join(next(strings,"No More Combos..."))
'aab'
>>> "".join(next(strings,"No More Combos..."))
'aac'
...

is how I would probably do it

if you want to cycle back to 'aaa' after the end you can just use itertools.cycle

strings = itertools.cycle(itertools.product(*[ascii_lowercase]*3))

edited Jun 21, 2016 at 23:41

answered Jun 21, 2016 at 23:36

Joran Beasley

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1 Comment

Maxim Neaga Over a year ago

Thank you, +1 for a very neat solution. Unfortunately does not support input (to get the next value based off of it), but shouldn't be hard to add.

ShadowRanger · Accepted Answer · 2016-06-21 23:47:36Z

2

You can simplify the loop a lot:

def next_code(code):
    code = list(code)
    for i, let in enumerate(code):
        if let != 'z':
            code[i] = chr(ord(let) + 1)
            break
        code[i] = 'a'
    return ''.join(code)

If the goal is to just produce all values one by one, starting with 'aaa', itertools.product can be used to make a generator:

from future_builtins import map  # Only on Python 2
from itertools import product

def allcodes():
    # You want the left side to vary faster, so reverse before joining
    return map(''.join, map(reversed, product(string.ascii_lowercase, repeat=3)))

for code in allcodes():
    print(code)

Or you make it a function that you call as needed to get the next code in the sequence without using it as an iterator:

nextcode = allcodes().__next__  # .next on Py2

And if the generator should be infinite (so it wraps from zzz to aaa), just change allcodes to either:

# Avoid cycle if storing all 26**3 codes in memory is a bad idea
def allcodes():
    while True:
        yield from map(''.join, map(reversed, product(string.ascii_lowercase, repeat=3)))
        # On Py2, change yield from line to:
        # for code in map(''.join, map(reversed, product(string.ascii_lowercase, repeat=3))): yield code

or at higher memory cost but greater simplicity:

from itertools import cycle

def allcodes():
    return cycle(map(''.join, map(reversed, product(string.ascii_lowercase, repeat=3))))

edited Jun 21, 2016 at 23:47

answered Jun 21, 2016 at 23:36

ShadowRanger

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2 Comments

Joran Beasley Over a year ago

wow great answer ... you covered so many bases for this guy +1 from me :)

ShadowRanger Over a year ago

@JoranBeasley: TIMTOWTDI? :-) And I like to find them all. I started off with Perl.

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