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I want the shortest possible way of representing an integer in a URL. For example, 11234 can be shortened to '2be2' using hexadecimal. Since base64 uses is a 64 character encoding, it should be possible to represent an integer in base64 using even less characters than hexadecimal. The problem is I can't figure out the cleanest way to convert an integer to base64 (and back again) using Python.

The base64 module has methods for dealing with bytestrings - so maybe one solution would be to convert an integer to its binary representation as a Python string... but I'm not sure how to do that either.

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Simon: please look at Øystein krog's answer. You want to use a "base 64" representation of your integer data, and NOT the base64 module, which is meant to encode arbitrary binary data and doesn't compress the text representation of numbers. See en.wikipedia.org/wiki/Base_64 ) –  Ricardo Reyes Feb 18 '09 at 17:25
I was hoping it was possible to reuse the existing base64 module for part of the work, but sadly it looks like that's not the case. Thanks everyone for all of the excellent responses. –  Simon Willison Feb 25 '09 at 0:55
For anyone who's interested, I ended up rolling my own code for doing this: djangosnippets.org/snippets/1431 –  Simon Willison Apr 28 '09 at 14:23
("foo.com/blah_blah_(wikipedia)") –  Alix Axel Jan 8 '10 at 4:39
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12 Answers

up vote 43 down vote accepted

This answer is similar in spirit to Douglas Leeder's, with the following changes:

  • It doesn't use actual Base64, so there's no padding characters
  • Instead of converting the number first to a byte-string (base 256), it converts it directly to base 64, which has the advantage of letting you represent negative numbers using a sign character.

    import string
    ALPHABET = string.ascii_uppercase + string.ascii_lowercase + \
               string.digits + '-_'
    ALPHABET_REVERSE = dict((c, i) for (i, c) in enumerate(ALPHABET))
    BASE = len(ALPHABET)
    def num_encode(n):
        if n < 0:
            return SIGN_CHARACTER + num_encode(-n)
        s = []
        while True:
            n, r = divmod(n, BASE)
            if n == 0: break
        return ''.join(reversed(s))
    def num_decode(s):
        if s[0] == SIGN_CHARACTER:
            return -num_decode(s[1:])
        n = 0
        for c in s:
            n = n * BASE + ALPHABET_REVERSE[c]
        return n

    >>> num_encode(0)
    >>> num_encode(64)
    >>> num_encode(-(64**5-1))

A few side notes:

  • You could (marginally) increase the human-readibility of the base-64 numbers by putting string.digits first in the alphabet (and making the sign character '-'); I chose the order that I did based on Python's urlsafe_b64encode.
  • If you're encoding a lot of negative numbers, you could increase the efficiency by using a sign bit or one's/two's complement instead of a sign character.
  • You should be able to easily adapt this code to different bases by changing the alphabet, either to restrict it to only alphanumeric characters or to add additional "URL-safe" characters.
  • I would recommend against using a representation other than base 10 in URIs in most cases—it adds complexity and makes debugging harder without significant savings compared to the overhead of HTTP—unless you're going for something TinyURL-esque.
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Voted up to have thought about negative numbers. But isn’t one byte for the sign a bit expensive? –  kmkaplan Feb 21 '09 at 9:56
Yes, it is, which I addressed somewhat in my second note; but if that's not a concern, the implementation using a sign character was the simplest ;) –  Miles Feb 21 '09 at 22:34
The initial place I want to use this is "recover your account" style URLs which include a user ID, a timestamp and an sha1 hash - and should ideally be less than 80 characters to ensure they can be safely e-mailed without text wrapping screwing them up. –  Simon Willison Feb 25 '09 at 0:57
That's really good code but, according to Alex Martelli (stackoverflow.com/questions/931092/reverse-a-string-in-python/…), s[::-1] would be a faster way to reverse a string –  hwiechers Jul 13 '10 at 16:33
@hwiechers: s isn't actually a string, it's a list, so I still have to join it; I could do ''.join(s[::-1]) or ''.join(s)[::-1], but those are only somewhat faster—far less than the order of magnitude seen in telliott99's microbenchmark for reversing a string. –  Miles Jul 15 '10 at 7:14
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You probably do not want real base64 encoding for this - it will add padding etc, potentially even resulting in larger strings than hex would for small numbers. If there's no need to interoperate with anything else, just use your own encoding. Eg. here's a function that will encode to any base (note the digits are actually stored least-significant first to avoid extra reverse() calls:

def make_encoder(baseString):
    size = len(baseString)
    d = dict((ch, i) for (i, ch) in enumerate(baseString)) # Map from char -> value
    if len(d) != size:
        raise Exception("Duplicate characters in encoding string")

    def encode(x):
        if x==0: return baseString[0]  # Only needed if don't want '' for 0
        while x>0:
            l.append(baseString[x % size])
            x //= size
        return ''.join(l)

    def decode(s):
        return sum(d[ch] * size**i for (i,ch) in enumerate(s))

    return encode, decode

# Base 64 version:
encode,decode = make_encoder("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/")

assert decode(encode(435346456456)) == 435346456456

This has the advantage that you can use whatever base you want, just by adding appropriate characters to the encoder's base string.

Note that the gains for larger bases are not going to be that big however. base 64 will only reduce the size to 2/3rds of base 16 (6 bits/char instead of 4). Each doubling only adds one more bit per character. Unless you've a real need to compact things, just using hex will probably be the simplest and fastest option.

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The easy bit is converting the byte string to web-safe base64:

import base64
output = base64.urlsafe_b64encode(s)

The tricky bit is the first step - convert the integer to a byte string.

If your integers are small you're better off hex encoding them - see saua

Otherwise (hacky recursive version):

def convertIntToByteString(i):
    if i == 0:
        return ""
        return convertIntToByteString(i >> 8) + chr(i & 255)
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To encode n:

data = ''
while n > 0:
    data = chr(n & 255) + data
    n = n >> 8
encoded = base64.urlsafe_b64encode(data).rstrip('=')

To decode s:

data = base64.urlsafe_b64decode(s + '===')
decoded = 0
while len(data) > 0:
    decoded = (decoded << 8) | ord(data[0])
    data = data[1:]

In the same spirit as other for some “optimal” encoding, you can use 73 characters according to RFC 1738 (actually 74 if you count “+” as usable):

alphabet = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz_`\"!$'()*,-."
encoded = ''
while n > 0:
    n, r = divmod(n, len(alphabet))
    encoded = alphabet[r] + encoded

and the decoding:

decoded = 0
while len(s) > 0:
    decoded = decoded * len(alphabet) + alphabet.find(s[0])
    s = s[1:]
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You don't want base64 encoding, you want to represent a base 10 numeral in numeral base X.

If you want your base 10 numeral represented in the 26 letters available you could use: http://en.wikipedia.org/wiki/Hexavigesimal. (You can extend that example for a much larger base by using all the legal url characters)

You should atleast be able to get base 38 (26 letters, 10 numbers, +, _)

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You are correct, but he can still use base 64 by using digits, lowercase, uppercase, and -_. –  Mike Boers Feb 18 '09 at 16:45
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All the answers given regarding Base64 are very reasonable solutions. But they're technically incorrect. To convert an integer to the shortest URL safe string possible, what you want is base 66 (there are 66 URL safe characters).

That code looks like this:

from io import StringIO
import urllib

BASE66_ALPHABET = u"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz-_.~"

def hexahexacontadecimal_encode_int(n):
    if n == 0:
        return BASE66_ALPHABET[0].encode('ascii')

    r = StringIO()
    while n:
        n, t = divmod(n, BASE)
    return r.getvalue().encode('ascii')[::-1]

Here's a full implementation with source and ready to go pip installable package:


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a little hacky, but it works:

def b64num(num_to_encode):
  h = hex(num_to_encode)[2:]     # hex(n) returns 0xhh, strip off the 0x
  h = len(h) & 1 and '0'+h or h  # if odd number of digits, prepend '0' which hex codec requires
  return h.decode('hex').encode('base64')

you could replace the call to .encode('base64') with something in the base64 module, such as urlsafe_b64encode()

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I tried that with 12345. It gave me: 'MDk=\n' That seems to have converted a 5-digit integer into a length 5 string. I can think of easier ways to achieve that :-) –  John Fouhy Feb 18 '09 at 20:58
the = and the \n are padding that you can strip off –  ʞɔıu Sep 29 '09 at 15:43
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I maintain a little library named zbase62: http://pypi.python.org/pypi/zbase62

With it you can convert from a Python 2 str object to a base-62 encoded string and vice versa:

Python 2.7.1+ (r271:86832, Apr 11 2011, 18:13:53) 
[GCC 4.5.2] on linux2
>>> import os
>>> d = os.urandom(32)
>>> d
>>> from zbase62 import zbase62
>>> encoded = zbase62.b2a(d)
>>> encoded
>>> zbase62.a2b(encoded)

However, you still need to convert from integer to str. This comes built-in to Python 3:

Python 3.2 (r32:88445, Mar 25 2011, 19:56:22)
[GCC 4.5.2] on linux2
>>> import os
>>> d = os.urandom(32)
>>> d
>>> int.from_bytes(d, 'big')
>>> x= _ 
>>> x.to_bytes(32, 'big')

To convert from int to bytes and vice versa in Python 2, there is not a convenient, standard way as far as I know. I guess maybe I should copy some implementation, such as this one: https://github.com/warner/foolscap/blob/46e3a041167950fa93e48f65dcf106a576ed110e/foolscap/banana.py#L41 into zbase62 for your convenience.

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Base64 takes 4 bytes/characters to encode 3 bytes and can only encode multiples of 3 bytes (and adds padding otherwise).

So representing 4 bytes (your average int) in Base64 would take 8 bytes. Encoding the same 4 bytes in hex would also take 8 bytes. So you wouldn't gain anything for a single int.

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@saua: You forget that each digit only encodes ~3.3 bits while each character of base64 encodes 6, ergo representing an integer in base64 (instead of base 10) will result in a string approx half as long. –  Mike Boers Feb 18 '09 at 16:39
@Mike I discussed the length of hex (base-16) encoding vs. base64, and due to the padding the length is the same for 4 bytes of data. Of course this changes for longer strings, but the question is explicitely about encoding an int. –  Joachim Sauer Feb 18 '09 at 16:42
@saua: But you don't nessesarily have an int that requires 4 whole bytes. Decimal 1 can still be B64 1, and then decimal 64 can be B64 10. –  Mike Boers Feb 18 '09 at 16:48
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If you are looking for a way to shorten the integer representation using base64, I think you need to look elsewhere. When you encode something with base64 it doesn't get shorter, in fact it gets longer.

E.g. 11234 encoded with base64 would yield MTEyMzQ=

When using base64 you have overlooked the fact that you are not converting just the digits (0-9) to a 64 character encoding. You are converting 3 bytes into 4 bytes so you are guaranteed your base64 encoded string would be 33.33% longer.

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The first step is to convert the integer to a byte string. –  Douglas Leeder Feb 18 '09 at 15:49
You are correct if you are encoding a string representation of a decimal number into base 64, but not if you want to encode the number itself into base 64. Each decimal digit encodes ~3.3 bits of info, while each char of base 64 encodes 6 bits of info. Ergo the base64 number will be shorter. –  Mike Boers Feb 18 '09 at 16:43
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I'd go the 'encode integer as binary string, then base64 encode that' method you suggest, and I'd do it using struct:

>>> import struct, base64
>>> base64.b64encode(struct.pack('l', 47))
>>> struct.unpack('l', base64.b64decode(_))

Edit again: To strip out the extra 0s on numbers that are too small to need full 32-bit precision, try this:

def pad(str, l=4):
    while len(str) < l:
        str = '\x00' + str
    return str

>>> base64.b64encode(struct.pack('!l', 47).replace('\x00', ''))
>>> struct.unpack('!l', pad(base64.b64decode('Lw==')))
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@Jorenko: This is far from the most efficient. 47 in base 64 can be represented by a single character (as 47 is less than 64.) –  Mike Boers Feb 18 '09 at 16:42
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I needed a signed integer, so I ended up going with:

import struct, base64

def b64encode_integer(i):
   return base64.urlsafe_b64encode(struct.pack('i', i)).rstrip('=\n')


>>> b64encode_integer(1)
>>> b64encode_integer(-1)
>>> b64encode_integer(256)
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