将一系列1和0压缩为尽可能短的ascii字符串

Question

How could you convert a series of 1 s and 0 s into the shortest possible form consisting of URL safe ascii characters?

eg.

s = '00100101000101111010101'
compress(s)

Resulting in something like:

Ysi8aaU

And obviously:

decompress(compress(s)) == s

(I ask this question purely out of curiousity)

Answer 1

Here's the solution I came up with (+ far too many comments):

# A set of 64 characters, which allows a maximum chunk length of 6 .. because
# int('111111', 2) == 63 (plus zero)
charset = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789-_'

def encode(bin_string):
    # Split the string of 1s and 0s into lengths of 6.
    chunks = [bin_string[i:i+6] for i in range(0, len(bin_string), 6)]
    # Store the length of the last chunk so that we can add that as the last bit
    # of data so that we know how much to pad the last chunk when decoding.
    last_chunk_length = len(chunks[-1])
    # Convert each chunk from binary into a decimal
    decimals = [int(chunk, 2) for chunk in chunks]
    # Add the length of our last chunk to our list of decimals.
    decimals.append(last_chunk_length)
    # Produce an ascii string by using each decimal as an index of our charset.
    ascii_string = ''.join([charset[i] for i in decimals])

    return ascii_string

def decode(ascii_string):
    # Convert each character to a decimal using its index in the charset.
    decimals = [charset.index(char) for char in ascii_string]
    # Take last decimal which is the final chunk length, and the second to last
    # decimal which is the final chunk, and keep them for later to be padded
    # appropriately and appended.
    last_chunk_length, last_decimal = decimals.pop(-1), decimals.pop(-1)
    # Take each decimal, convert it to a binary string (removing the 0b from the
    # beginning, and pad it to 6 digits long.
    bin_string = ''.join([bin(decimal)[2:].zfill(6) for decimal in decimals])
    # Add the last decimal converted to binary padded to the appropriate length
    bin_string += bin(last_decimal)[2:].zfill(last_chunk_length)

    return bin_string

So:

>>> bin_string = '000111000010101010101000101001110'
>>> encode(bin_string)
'hcQOPgd'
>>> decode(encode(bin_string))
'000111000010101010101000101001110'

And here it is in CoffeeScript:

class Urlify
    constructor: ->
        @charset = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789-_'

    encode: (bits) ->
        chunks = (bits[i...i+6] for i in [0...bits.length] by 6)
        last_chunk_length = chunks[chunks.length-1].length
        decimals = (parseInt(chunk, 2) for chunk in chunks)
        decimals.push(last_chunk_length)
        encoded = (@charset[i] for i in decimals).join('')

        return encoded

    decode: (encoded) ->
        decimals = (@charset.indexOf(char) for char in encoded)
        [last_chunk_length, last_decimal] = [decimals.pop(), decimals.pop()]
        decoded = (('00000'+d.toString(2)).slice(-6) for d in decimals).join('')
        last_chunk = ('00000'+last_decimal.toString(2)).slice(-last_chunk_length)
        decoded += last_chunk

        return decoded

Answer 2

As one of the comments mentioned, using base64 would probably be the way to go. However, you don't want to stick the binary in without some converting.

Two options are converting to int first then packing:

import base64

s = '0110110'
n = int(s, 2)

result = base64.urlsafe_b64encode(str(n)).rstrip('=')

The other option would be to use the struct module to pack the value into a binary format and use this. (The code below is from http://www.fuyun.org/2009/10/how-to-convert-an-integer-to-base64-in-python/ )

import base64
import struct

def encode(n):
  data = struct.pack('<Q', n).rstrip('\x00')
  if len(data)==0:
    data = '\x00'
  s = base64.urlsafe_b64encode(data).rstrip('=')
  return s

def decode(s):
  data = base64.urlsafe_b64decode(s + '==')
  n = struct.unpack('<Q', data + '\x00'* (8-len(data)) )
  return n[0]

Answer 3

我会使用查找表将这些0和1中的8个转换为字节，然后使用base64对这些字节进行编码。