基62转换

这方面没有标准的模块,但我已经编写了自己的函数来实现这一点。

BASE62 = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"

def encode(num, alphabet=BASE62):
    """Encode a positive number in Base X

    Arguments:
    - `num`: The number to encode
    - `alphabet`: The alphabet to use for encoding
    """
    if num == 0:
        return alphabet[0]
    arr = []
    base = len(alphabet)
    while num:
        num, rem = divmod(num, base)
        arr.append(alphabet[rem])
    arr.reverse()
    return ''.join(arr)

def decode(string, alphabet=BASE62):
    """Decode a Base X encoded string into the number

    Arguments:
    - `string`: The encoded string
    - `alphabet`: The alphabet to use for encoding
    """
    base = len(alphabet)
    strlen = len(string)
    num = 0

    idx = 0
    for char in string:
        power = (strlen - (idx + 1))
        num += alphabet.index(char) * (base ** power)
        idx += 1

    return num

请注意,您可以为它提供任何用于编码和解码的字母表。如果你离开 alphabet 论证一下,你将得到在代码的第一行定义的62个字符的字母表,从而对62个碱基进行编码/解码。

希望这有帮助。

ps-对于URL缩短器,我发现最好去掉一些令人困惑的字符,如0ol1oi等。因此,我使用这个字母表来满足我的URL缩短需求。- "23456789abcdefghijkmnpqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ"

玩得高兴。

我曾经写过一个脚本来做这个,我认为它相当优雅:)

import string
BASE_LIST = string.digits + string.letters + '_@'
BASE_DICT = dict((c, i) for i, c in enumerate(BASE_LIST))

def base_decode(string, reverse_base=BASE_DICT):
    length = len(reverse_base)
    ret = 0
    for i, c in enumerate(string[::-1]):
        ret += (length ** i) * reverse_base[c]

    return ret

def base_encode(integer, base=BASE_LIST):
    if integer == 0:
        return base[0]

    length = len(base)
    ret = ''
    while integer != 0:
        ret = base[integer % length] + ret
        integer /= length

    return ret

示例用法:

for i in range(100):                                    
    print i, base_decode(base_encode(i)), base_encode(i)

下面的解码器生成器使用任何合理的基础,具有更整齐的循环,并在遇到无效字符时给出明确的错误消息。

def base_n_decoder(alphabet):
    """Return a decoder for a base-n encoded string
    Argument:
    - `alphabet`: The alphabet used for encoding
    """
    base = len(alphabet)
    char_value = dict(((c, v) for v, c in enumerate(alphabet)))
    def f(string):
        num = 0
        try:
            for char in string:
                num = num * base + char_value[char]
        except KeyError:
            raise ValueError('Unexpected character %r' % char)
        return num
    return f

if __name__ == "__main__":
    func = base_n_decoder('0123456789abcdef')
    for test in ('0', 'f', '2020', 'ffff', 'abqdef'):
        print test
        print func(test)

如果你在寻找最高的效率(如Django),你会想要如下的东西。这是一个有效的方法结合白沙姆帕扬戈泽和沃尔夫和约翰马金。

# Edit this list of characters as desired.
BASE_ALPH = tuple("0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz")
BASE_DICT = dict((c, v) for v, c in enumerate(BASE_ALPH))
BASE_LEN = len(BASE_ALPH)

def base_decode(string):
    num = 0
    for char in string:
        num = num * BASE_LEN + BASE_DICT[char]
    return num

def base_encode(num):
    if not num:
        return BASE_ALPH[0]

    encoding = ""
    while num:
        num, rem = divmod(num, BASE_LEN)
        encoding = BASE_ALPH[rem] + encoding
    return encoding

你也可以提前计算你的字典。(注意:使用字符串编码比使用列表编码效率高,即使使用很长的数字也是如此。)

>>> timeit.timeit("for i in xrange(1000000): base.base_decode(base.base_encode(i))", setup="import base", number=1)
2.3302059173583984

在2.5秒内编码和解码了100万个数字。(2.2GHz i7-2670qm)

你可能想要base64,而不是base62。它有一个与URL兼容的版本,所以额外的两个填充字符不应该是问题。

这个过程相当简单;考虑base64代表6位,常规字节代表8。为所选的64个字符中的每一个指定一个从000000到111111的值,并将4个值放在一起以匹配一组3位256字节。对每组3个字节重复此操作,在末尾填充您选择的填充字符(0通常很有用)。

如果您所需要的只是生成一个简短的ID(因为您提到了URL缩短器),而不是编码/解码某些内容,那么此模块可能会帮助:

https://github.com/stochastic-technologies/shortuuid/

您可以从下载zbase62模块 pypi

如

>>> import zbase62
>>> zbase62.b2a("abcd")
'1mZPsa'

我从别人的帖子中获益匪浅。我最初需要在Django项目中使用python代码,但从那时起我就转向node.js,所以这里是 javascript版本 Baishampayan Ghose提供的代码(编码部分)。

var ALPHABET = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";

function base62_encode(n, alpha) {
  var num = n || 0;
  var alphabet = alpha || ALPHABET;

  if (num == 0) return alphabet[0];
  var arr = [];
  var base = alphabet.length;

  while(num) {
    rem = num % base;
    num = (num - rem)/base;
    arr.push(alphabet.substring(rem,rem+1));
  }

  return arr.reverse().join('');
}

console.log(base62_encode(2390687438976, "123456789ABCDEFGHIJKLMNPQRSTUVWXYZ"));

我希望下面的片段能有所帮助。

def num2sym(num, sym, join_symbol=''):
    if num == 0:
        return sym[0]
    if num < 0 or type(num) not in (int, long):
        raise ValueError('num must be positive integer')

    l = len(sym)  # target number base
    r = []
    div = num
    while div != 0: # base conversion
        div, mod = divmod(div, l)
        r.append(sym[mod])

    return join_symbol.join([x for x in reversed(r)])

您的案例用法:

number = 367891
alphabet = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
print num2sym(number, alphabet)  # will print '1xHJ'

显然,您可以指定另一个字母表,由较小或较大数量的符号组成,然后它将您的数字转换为较小或较大的数字基数。例如,提供“01”作为字母表将输出表示输入数字为二进制的字符串。

你可以先改变字母表,使其具有唯一的数字表示。如果你在做网址缩短服务,它会很有帮助。

我的解决方案是:

def base62(a):
    baseit = (lambda a=a, b=62: (not a) and '0' or
        baseit(a-a%b, b*62) + '0123456789abcdefghijklmnopqrstuvwxyz'
                              'ABCDEFGHIJKLMNOPQRSTUVWXYZ'[a%b%61 or -1*bool(a%b)])
    return baseit()

解释

在任何基数中,每个数字都等于 a1+a2*base**2+a3*base**3... 所以我们的目标是找到 a S.

对于每一个 N=1,2,3... 代码隔离了 aN*base**N 按“模块化”按 b 对于 b=base**(N+1) 哪一片全切了 一 比S大 N 把所有的 一 所以他们的序列号小于 n 通过减少 一 每次函数被当前 A*BASE**N .

Base%(base-1)==1 因此 base**p%(base-1)==1 因此 q*base^p%(base-1)==q 只有一个例外,当 q==base-1 哪些回报 0 . 要解决这个问题,它会返回 零 . 功能检查 零 从一开始。

优势

在这个示例中,只有一个乘法(而不是除法)和一些模运算,它们都比较快。

如果使用django框架,则可以使用django.utils.baseconv模块。

>>> from django.utils import baseconv
>>> baseconv.base62.encode(1234567890)
1LY7VK

除了base62,baseconv还定义base2/base16/base36/base56/base64。

就我个人而言,我喜欢白沙巴扬的解决方案,主要是因为去掉了令人困惑的角色。

为了完整性和更好的性能解决方案, this post 显示了使用python base64模块的方法。

我以前写过这篇文章,效果很好(包括底片)

def code(number,base):
    try:
        int(number),int(base)
    except ValueError:
        raise ValueError('code(number,base): number and base must be in base10')
    else:
        number,base = int(number),int(base)
    if base < 2:
        base = 2
    if base > 62:
        base = 62
    numbers = [0,1,2,3,4,5,6,7,8,9,"a","b","c","d","e","f","g","h","i","j",
               "k","l","m","n","o","p","q","r","s","t","u","v","w","x","y",
               "z","A","B","C","D","E","F","G","H","I","J","K","L","M","N",
               "O","P","Q","R","S","T","U","V","W","X","Y","Z"]
    final = ""
    loc = 0
    if number < 0:
        final = "-"
        number = abs(number)
    while base**loc <= number:
        loc = loc + 1
    for x in range(loc-1,-1,-1):
        for y in range(base-1,-1,-1):
            if y*(base**x) <= number:
                final = "{}{}".format(final,numbers[y])
                number = number - y*(base**x)
                break
    return final

def decode(number,base):
    try:
        int(base)
    except ValueError:
        raise ValueError('decode(value,base): base must be in base10')
    else:
        base = int(base)
    number = str(number)
    if base < 2:
        base = 2
    if base > 62:
        base = 62
    numbers = ["0","1","2","3","4","5","6","7","8","9","a","b","c","d","e","f",
               "g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v",
               "w","x","y","z","A","B","C","D","E","F","G","H","I","J","K","L",
               "M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"]
    final = 0
    if number.startswith("-"):
        neg = True
        number = list(number)
        del(number[0])
        temp = number
        number = ""
        for x in temp:
            number = "{}{}".format(number,x)
    else:
        neg = False
    loc = len(number)-1
    number = str(number)
    for x in number:
        if numbers.index(x) > base:
            raise ValueError('{} is out of base{} range'.format(x,str(base)))
        final = final+(numbers.index(x)*(base**loc))
        loc = loc - 1
    if neg:
        return -final
    else:
        return final

对不起,这一切的长度

BASE_LIST = tuple("23456789ABCDEFGHJKLMNOPQRSTUVWXYZabcdefghjkmnpqrstuvwxyz")
BASE_DICT = dict((c, v) for v, c in enumerate(BASE_LIST))
BASE_LEN = len(BASE_LIST)

def nice_decode(str):
    num = 0
    for char in str[::-1]:
        num = num * BASE_LEN + BASE_DICT[char]
    return num

def nice_encode(num):
    if not num:
        return BASE_LIST[0]

    encoding = ""
    while num:
        num, rem = divmod(num, BASE_LEN)
        encoding += BASE_LIST[rem]
    return encoding

这是一种反复迭代的方法。根据执行的计数,迭代的速度要快一点。

def base62_encode_r(dec):
    s = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
    return s[dec] if dec < 62 else base62_encode_r(dec / 62) + s[dec % 62]
print base62_encode_r(2347878234)

def base62_encode_i(dec):
    s = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
    ret = ''
    while dec > 0:
        ret = s[dec % 62] + ret
        dec /= 62
    return ret
print base62_encode_i(2347878234)

def base62_decode_r(b62):
    s = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
    if len(b62) == 1:
        return s.index(b62)
    x = base62_decode_r(b62[:-1]) * 62 + s.index(b62[-1:]) % 62
    return x
print base62_decode_r("2yTsnM")

def base62_decode_i(b62):
    s = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
    ret = 0
    for i in xrange(len(b62)-1,-1,-1):
        ret = ret + s.index(b62[i]) * (62**(len(b62)-i-1))
    return ret
print base62_decode_i("2yTsnM")

if __name__ == '__main__':
    import timeit
    print(timeit.timeit(stmt="base62_encode_r(2347878234)", setup="from __main__ import base62_encode_r", number=100000))
    print(timeit.timeit(stmt="base62_encode_i(2347878234)", setup="from __main__ import base62_encode_i", number=100000))
    print(timeit.timeit(stmt="base62_decode_r('2yTsnM')", setup="from __main__ import base62_decode_r", number=100000))
    print(timeit.timeit(stmt="base62_decode_i('2yTsnM')", setup="from __main__ import base62_decode_i", number=100000))

0.270266867033
0.260915645986
0.344734796766
0.311662500262

现在有了一个Python库。

我正在为这个做一个PIP包。

我建议你用我的基地。 https://github.com/kamijoutouma/bases.py 灵感来源于bases.js

from bases import Bases
bases = Bases()

bases.toBase16(200)                // => 'c8'
bases.toBase(200, 16)              // => 'c8'
bases.toBase62(99999)              // => 'q0T'
bases.toBase(200, 62)              // => 'q0T'
bases.toAlphabet(300, 'aAbBcC')    // => 'Abba'

bases.fromBase16('c8')               // => 200
bases.fromBase('c8', 16)             // => 200
bases.fromBase62('q0T')              // => 99999
bases.fromBase('q0T', 62)            // => 99999
bases.fromAlphabet('Abba', 'aAbBcC') // => 300

参照 https://github.com/kamijoutouma/bases.py#known-basesalphabets 什么样的基础是有用的

对不起,我帮不了你办图书馆。如果可能的话,我更喜欢使用base64,只需添加额外的字符就可以了!

然后您可以使用base64模块。

如果这真的,真的不可能:

您可以自己这样做(这是伪代码):

base62vals = []
myBase = 62
while num > 0:
   reminder = num % myBase
   num = num / myBase
   base62vals.insert(0, reminder)