Base 62转换

如何将整数转换为以62为底的整数（如十六进制，但具有以下数字：“ 0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ”）。

我一直在为它寻找一个好的Python库，但是它们似乎都被转换字符串所占据。Python base64模块仅接受字符串并将一个数字转换为四个字符。我正在寻找类似于URL缩短器使用的东西。

没有为此的标准模块，但是我编写了自己的函数来实现这一点。

BASE62 = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"

def encode(num, alphabet):
    """Encode a positive number into Base X and return the string.

    Arguments:
    - `num`: The number to encode
    - `alphabet`: The alphabet to use for encoding
    """
    if num == 0:
        return alphabet[0]
    arr = []
    arr_append = arr.append  # Extract bound-method for faster access.
    _divmod = divmod  # Access to locals is faster.
    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 decoding
    """
    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，您可以给它提供任何字母以用于编码和解码的事实。如果您忽略该alphabet参数，则将获得在第一行代码中定义的62个字符的字母，从而对62个基数进行编码/解码。

希望这可以帮助。

PS-对于URL缩短器，我发现最好省略一些令人困惑的字符，例如0Ol1oI等。因此，为了满足URL缩短的需要，我使用了该字母- "23456789abcdefghijkmnpqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ"

玩得开心。

我曾经写过一个脚本也可以做到这一点，我认为它很优雅:)

import string
# Remove the `_@` below for base62, now it has 64 characters
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

用法示例：

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

以下解码器制造商可以使用任何合理的基础，并具有更整洁的循环，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)

如果您正在寻找最高效率（例如django），则需要以下内容。该代码结合了Baishampayan Ghose和WoLpH和John Machin的高效方法。

# 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

您可能还需要提前计算字典。（注意：即使是很长的数字，使用字符串编码也比使用列表显示效率更高。）

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

在2.5秒内对100万个数字进行编码和解码。（2.2Ghz i7-2670QM）

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还定义了base2 / base16 / base36 / base56 / base64。

您可能需要base64，而不是base62。它有一个与URL兼容的版本，因此多余的两个填充符应该不是问题。

这个过程非常简单。考虑到base64代表6位，常规字节代表8。为选择的64个字符中的每个字符分配一个从000000到111111的值，并将这四个值放在一起以匹配一组3个base256字节。对于每3个字节的集合重复一次，并在末尾选择填充字符（通常是0）。

如果您只需要生成一个简短的ID（因为您提到了URL缩短器）而不是对某些内容进行编码/解码，那么该模块可能会有所帮助：

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

您可以从pypi下载zbase62模块

例如

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

我从这里其他人的帖子中受益匪浅。我最初需要用于Django项目的python代码，但是从那时起，我转向了node.js，所以这是Baishampayan Ghose提供的代码的javascript版本（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"));

我希望以下代码片段可以有所帮助。

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)])

您的情况的用法：

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

显然，您可以指定另一个字母，该字母由更少或更多的符号组成，然后它将您的数字转换为更少或更多的基数。例如，提供“ 01”作为字母将输出表示输入数字为二进制的字符串。

最初，您可以对字母进行混洗以使数字具有唯一性。如果您要进行URL缩短服务，这可能会有所帮助。

现在有一个python库。

我正在为此做一个点子包。

我建议您使用受bases.js启发的bases.py https://github.com/kamijoutouma/bases.py

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

请参阅https://github.com/kamijoutouma/bases.py#known-basesalphabets 了解可用的碱基

这是我的解决方案：

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()

说明

在任何基数中，每个数字均等于。 a1+a2*base**2+a3*base**3...因此，目标是找到所有as。

对于每个N=1,2,3...代码，aN*base**N通过“取模” 将所有s大于s的切片隔离开来，然后b对b=base**(N+1)所有as进行N切片，a以使它们的序列小于每次通过当前递归调用该函数N而减少时的序列。aaN*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，即何时q==base-1返回0。为了解决这种情况，它返回0。该功能0从头开始检查。

优点

在此示例中，只有一个乘法（而不是除法）和一些模运算，这些运算都相对较快。

我个人喜欢Baishampayan的解决方案，主要是因为去除了令人困惑的字符。

为了完整性和更好的性能解决方案，本文介绍了一种使用Python base64模块的方法。

我前一段时间写了这篇文章，而且效果很好（包括负数在内）

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

对这一切的长度感到抱歉

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

这是一种递归和迭代的方法。迭代的速度要快一些，具体取决于执行次数。

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

蟒蛇 3.7.x

当寻找现有的base62脚本时，我找到了一些算法的博士学位的github 。目前，它不适用于当前的Python 3的max-version版本，因此我继续进行修复，并在需要的地方进行了一些重构。我通常不使用Python，并且总是临时使用它，因此YMMV。赖志华博士全归功于他。我只是解决了这个版本的Python。

文件 base62.py

#modified from Dr. Zhihua Lai's original on GitHub
from math import floor
base = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';
b = 62;
def toBase10(b62: str) -> int:
    limit = len(b62)
    res = 0
    for i in range(limit):
        res = b * res + base.find(b62[i])
    return res
def toBase62(b10: int) -> str:
    if b <= 0 or b > 62:
        return 0
    r = b10 % b
    res = base[r];
    q = floor(b10 / b)
    while q:
        r = q % b
        q = floor(q / b)
        res = base[int(r)] + res
    return res

文件 try_base62.py

import base62
print("Base10 ==> Base62")
for i in range(999):
    print(f'{i} => {base62.toBase62(i)}')
base62_samples = ["gud", "GA", "mE", "lo", "lz", "OMFGWTFLMFAOENCODING"]
print("Base62 ==> Base10")
for i in range(len(base62_samples)):
    print(f'{base62_samples[i]} => {base62.toBase10(base62_samples[i])}')

输出 try_base62.py

Base10 ==> Base62
0 => 0
[...]
998 => g6
Base62 ==> Base10
gud => 63377
GA => 2640
mE => 1404
lo => 1326
lz => 1337
OMFGWTFLMFAOENCODING => 577002768656147353068189971419611424

_{^{由于回购中没有许可信息，因此我确实提交了PR，因此原始作者至少知道其他人正在使用和修改他们的代码。}}

抱歉，这里没有图书馆可以帮助您。我更喜欢使用base64并添加额外的字符到您的选择中-如果可能的话！

然后，您可以使用base64模块。

如果确实不能，那么：

您可以用这种方式自己做（这是伪代码）：

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

简单的递归

"""
This module contains functions to transform a number to string and vice-versa
"""
BASE = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
LEN_BASE = len(BASE)


def encode(num):
    """
    This function encodes the given number into alpha numeric string
    """

    if num < LEN_BASE:
        return BASE[num]

    return BASE[num % LEN_BASE] + encode(num//LEN_BASE)


def decode_recursive(string, index):
    """
    recursive util function for decode
    """

    if not string or index >= len(string):
        return 0

    return (BASE.index(string[index]) * LEN_BASE ** index) + decode_recursive(string, index + 1)


def decode(string):
    """
    This function decodes given string to number
    """

    return decode_recursive(string, 0)

有史以来最简单的。

BASE62 = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
def encode_base62(num):
    s = ""
    while num>0:
      num,r = divmod(num,62)
      s = BASE62[r]+s
    return s


def decode_base62(num):
   x,s = 1,0
   for i in range(len(num)-1,-1,-1):
      s = int(BASE62.index(num[i])) *x + s
      x*=62
   return s

print(encode_base62(123))
print(decode_base62("1Z"))