令人惊讶的是，人们给出的答案只能转换成小基数(比英语字母表的长度还小)。没有人试图给出一个可以转换为2到无穷任意底数的解。

这里有一个超级简单的解决方案:

def numberToBase(n, b):
    if n == 0:
        return [0]
    digits = []
    while n:
        digits.append(int(n % b))
        n //= b
    return digits[::-1]

所以如果你需要把一个超级大的数转换成577的底数，

numberToBase(67854 ** 15 - 102,577)，将为您提供正确的解决方案: [4, 473, 131, 96, 431, 285, 524, 486, 28, 23, 16, 82, 292, 538, 149, 25, 41, 483, 100, 517, 131, 28, 0, 435, 197, 264, 455]，

你以后可以把它转换成任何你想要的基数

at some point of time you will notice that sometimes there is no built-in library function to do things that you want, so you need to write your own. If you disagree, post you own solution with a built-in function which can convert a base 10 number to base 577. this is due to lack of understanding what a number in some base means. I encourage you to think for a little bit why base in your method works only for n <= 36. Once you are done, it will be obvious why my function returns a list and has the signature it has.

http://code.activestate.com/recipes/65212/

def base10toN(num,n):
    """Change a  to a base-n number.
    Up to base-36 is supported without special notation."""
    num_rep={10:'a',
         11:'b',
         12:'c',
         13:'d',
         14:'e',
         15:'f',
         16:'g',
         17:'h',
         18:'i',
         19:'j',
         20:'k',
         21:'l',
         22:'m',
         23:'n',
         24:'o',
         25:'p',
         26:'q',
         27:'r',
         28:'s',
         29:'t',
         30:'u',
         31:'v',
         32:'w',
         33:'x',
         34:'y',
         35:'z'}
    new_num_string=''
    current=num
    while current!=0:
        remainder=current%n
        if 36>remainder>9:
            remainder_string=num_rep[remainder]
        elif remainder>=36:
            remainder_string='('+str(remainder)+')'
        else:
            remainder_string=str(remainder)
        new_num_string=remainder_string+new_num_string
        current=current/n
    return new_num_string

这是来自同一个链接的另一个

def baseconvert(n, base):
    """convert positive decimal integer n to equivalent in another base (2-36)"""

    digits = "0123456789abcdefghijklmnopqrstuvwxyz"

    try:
        n = int(n)
        base = int(base)
    except:
        return ""

    if n < 0 or base < 2 or base > 36:
        return ""

    s = ""
    while 1:
        r = n % base
        s = digits[r] + s
        n = n / base
        if n == 0:
            break

    return s

我个人使用我写的这个函数

import string

def to_base(value, base, digits=string.digits+string.ascii_letters):    # converts decimal to base n

    digits_slice = digits[0:base]

    temporary_var = value
    data = [temporary_var]

    while True:
        temporary_var = temporary_var // base
        data.append(temporary_var)
        if temporary_var < base:
            break

    result = ''
    for each_data in data:
        result += digits_slice[each_data % base]
    result = result[::-1]

    return result

你可以这样使用它

基础= 2)打印(to_base(7日)

输出: “111”

基础= 3)打印(to_base(23日)

输出: “212”

请随时对我的代码提出改进建议。

另一个简短的(在我看来更容易理解):

def int_to_str(n, b, symbols='0123456789abcdefghijklmnopqrstuvwxyz'):
    return (int_to_str(n/b, b, symbols) if n >= b else "") + symbols[n%b]

通过适当的异常处理:

def int_to_str(n, b, symbols='0123456789abcdefghijklmnopqrstuvwxyz'):
    try:
        return (int_to_str(n/b, b) if n >= b else "") + symbols[n%b]
    except IndexError:
        raise ValueError(
            "The symbols provided are not enough to represent this number in "
            "this base")

如果你需要兼容Python的古老版本，你可以使用gmpy(它包含一个快速的，完全通用的int-to-string转换函数，可以为这样的古老版本构建-你可能需要尝试更老的版本，因为最近的版本还没有针对古老的Python和GMP版本进行测试，只有一些最近的版本)，或者，为了速度较慢但更方便，使用Python代码-例如，对于Python 2，最简单的方法是:

import string
digs = string.digits + string.ascii_letters


def int2base(x, base):
    if x < 0:
        sign = -1
    elif x == 0:
        return digs[0]
    else:
        sign = 1

    x *= sign
    digits = []

    while x:
        digits.append(digs[int(x % base)])
        x = int(x / base)

    if sign < 0:
        digits.append('-')

    digits.reverse()

    return ''.join(digits)

对于Python 3, int(x / base)会导致不正确的结果，必须将其更改为x // base:

import string
digs = string.digits + string.ascii_letters


def int2base(x, base):
    if x < 0:
        sign = -1
    elif x == 0:
        return digs[0]
    else:
        sign = 1

    x *= sign
    digits = []

    while x:
        digits.append(digs[x % base])
        x = x // base

    if sign < 0:
        digits.append('-')

    digits.reverse()

    return ''.join(digits)

def base_changer(number,base):
    buff=97+abs(base-10)
    dic={};buff2='';buff3=10
    for i in range(97,buff+1):
        dic[buff3]=chr(i)
        buff3+=1   
    while(number>=base):
        mod=int(number%base)
        number=int(number//base)
        if (mod) in dic.keys():
            buff2+=dic[mod]
            continue
        buff2+=str(mod)
    if (number) in dic.keys():
        buff2+=dic[number]
    else:
        buff2+=str(number)

    return buff2[::-1]