这就是我的方法。首先转换数字，然后将其转换为字符串。

    def to_base(n, base):
        if base == 10:
            return n
        
        result = 0
        counter = 0
        
        while n:
            r = n % base
            n //= base
            result += r * 10**counter
            counter+=1
        return str(result)

您可以使用我的项目中的baseconv.py: https://github.com/semente/python-baseconv

示例用法:

>>> from baseconv import BaseConverter
>>> base20 = BaseConverter('0123456789abcdefghij')
>>> base20.encode(1234)
'31e'
>>> base20.decode('31e')
'1234'
>>> base20.encode(-1234)
'-31e'
>>> base20.decode('-31e')
'-1234'
>>> base11 = BaseConverter('0123456789-', sign='$')
>>> base11.encode('$1234')
'$-22'
>>> base11.decode('$-22')
'$1234'

有一些bultin转换器，例如baseconv。base2 baseconv。Base16和baseconv.base64。

下面是一个处理有符号整数和自定义数字的递归版本。

import string

def base_convert(x, base, digits=None):
    """Convert integer `x` from base 10 to base `base` using `digits` characters as digits.
    If `digits` is omitted, it will use decimal digits + lowercase letters + uppercase letters.
    """
    digits = digits or (string.digits + string.ascii_letters)
    assert 2 <= base <= len(digits), "Unsupported base: {}".format(base)
    if x == 0:
        return digits[0]
    sign = '-' if x < 0 else ''
    x = abs(x)
    first_digits = base_convert(x // base, base, digits).lstrip(digits[0])
    return sign + first_digits + digits[x % base]

def baseConverter(x, b):
    s = ""
    d = string.printable.upper()
    while x > 0:
        s += d[x%b]
        x = x / b
    return s[::-1]

The below provided Python code converts a Python integer to a string in arbitrary base ( from 2 up to infinity ) and works in both directions. So all the created strings can be converted back to Python integers by providing a string for N instead of an integer. The code works only on positive numbers by intention (there is in my eyes some hassle about negative values and their bit representations I don't want to dig into). Just pick from this code what you need, want or like, or just have fun learning about available options. Much is there only for the purpose of documenting all the various available approaches ( e.g. the Oneliner seems not to be fast, even if promised to be ).

我喜欢萨尔瓦多·达利提出的无限大基地的格式。一个很好的建议，它在光学上工作得很好，即使是简单的二进制位表示。注意，在infiniteBase=True格式的字符串的情况下，width=x填充参数适用于数字，而不是整个数字。似乎，代码处理无穷大数字格式运行甚至比其他选项快一点-使用它的另一个原因?

我不喜欢使用Unicode来扩展数字可用的符号数量的想法，所以不要在下面的代码中寻找它，因为它不存在。请使用建议的infiniteBase格式，或者将整数存储为字节以进行压缩。

    def inumToStr( N, base=2, width=1, infiniteBase=False,\
    useNumpy=False, useRecursion=False, useOneliner=False, \
    useGmpy=False, verbose=True):
    ''' Positive numbers only, but works in BOTH directions.
    For strings in infiniteBase notation set for bases <= 62 
    infiniteBase=True . Examples of use:
    inumToStr( 17,  2, 1, 1)             # [1,0,0,0,1]
    inumToStr( 17,  3, 5)                #       00122
    inumToStr(245, 16, 4)                #        00F5
    inumToStr(245, 36, 4,0,1)            #        006T
    inumToStr(245245245245,36,10,0,1)    #  0034NWOQBH
    inumToStr(245245245245,62)           #     4JhA3Th 
        245245245245 == int(gmpy2.mpz('4JhA3Th',62))
    inumToStr(245245245245,99,2) # [25,78, 5,23,70,44]
    ----------------------------------------------------
    inumToStr( '[1,0,0,0,1]',2, infiniteBase=True ) # 17 
    inumToStr( '[25,78, 5,23,70,44]', 99) # 245245245245
    inumToStr( '0034NWOQBH', 36 )         # 245245245245 
    inumToStr( '4JhA3Th'   , 62 )         # 245245245245
    ----------------------------------------------------
    --- Timings for N = 2**4096, base=36: 
                                      standard: 0.0023
                                      infinite: 0.0017
                                      numpy   : 0.1277
                                      recursio; 0.0022
                                      oneliner: 0.0146
                For N = 2**8192: 
                                      standard: 0.0075
                                      infinite: 0.0053
                                      numpy   : 0.1369
    max. recursion depth exceeded:    recursio/oneliner
    '''
    show = print
    if type(N) is str and ( infiniteBase is True or base > 62 ):
        lstN = eval(N)
        if verbose: show(' converting a non-standard infiniteBase bits string to Python integer')
        return sum( [ item*base**pow for pow, item in enumerate(lstN[::-1]) ] )
    if type(N) is str and base <= 36:
        if verbose: show('base <= 36. Returning Python int(N, base)')
        return int(N, base)
    if type(N) is str and base <= 62:
        if useGmpy: 
            if verbose: show(' base <= 62, useGmpy=True, returning int(gmpy2.mpz(N,base))')
            return int(gmpy2.mpz(N,base))
        else:
            if verbose: show(' base <= 62, useGmpy=False, self-calculating return value)')
            lstStrOfDigits="0123456789"+ \
                "abcdefghijklmnopqrstuvwxyz".upper() + \
                "abcdefghijklmnopqrstuvwxyz"
            dictCharToPow = {}
            for index, char in enumerate(lstStrOfDigits):
                dictCharToPow.update({char : index}) 
            return sum( dictCharToPow[item]*base**pow for pow, item in enumerate(N[::-1]) )
        #:if
    #:if        
        
    if useOneliner and base <= 36:  
        if verbose: show(' base <= 36, useOneliner=True, running the Oneliner code')
        d="0123456789abcdefghijklmnopqrstuvwxyz"
        baseit = lambda a=N, b=base: (not a) and d[0]  or \
        baseit(a-a%b,b*base)+d[a%b%(base-1) or (a%b) and (base-1)]
        return baseit().rjust(width, d[0])[1:]

    if useRecursion and base <= 36: 
        if verbose: show(' base <= 36, useRecursion=True, running recursion algorythm')
        BS="0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
        def to_base(n, b): 
            return "0" if not n else to_base(n//b, b).lstrip("0") + BS[n%b]
        return to_base(N, base).rjust(width,BS[0])
        
    if base > 62 or infiniteBase:
        if verbose: show(' base > 62 or infiniteBase=True, returning a non-standard digits string')
        # Allows arbitrary large base with 'width=...' 
        # applied to each digit (useful also for bits )
        N, digit = divmod(N, base)
        strN = str(digit).rjust(width, ' ')+']'
        while N:
            N, digit = divmod(N, base)
            strN = str(digit).rjust(width, ' ') + ',' + strN
        return '[' + strN
    #:if        
    
    if base == 2:
        if verbose: show(" base = 2, returning Python str(f'{N:0{width}b}')")
        return str(f'{N:0{width}b}')
    if base == 8:
        if verbose: show(" base = 8, returning Python str(f'{N:0{width}o}')")
        return str(f'{N:0{width}o}')
    if base == 16:
        if verbose: show(" base = 16, returning Python str(f'{N:0{width}X}')")
        return str(f'{N:0{width}X}')

    if base <= 36:
        if useNumpy: 
            if verbose: show(" base <= 36, useNumpy=True, returning np.base_repr(N, base)")
            import numpy as np
            strN = np.base_repr(N, base)
            return strN.rjust(width, '0') 
        else:
            if verbose: show(' base <= 36, useNumpy=False, self-calculating return value)')
            lstStrOfDigits="0123456789"+"abcdefghijklmnopqrstuvwxyz".upper()
            strN = lstStrOfDigits[N % base] # rightmost digit
            while N >= base:
                N //= base # consume already converted digit
                strN = lstStrOfDigits[N % base] + strN # add digits to the left
            #:while
            return strN.rjust(width, lstStrOfDigits[0])
        #:if
    #:if
    
    if base <= 62:
        if useGmpy: 
            if verbose: show(" base <= 62, useGmpy=True, returning gmpy2.digits(N, base)")
            import gmpy2
            strN = gmpy2.digits(N, base)
            return strN.rjust(width, '0') 
            # back to Python int from gmpy2.mpz with 
            #     int(gmpy2.mpz('4JhA3Th',62))
        else:
            if verbose: show(' base <= 62, useGmpy=False, self-calculating return value)')
            lstStrOfDigits= "0123456789" + \
                "abcdefghijklmnopqrstuvwxyz".upper() + \
                "abcdefghijklmnopqrstuvwxyz"
            strN = lstStrOfDigits[N % base] # rightmost digit
            while N >= base:
                N //= base # consume already converted digit
                strN = lstStrOfDigits[N % base] + strN # add digits to the left
            #:while
            return strN.rjust(width, lstStrOfDigits[0])
        #:if
    #:if    
#:def

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.