假设我们想把14转换成2进制。我们反复应用除法算法，直到商为0:

14 = 2 × 7

7 = 2 × 3 + 1

3 = 2 × 1 + 1

1 = 2 × 0 + 1

二进制表示就是从下往上读的余数。这可以通过展开来证明

14 = 2 × 7 = 2 × (2 × 3 + 1) = 2 × (2 × (2 × 1 + 1) + 1) = 2 × (2 × (2 × 0 + 1) + 1) = 2^3 + 2^2 + 2

本代码是上述算法的实现。

def toBaseX(n, X):
strbin = ""
while n != 0:
    strbin += str(n % X)
    n = n // X
return strbin[::-1]

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

我知道这是一个老帖子，但我只是把我的解决方案留在这里以防万一。

def decimal_to_given_base(integer_to_convert, base):
     remainder = integer_to_convert // base
     digit = integer_to_convert % base
     if integer_to_convert == 0:
         return '0'
     elif remainder == 0:
         return str(digit)
     else:
         return decimal_to_given_base(remainder, base) + str(digit)

def baseN(num,b,numerals="0123456789abcdefghijklmnopqrstuvwxyz"):
    return ((num == 0) and numerals[0]) or (baseN(num // b, b, numerals).lstrip(numerals[0]) + numerals[num % b])

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

请注意这可能会导致

RuntimeError: maximum recursion depth exceeded in cmp

对于非常大的整数。

字符串不是表示数字的唯一选择:您可以使用一个整数列表来表示每个数字的顺序。这些可以很容易地转换为字符串。

没有一个答案拒绝底数< 2;对于非常大的数字(如56789 ** 43210)，大多数将运行非常缓慢或因堆栈溢出而崩溃。为了避免这种失败，可以像这样快速减少:

def n_to_base(n, b):
    if b < 2: raise # invalid base
    if abs(n) < b: return [n]
    ret = [y for d in n_to_base(n, b*b) for y in divmod(d, b)]
    return ret[1:] if ret[0] == 0 else ret # remove leading zeros

def base_to_n(v, b):
    h = len(v) // 2
    if h == 0: return v[0]
    return base_to_n(v[:-h], b) * (b**h) + base_to_n(v[-h:], b)

assert ''.join(['0123456789'[x] for x in n_to_base(56789**43210,10)])==str(56789**43210)

在速度方面，n_to_base对于较大的数字(在我的机器上约为0.3秒)与str相当，但如果与十六进制进行比较，您可能会感到惊讶(在我的机器上约为0.3毫秒，或快1000倍)。这是因为大整数以256(字节)为基数存储在内存中。每个字节可以简单地转换为两个字符的十六进制字符串。这种对齐只发生在底数为2的幂的情况下，这就是为什么有2、8和16(以及base64, ascii, utf16, utf32)的特殊情况。

Consider the last digit of a decimal string. How does it relate to the sequence of bytes that forms its integer? Let's label the bytes s[i] with s[0] being the least significant (little endian). Then the last digit is sum([s[i]*(256**i) % 10 for i in range(n)]). Well, it happens that 256**i ends with a 6 for i > 0 (6*6=36) so that last digit is (s[0]*5 + sum(s)*6)%10. From this, you can see that the last digit depends on the sum of all the bytes. This nonlocal property is what makes converting to decimal harder.

"{0:b}".format(100) # bin: 1100100
"{0:x}".format(100) # hex: 64
"{0:o}".format(100) # oct: 144