我提出了一个“非优化”的2到9基的解决方案:

  def to_base(N, base=2):
    N_in_base = ''
    while True:
        N_in_base = str(N % base) + N_in_base
        N //= base
        if N == 0:
            break
    return N_in_base

这个解决方案不需要反转最终结果，但实际上并没有优化。请参考以下答案了解原因:https://stackoverflow.com/a/37133870/7896998

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.

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。

这个函数将任意整数从任意进制转换为任意进制

def baseconvert(number, srcbase, destbase):
    if srcbase != 10:
        sum = 0
        for _ in range(len(str(number))):
            sum += int(str(number)[_]) * pow(srcbase, len(str(number)) - _ - 1)
        b10 = sum
        return baseconvert(b10, 10, destbase)
    end = ''
    q = number
    while(True):
        r = q % destbase
        q = q // destbase
        end = str(r) + end
        if(q<destbase):
            end = str(q) + end
            return int(end)

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。

虽然目前排名第一的答案绝对是一个很棒的解决方案，但仍然有更多用户可能喜欢的定制。

Basencode添加了其中的一些特性，包括浮点数的转换、修改数字(在链接的答案中，只能使用数字)。

下面是一个可能的用例:

>>> from basencode import *
>>> n1 = Number(12345)
>> n1.repr_in_base(64) # convert to base 64
'30V'
>>> Number('30V', 64) # construct Integer from base 64
Integer(12345)
>>> n1.repr_in_base(8)
'30071'
>>> n1.repr_in_octal() # shortcuts
'30071'
>>> n1.repr_in_bin() # equivelant to `n1.repr_in_base(2)`
'11000000111001'
>>> n1.repr_in_base(2, digits=list('-+')) # override default digits: use `-` and `+` in place of `0` and `1`
'++------+++--+'
>>> n1.repr_in_base(33) # yet another base - all bases from 2 to 64 are supported from the start
'bb3'

你怎么添加你想要的碱基?让我复制一下目前投票最多的答案的例子:digits参数允许您覆盖从2到64的默认数字，并为任何高于该基数的数字提供数字。mode参数决定了表示的值如何决定(列表或字符串)如何返回答案。

>>> n2 = Number(67854 ** 15 - 102)
>>> n2.repr_in_base(577, digits=[str(i) for i in range(577)], mode="l")
['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']
>>> n2.repr_in_base(577, mode="l") # the program remembers the digits for base 577 now
['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']

可以执行以下操作:Number类返回basencode的一个实例。如果提供的数字是Integer，则返回一个基本编码。浮动

>>> n3 = Number(54321) # the Number class returns an instance of `basencode.Integer` if the provided number is an Integer, otherwise it returns a `basencode.Float`.
>>> n1 + n3
Integer(66666)
>>> n3 - n1
Integer(41976)
>>> n1 * n3
Integer(670592745)
>>> n3 // n1
Integer(4)
>>> n3 / n1 # a basencode.Float class allows conversion of floating point numbers
Float(4.400243013365735)
>>> (n3 / n1).repr_in_base(32)
'4.cpr56v6rnc4oitoblha2r11sus0dheqd4pgechfcjklo74b2bgom7j8ih86mipdvss0068sehi9f3791mdo4uotfujq66cf0jkgo'
>>> n4 = Number(0.5) # returns a basencode.Float
>>> n4.repr_in_bin() # binary version of 0.5
'0.1'

免责声明:此项目正在积极维护中，我是贡献者。