How to convert an integer in any base to a string?

Question

Python allows easy creation of an integer from a string of a given base via

int(str, base). 

I want to perform the inverse: creation of a string from an integer, i.e. I want some function int2base(num, base), such that:

int(int2base(x, b), b) == x

The function name/argument order is unimportant.

For any number x and base b that int() will accept.

This is an easy function to write: in fact it's easier than describing it in this question. However, I feel like I must be missing something.

I know about the functions bin, oct, hex, but I cannot use them for a few reasons:

• Those functions are not available on older versions of Python, with which I need compatibility with (2.2)

• I want a general solution that can be called the same way for different bases

• I want to allow bases other than 2, 8, 16

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Answer 1

Another short one (and easier to understand imo):

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

And with proper exception handling:

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")

Answer 2

If you need compatibility with ancient versions of Python, you can either use gmpy (which does include a fast, completely general int-to-string conversion function, and can be built for such ancient versions -- you may need to try older releases since the recent ones have not been tested for venerable Python and GMP releases, only somewhat recent ones), or, for less speed but more convenience, use Python code -- e.g., most simply:

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)

Answer 3

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

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

Please be aware that this may lead to

RuntimeError: maximum recursion depth exceeded in cmp

for very big integers.

Answer 4

num = input("number")
power = 0
num = int(num)
while num > 10:
    num = num / 10
    power += 1

print(str(round(num, 2)) + "^" + str(power))

Answer 5

Strings aren't the only choice for representing numbers: you can use a list of integers to represent the order of each digit. Those can easily be converted to a string.

None of the answers reject base < 2; and most will run very slowly or crash with stack overflows for very large numbers (such as 56789 ** 43210). To avoid such failures, reduce quickly like this:

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)

Speedwise, n_to_base is comparable with str for large numbers (about 0.3s on my machine), but if you compare against hex you may be surprised (about 0.3ms on my machine, or 1000x faster). The reason is because the large integer is stored in memory in base 256 (bytes). Each byte can simply be converted to a two-character hex string. This alignment only happens for bases that are powers of two, which is why there are special cases for 2,8, and 16 (and base64, ascii, utf16, utf32).

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.

Answer 6

You could use baseconv.py from my project: https://github.com/semente/python-baseconv

Sample usage:

>>> 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'

There is some bultin converters as for example baseconv.base2, baseconv.base16 and baseconv.base64.