Question

I have this Python code to do this:

from struct import pack as _pack

def packl(lnum, pad = 1):
    if lnum < 0:
        raise RangeError("Cannot use packl to convert a negative integer "
                         "to a string.")
    count = 0
    l = []
    while lnum > 0:
        l.append(lnum & 0xffffffffffffffffL)
        count += 1
        lnum >>= 64
    if count <= 0:
        return '\0' * pad
    elif pad >= 8:
        lens = 8 * count % pad
        pad = ((lens != 0) and (pad - lens)) or 0
        l.append('>' + 'x' * pad + 'Q' * count)
        l.reverse()
        return _pack(*l)
    else:
        l.append('>' + 'Q' * count)
        l.reverse()
        s = _pack(*l).lstrip('\0')
        lens = len(s)
        if (lens % pad) != 0:
            return '\0' * (pad - lens % pad) + s
        else:
            return s

This takes approximately 174 usec to convert 2**9700 - 1 to a string of bytes on my machine. If I’m willing to use the Python 2.7 and Python 3.x specific bit_length method, I can shorten that to 159 usecs by pre-allocating the l array to be the exact right size at the very beginning and using l[something] = syntax instead of l.append.

Is there anything I can do that will make this faster? This will be used to convert large prime numbers used in cryptography as well as some (but not many) smaller numbers.

Edit

This is currently the fastest option in Python < 3.2, it takes about half the time either direction as the accepted answer:

def packl(lnum, padmultiple=1):
    """Packs the lnum (which must be convertable to a long) into a
       byte string 0 padded to a multiple of padmultiple bytes in size. 0
       means no padding whatsoever, so that packing 0 result in an empty
       string.  The resulting byte string is the big-endian two's
       complement representation of the passed in long."""

    if lnum == 0:
        return b'\0' * padmultiple
    elif lnum < 0:
        raise ValueError("Can only convert non-negative numbers.")
    s = hex(lnum)[2:]
    s = s.rstrip('L')
    if len(s) & 1:
        s = '0' + s
    s = binascii.unhexlify(s)
    if (padmultiple != 1) and (padmultiple != 0):
        filled_so_far = len(s) % padmultiple
        if filled_so_far != 0:
            s = b'\0' * (padmultiple - filled_so_far) + s
    return s

def unpackl(bytestr):
    """Treats a byte string as a sequence of base 256 digits
    representing an unsigned integer in big-endian format and converts
    that representation into a Python integer."""

    return int(binascii.hexlify(bytestr), 16) if len(bytestr) > 0 else 0

In Python 3.2 the int class has to_bytes and from_bytes functions that can accomplish this much more quickly that the method given above.

Answer 1

For completeness and for future readers of this question:

Starting in Python 3.2, there are functions int.from_bytes() and int.to_bytes() that perform the conversion between bytes and int objects in a choice of byte orders.