To get the value equivalent to your C cast, just bitwise and with the appropriate mask. e.g. if unsigned long is 32 bit:

>>> i = -6884376
>>> i & 0xffffffff
4288082920

or if it is 64 bit:

>>> i & 0xffffffffffffffff
18446744073702667240

Do be aware though that although that gives you the value you would have in C, it is still a signed value, so any subsequent calculations may give a negative result and you'll have to continue to apply the mask to simulate a 32 or 64 bit calculation.

This works because although Python looks like it stores all numbers as sign and magnitude, the bitwise operations are defined as working on two's complement values. C stores integers in twos complement but with a fixed number of bits. Python bitwise operators act on twos complement values but as though they had an infinite number of bits: for positive numbers they extend leftwards to infinity with zeros, but negative numbers extend left with ones. The & operator will change that leftward string of ones into zeros and leave you with just the bits that would have fit into the C value.

Displaying the values in hex may make this clearer (and I rewrote to string of f's as an expression to show we are interested in either 32 or 64 bits):

>>> hex(i)
'-0x690c18'
>>> hex (i & ((1 << 32) - 1))
'0xff96f3e8'
>>> hex (i & ((1 << 64) - 1)
'0xffffffffff96f3e8L'

For a 32 bit value in C, positive numbers go up to 2147483647 (0x7fffffff), and negative numbers have the top bit set going from -1 (0xffffffff) down to -2147483648 (0x80000000). For values that fit entirely in the mask, we can reverse the process in Python by using a smaller mask to remove the sign bit and then subtracting the sign bit:

>>> u = i & ((1 << 32) - 1)
>>> (u & ((1 << 31) - 1)) - (u & (1 << 31))
-6884376

Or for the 64 bit version:

>>> u = 18446744073702667240
>>> (u & ((1 << 63) - 1)) - (u & (1 << 63))
-6884376

This inverse process will leave the value unchanged if the sign bit is 0, but obviously it isn't a true inverse because if you started with a value that wouldn't fit within the mask size then those bits are gone.

Essentially, the problem is to sign extend from 32 bits to... an infinite number of bits, because Python has arbitrarily large integers. Normally, sign extension is done automatically by CPU instructions when casting, so it's interesting that this is harder in Python than it would be in, say, C.

By playing around, I found something similar to BreizhGatch's function, but that doesn't require a conditional statement. n & 0x80000000 extracts the 32-bit sign bit; then, the - keeps the same 32-bit representation but sign-extends it; finally, the extended sign bits are set on n.

def toSigned32(n):
    n = n & 0xffffffff
    return n | (-(n & 0x80000000))

Bit Twiddling Hacks suggests another solution that perhaps works more generally. n ^ 0x80000000 flips the 32-bit sign bit; then - 0x80000000 will sign-extend the opposite bit. Another way to think about it is that initially, negative numbers are above positive numbers (separated by 0x80000000); the ^ swaps their positions; then the - shifts negative numbers to below 0.

def toSigned32(n):
    n = n & 0xffffffff
    return (n ^ 0x80000000) - 0x80000000

def convert(x):
    return x - (x >> 9) * 1024

Using ctypes:

from ctypes import c_short

def convert(x):
    return c_short(x << 6).value >> 6

using numpy for example:

import numpy
result = numpy.uint32( numpy.int32(myval) )

or even on arrays,

arr = numpy.array(range(10))
result = numpy.uint32( numpy.int32(arr) )

Dealing with signed/unsigned integer, or converting between 16-bit/32-bit/64-bit integers in Python is a nightmare.

You can use bitstring module to convert between these types. For example:

a = bitstring.BitArray(uint=65365, length=32)
print a.float

9.159587412059167e-41