Python: convert numpy array of signs to int and back

Question

I'm trying to convert from a numpy array of signs (i.e., a numpy array whose entries are either 1. or -1.) to an integer and back through a binary representation. I have something that works, but it's not Pythonic, and I expect it'll be slow.

def sign2int(s):
    s[s==-1.] = 0.
    bstr = ''
    for i in range(len(s)):
        bstr = bstr + str(int(s[i]))
    return int(bstr, 2)

def int2sign(i, m):
    bstr = bin(i)[2:].zfill(m)
    s = []
    for d in bstr:
        s.append(float(d))
    s = np.array(s)
    s[s==0.] = -1.
    return s

Then

>>> m = 4
>>> s0 = np.array([1., -1., 1., 1.])
>>> i = sign2int(s0)
>>> print i
11
>>> s = int2sign(i, m)
>>> print s
[ 1. -1.  1.  1.]

I'm concerned about (1) the for loops in each and (2) having to build an intermediate representation as a string.

Ultimately, I will want something that works with a 2-d numpy array, too---e.g.,

>>> s = np.array([[1., -1., 1.], [1., 1., 1.]])
>>> print sign2int(s)
[5, 7]

Answer 1

For 1d arrays you can use this one linear Numpythonic approach, using np.packbits:

>>> np.packbits(np.pad((s0+1).astype(bool).astype(int), (8-s0.size, 0), 'constant'))
array([11], dtype=uint8)

And for reversing:

>>> unpack = (np.unpackbits(np.array([11], dtype=np.uint8))[-4:]).astype(float)
>>> unpack[unpack==0] = -1
>>> unpack
array([ 1., -1.,  1.,  1.])

And for 2d array:

>>> x, y = s.shape
>>> np.packbits(np.pad((s+1).astype(bool).astype(int), (8-y, 0), 'constant')[-2:])
array([5, 7], dtype=uint8)

And for reversing:

>>> unpack = (np.unpackbits(np.array([5, 7], dtype='uint8'))).astype(float).reshape(x, 8)[:,-y:]
>>> unpack[unpack==0] = -1
>>> unpack
array([[ 1., -1.,  1.],
       [ 1.,  1.,  1.]])

Answer 2

I'll start with sig2int.. Convert from a sign representation to binary

>>> a
array([ 1., -1.,  1., -1.])
>>> (a + 1) / 2
array([ 1.,  0.,  1.,  0.])
>>> 

Then you can simply create an array of powers of two, multiply it by the binary and sum.

>>> powers = np.arange(a.shape[-1])[::-1]
>>> np.power(2, powers)
array([8, 4, 2, 1])
>>> a = (a + 1) / 2
>>> powers = np.power(2, powers)
>>> a * powers
array([ 8.,  0.,  2.,  0.])
>>> np.sum(a * powers)
10.0
>>> 

Then make it operate on rows by adding axis information and rely on broadcasting.

def sign2int(a):
    # powers of two
    powers = np.arange(a.shape[-1])[::-1]
    np.power(2, powers, powers)
    # sign to "binary" - add one and divide by two
    np.add(a, 1, a)
    np.divide(a, 2, a)
    # scale by powers of two and sum
    np.multiply(a, powers, a)
    return np.sum(a, axis = -1)
>>> b = np.array([a, a, a, a, a])
>>> sign2int(b)
array([ 11.,  11.,  11.,  11.,  11.])
>>> 

I tried it on a 4 by 100 bit array and it seemed fast

>>> a = a.repeat(100)
>>> b = np.array([a, a, a, a, a])
>>> b
array([[ 1.,  1.,  1., ...,  1.,  1.,  1.],
       [ 1.,  1.,  1., ...,  1.,  1.,  1.],
       [ 1.,  1.,  1., ...,  1.,  1.,  1.],
       [ 1.,  1.,  1., ...,  1.,  1.,  1.],
       [ 1.,  1.,  1., ...,  1.,  1.,  1.]])
>>> sign2int(b)
array([  2.58224988e+120,   2.58224988e+120,   2.58224988e+120,
         2.58224988e+120,   2.58224988e+120])
>>> 

---

I'll add the reverse if i can figure it. - the best I could do relies on some plain Python without any numpy vectoriztion magic and I haven't figured how to make it work with a sequence of ints other than to iterate over them and convert them one at a time - but the time still seems acceptable.

def foo(n):
    '''yields bits in increasing powers of two

    bit sequence from lsb --> msb
    '''
    while n > 0:
        n, r = divmod(n, 2)
        yield r

def int2sign(n):
    n = int(n)
    a = np.fromiter(foo(n), dtype = np.int8, count = n.bit_length())
    np.multiply(a, 2, a)
    np.subtract(a, 1, a)
    return a[::-1]

Works on 1324:

>>> bin(1324)
'0b10100101100'
>>> a = int2sign(1324)
>>> a
array([ 1, -1,  1, -1, -1,  1, -1,  1,  1, -1, -1], dtype=int8)

Seems to work with 1.2e305:

>>> n = int(1.2e305)
>>> n.bit_length()
1014
>>> a = int2sign(n)
>>> a.shape
(1014,)

>>> s = bin(n)
>>> s = s[2:]
>>> all(2 * int(x) -1 == y for x, y in zip(s, a))
True
>>>