How to rotate a binary vector to minimum in Python

Question

If I have an arbitrary binary vector (numpy array) in Python, e.g.

import numpy as np

vector = np.zeros((8,1))
vector[2,1] = 1
vector[3,1] = 1 

This would give me the binary array 00001100. I could also have 00000000 or 00010100 etc. How to make such a script that when I give this binary vector as an input, the script gives the minimum right-rotated binary numpy array as output? Few examples:

00010000 --> 00000001
10100000 --> 00000101
11000001 --> 00000111
00000000 --> 00000000
11111111 --> 11111111
10101010 --> 01010101
11110000 --> 00001111
00111000 --> 00000111
10001111 --> 00011111

etc. Any suggestions / good optimized Python implementations in mind? =) Thank you for any assistance. I need this for Local Binary Pattern implementation =)

Answer 1

The fastest way to do this is create a table first and then you can use ndarray indexing to get the result, here is the code:

You need create the table yourself, the code here is just a demo

import numpy as np
np.random.seed(0)

#create the table
def rotated(s):
    for i in range(len(s)):
        s2 = s[i:] + s[:i]
        if s2[-1] == "1":
            yield int(s2, 2)

bitmap = []
for i in range(256):
    s = "{:08b}".format(i)
    try:
        r = min(rotated(s))
    except ValueError:
        r = i
    bitmap.append(r)

bitmap = np.array(bitmap, np.uint8)

Then we can use bitmap and numpy.packbits() and numpy.unpackbits():

a = np.random.randint(0, 2, (10, 8))
a = np.vstack((a, np.array([[1,1,0,0,0,0,0,1]])))
b = np.unpackbits(bitmap[np.packbits(a, axis=1)], axis=1)
print a
print
print b

here is the output:

[[0 1 1 0 1 1 1 1]
 [1 1 1 0 0 1 0 0]
 [0 0 0 1 0 1 1 0]
 [0 1 1 1 1 0 1 0]
 [1 0 1 1 0 1 1 0]
 [0 1 0 1 1 1 1 1]
 [0 1 0 1 1 1 1 0]
 [1 0 0 1 1 0 1 0]
 [1 0 0 0 0 0 1 1]
 [0 0 0 1 1 0 1 0]
 [1 1 0 0 0 0 0 1]]

[[0 1 1 0 1 1 1 1]
 [0 0 1 0 0 1 1 1]
 [0 0 0 0 1 0 1 1]
 [0 0 1 1 1 1 0 1]
 [0 1 0 1 1 0 1 1]
 [0 1 0 1 1 1 1 1]
 [0 0 1 0 1 1 1 1]
 [0 0 1 1 0 1 0 1]
 [0 0 0 0 0 1 1 1]
 [0 0 0 0 1 1 0 1]
 [0 0 0 0 0 1 1 1]]