Binary matrix to integer

Question

I'm having a binary MxN matrix like the following:

matrix([[0, 0, 0, ..., 0, 0, 0],
 [0, 1, 0, ..., 0, 0, 0],
 [0, 0, 1, ..., 0, 0, 0],
 ..., 
 [0, 0, 0, ..., 1, 0, 0],
 [0, 0, 0, ..., 0, 0, 0],
 [0, 0, 0, ..., 0, 0, 1]])

How can I convert this matrix row-wise to the integer representation? The goal is to get a Mx1 matrix. Each row contains the integer representation of the previous binary row.

• Warren Weckesser: possible duplicate of Binary numpy array to list of integers?

Answer 1

Another approach is using packbits from numpy. For this, however, numpy will fill with zeros the columns, until they get byte shape (length % 8). In order to avoid numpy from filling with 0's, we could add zeros at the beginning (to not change the binary value of the array).

Using the data from @Oliver W.

>>> import numpy as np
>>> b = np.random.random_integers(0,1,(3,5))
>>> b
array([[0, 1, 1, 1, 0],
 [1, 0, 1, 0, 1],
 [0, 1, 0, 0, 1]])
# Fill the begining with zeros to form bytes
>>> diff = 8 - b.shape[1] % 8
>>> if diff > 0: b = np.c_[np.zeros((b.shape[0], diff), int), b]
>>> b
array([[0, 0, 0, 0, 1, 1, 1, 0],
 [0, 0, 0, 1, 0, 1, 0, 1],
 [0, 0, 0, 0, 1, 0, 0, 1]])
# Get the integer number
>>> p = np.packbits(b, axis=1)
>>> p
array([[14],
 [21],
 [ 9]], dtype=uint8)

• Oliver W.: Works (+1), but be warned: it only works as intended for values <256, because it packs the data in 8 bits. So if you have more columns than 8, you'll get back several columns rather than Mx1.
• Imanol Luengo: Ops, thank you for that @Oliver W., I didn't know it. I guess if the columns have some sort of structure (i.e. represent a 8, 16, 32, 64 or 128 bits integers) it could be fixed by adding p.view('u4') (for a 32 bit integer, similar for others). It would return a new view for the columns

Answer 2

You could multiply each element of the row with its corresponding multiplication factor 2**exponent where the exponent depends on the position of the digit within the row (this works for any positional number system by the way, such as base-60 and hexadecimal):

>>> import numpy as np
>>> 
>>> b = np.random.random_integers(0,1,(3,5)) # example array
>>> b
array([[0, 1, 1, 1, 0],
 [1, 0, 1, 0, 1],
 [0, 1, 0, 0, 1]])
>>> c = 2**np.arange(b.shape[1])[::-1] # reverse order: assumes the last column corresponds with the decimal number 1
>>> b.dot(c) # matrix multiplication
array([14, 21, 9])

Same result if you have instances of numpy's matrix class.

Answer 3

You can join all 0s and 1s of every row as strings and from there it's easy to get the integer representation. I had to convert the matrix to an array to get this to work, but maybe there is a straight solution:

import numpy as np
A = np.matrix([[0, 1, 1, 1, 0],
 [1, 0, 1, 0, 1],
 [0, 1, 0, 0, 1]])
B = np.asarray(A)
#join all 0/1s together as string and convert to integer
C = np.matrix([int(''.join(str(x) for x in column),2) for column in B]).reshape((B.shape[0],1))
print C
>>>
[[14]
 [21]
 [ 9]]