'Conversion between binary vector and 128 bit number
Is there a way to convert back and forth between a binary vector and a 128-bit number? I have the following binary vector:
import numpy as np
bits = np.array([1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1,
0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0,
1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1,
1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1], dtype=np.uint8)
which is a MD5 hash that I am trying use as a feature for a scikit-learn machine learning classifier (I need to represent the hash as a single feature).
Solution 1:[1]
As commented above, numpy only goes up to 64bits, but python has variable length ints, so we can do 128bits int no problem.
The following will go from binary in np.array to python int back to binary in np.array.
import numpy as np
bits = np.array(
[
1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1,
0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0,
1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1,
1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1
],
dtype=np.uint8
)
s = "".join(bits.astype("str")) # move from array to string
n = int(s, 2) # convert to int from string of base 2
print(n)
s = bin(n)[2:] # get binary of int, cut "0b" prefix
np.array(list(s), dtype=np.uint8) # put back in np.array
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | ljmc |
