What's the easiest way to produce a numpy Boolean array representation of an integer? For example, map 6 to np.array([False, True, True], dtype=np.bool).
4 Answers
If n is the integer, the expression (n & (1 << np.arange(int(floor(log(n, 2) + 1))))) > 0 will create a boolean array representing the bits, with the least significant bit in the first position.
For example,
In [224]: n = 5
In [225]: from math import floor, log
In [226]: n = 5
In [227]: (n & (1 << np.arange(int(floor(log(n, 2) + 1))))) > 0
Out[227]: array([ True, False, True], dtype=bool)
In [228]: n = 8
In [229]: (n & (1 << np.arange(int(floor(log(n, 2) + 1))))) > 0
Out[229]: array([False, False, False, True], dtype=bool)
In [230]: n = 514
In [231]: (n & (1 << np.arange(int(floor(log(n, 2) + 1))))) > 0
Out[231]: array([False, True, False, False, False, False, False, False, False, True], dtype=bool)
1 Comment
Neil G
Clever! I'm actually specifying the
m = int(floor(log(n, 2) + 1)), so this solution is the shortest for me.First, use np.binary_repr to get the binary representation of the integer value. Remember to specify the bit-width you desire for your bit array here. I have included a way to programmatically determine the number of bits required to represent the given integer:
>>> n = 5
>>> bin_n = bin(n)
'0b101'
>>> smallest_length = n.bit_length()
3
>>> bin_array = np.binary_repr(5, width=smallest_length)
bin_array = '101'
Then, reverse the order of the bits in the output binary string:
>> reversed_bin_array[::-1]
reversed_bin_array = '101'
Then, finally to get an array of bits with the earlier-specified bit-width, use bitarray. I have found that the bitarray object is a very useful means to conveniently store and operate on bit arrays:
>>> bitarray('11110010') # from a string
I hope this helps, happy coding!
4 Comments
user2357112
np.invert does not reverse the order of the bits.
Vladislav Martin
Wow... what was I thinking? Well, fixed it. Does that look better @user2357112
user2357112
Sort of, but you have a lingering reference to the order of the bits being "already reversed by the previous step". Also, it looks like the questioner wants to autocompute the smallest required width instead of specifying a width manually, and for completeness, you should probably show how to get the NumPy array the questioner asked for out of this.
Vladislav Martin
@user2357112 I see. All of those are good points, I've made changes to my answer to address each of them - please let me know if I've yet to present my answer is a satisfactory manner. It's important that anyone who sees this would get some use out of it (especially the OP!) :)
Here's an approach using np.binary_repr and np.fromstring -
np.fromstring(np.binary_repr(num), dtype=np.uint8)==49
Sample run -
In [39]: num
Out[39]: 17
In [40]: np.fromstring(np.binary_repr(num), dtype=np.uint8)==49
Out[40]: array([ True, False, False, False, True], dtype=bool)
Comments
Maybe it is not the easiest, but a compact way is
from numpy import array
array([i for i in bin(5)[2:]]) == '1'
np.int8, which is ugly, and the output puts most significant bits first, and groups the bits into groups of 8 — neither are things I want.