Given a 1d integer array, e.g.:
[1, 0, -1]
looking for a binary representation matrix, desired output:
[[0 1], [0 0], [1 1]]
possibly using np.binary_repr with a given fixed width. Currently np.binary_repr gives back a string and only applies to a single number at a time.
tobin = np.vectorize(np.binary_repr)
tobin(np.arange(4))
# ['0000' '0001' '0010' '0011']
You can use np.unpackbits:
a=np.array([-1,0,1]) # dtype is np.int32
You have to input your data as np.uint8 because that is the only data type supported by unpackbits:
bi = np.unpackbits(a[:,None].view(np.uint8), axis=1)
Original input data is 32bit, so you get 32 values per input element, crop accordingly (keeping in mind min/max values in a):
result = bi[:, :8]
array([[1, 1, 1, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1]], dtype=uint8)
This works well for small numbers like the one in the answer. If you need more than 8 bits, you should read the 1st 8 elements of bi and then the 16th to 8th elements. It's a bit messy.
For a more generic solution, just better flip the array view. And cropping before unpacking will give you a some performance improvement:
def int_to_bin(arr, n_bytes=1):
# arr is 1-D
arr_ = np.fliplr(arr[:, None].view(np.uint8))[:, -n_bytes:]
return np.unpackbits(arr_, axis=1)
You can crop the output further if you want, say, only 4 bits. This takes about 10 ms for one million arr of int32.
int is one byte. I think that's because of hardware, so under the hood, you are getting 8 bits whether you want them all or not. For arbitrary bits, you can simply index what you need: int_to_bin(a)[:, -4:] for 4 bitsHere is an explicit for loop based solution I currently have, would be keen on getting a vectorised solution.
def int_to_bin_matrix(arr: np.ndarray, width: int):
"""Return the binary representation of matrix of integer array."""
return np.array([[int(c) for c in np.binary_repr(i, width=width)] for i in arr], dtype=np.int32)
print(int_to_bin_matrix(np.arange(-2,2), 4))
[[1 1 1 0]
[1 1 1 1]
[0 0 0 0]
[0 0 0 1]]