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)

Edit:

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.

Here 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]]