Generate Binary Data Array From a Range of Numbers

Question

I need to create Boolean/Binary arrays with rows representing the binary form of each number in given a range. Each column hold values for a given bit position. The range is determined by the bit-width and binary encoding scheme. Then I need to count the number of cells in a column that do not match the preceding cell.

I believe the counting process is simply XORing the array with itself shifted down one row. The sum of each column in the XOR result is my transition count.

I need to do this for each combination of [signed, unsigned] values encoded as [hexadecimal, octal, decimal]. Mismatches between the encoding scheme and available bits results in unusable bits.
e.g. A 4-bit unsigned octal array would still only have 8 rows. However, if it is signed then that 4th bit is used for sign and the array is the full 16 rows.

The ultimate goal is to create labeled set of summary n-grams for the number of transitions at each bit position for various bit widths, encoding, and use of sign. The summary n-gram for example 1 would be {3-bit, octal, unsigned, [1, 3, 7]}. The summary n-gram for example 3 is {4-bit, octal, signed, [1, 2, 6, 13]}. Example 4 is {4-bit, hexadecimal, signed, [1, 2, 6, 13]}.

Examples:
1) 8x3 Binary array of all possible 3-bit unsigned octal values:
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1

2) 8x4 Binary array of all possible 4-bit unsigned octal values:
Same as #1 except with an additional leftmost column of all zeros

3) 16x4 Binary array of all possible 4-bit signed octal values:
Note: It's ok to double count zero (-0 and +0) 1 1 1 1
1 1 1 0
1 1 0 1
1 1 0 0
1 0 1 1
1 0 1 0
1 0 0 1
1 0 0 0
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1

4) 16x4 Binary array of all possible 4-bit signed hexadecimal values:
Same as #3

Paul Panzer · Accepted Answer · 2017-12-06 23:37:55Z

3

You can use np.unpackbits, for example

np.unpackbits(np.arange(8).astype(np.uint8)[:, None], axis = 1)

Output:

array([[0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1],
       [0, 0, 0, 0, 0, 0, 1, 0],
       [0, 0, 0, 0, 0, 0, 1, 1],
       [0, 0, 0, 0, 0, 1, 0, 0],
       [0, 0, 0, 0, 0, 1, 0, 1],
       [0, 0, 0, 0, 0, 1, 1, 0],
       [0, 0, 0, 0, 0, 1, 1, 1]], dtype=uint8)

Slice off unwanted zeros:

np.unpackbits(np.arange(8).astype(np.uint8)[:, None], axis = 1)[:, -3:]

Result:

array([[0, 0, 0],
       [0, 0, 1],
       [0, 1, 0],
       [0, 1, 1],
       [1, 0, 0],
       [1, 0, 1],
       [1, 1, 0],
       [1, 1, 1]], dtype=uint8)

Use modulo for signed values:

np.unpackbits((np.arange(-4, 4)%8).astype(np.uint8)[:, None], axis = 1)[:, -3:]

Or view casting from signed to unsigned:

np.unpackbits(np.arange(-4, 4).astype(np.int8).view(np.uint8)[:, None], axis = 1)[:, -3:]

Result:

array([[1, 0, 0],
       [1, 0, 1],
       [1, 1, 0],
       [1, 1, 1],
       [0, 0, 0],
       [0, 0, 1],
       [0, 1, 0],
       [0, 1, 1]], dtype=uint8)

Transition count:

x = np.unpackbits(np.arange(-4, 4).astype(np.int8).view(np.uint8)[:, None], axis = 1)[:, -3:]
np.abs(np.diff(x.view(np.int8), axis=0)).sum(axis=1)

Result:

array([1, 2, 1, 3, 1, 2, 1])

Or using your xor method:

(x[1:]^x[:-1]).sum(axis=1)

Result:

array([1, 2, 1, 3, 1, 2, 1], dtype=uint64)

edited Dec 6, 2017 at 23:37

answered Dec 6, 2017 at 22:55

Paul Panzer

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1 Comment

Brent Over a year ago

Nice. I didn't know about unpack bits.

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