Decipher contents of a Hollerith-style punched card

hollerith_card.py

# As read on the card, top-down. For convenience, leading zeros can be skipped
rows = [
    "11000001100001001110010101010101001011000110",
    "11000000110000000101010001000100100100001",
    "100011000110001000000100000000000001000",
    "100001000000000000000100001001000000",
    "0",
    "100001000100101110110100000000100001000",
    "10010000000000000000000000000000000",
    "10001000000001000001000010000100000100011",
    "1001000000100000000000000000000000010000000",
    "0",
    "1000000011101011100000000000000100100",
    "10000000000000000010000001001001000000000000",
]

maxlen = max(len(row) for row in rows)
# Restore leading zeros
rows = ["0" * (maxlen - len(row)) + row for row in rows]

# Transpose: each column is a (badly named as it doesn't have exactly 8 bits) "octet"
octets = list(zip(*rows))
# Convert back to list of strings, MSB (highest bit is first)
bitstrings = ["".join(bits) for bits in octets]

# Source: http://www.columbia.edu/cu/computinghistory/026.html plus some educated guessing
hollerith = {
    0: ' ',
    0x842: '.', 0x242: ',',
    0x812: '(', 0x412: ')',
    0x200: '0', 0x100: '1', 0x080: '2', 0x040: '3', 0x020: '4', 0x010: '5',
    0x008: '6', 0x004: '7', 0x002: '8', 0x001: '9',
    0x900: 'A', 0x880: 'B', 0x840: 'C', 0x820: 'D', 0x810: 'E', 0x808: 'F', 0x804: 'G', 0x802: 'H', 0x801: 'I',
    0x500: 'J', 0x480: 'K', 0x440: 'L', 0x420: 'M', 0x410: 'N', 0x408: 'O', 0x404: 'P', 0x402: 'Q', 0x401: 'R',
    0x280: 'S', 0x240: 'T', 0x220: 'U', 0x210: 'V', 0x208: 'W', 0x204: 'X', 0x202: 'Y', 0x201: 'Z',
}

# Convert octets to integers
values = [int(bitstring, 2) for bitstring in bitstrings]

# Print any unknown values and their positions
for i, v in enumerate(values):
    if v not in hollerith:
        print("{}:{}".format(i, hex(v)))

# Decode the card
print("".join(hollerith.get(b, '?') for b in values))