Creating a half Byte dictionary of hexadecimals - without built in functions

Question

base_destination (dict_create - the function receives as input the target base (8 or 16) as an integer And returns a dictionary so that the keys in the dictionary are the binary representation of the number and the values are the representation at the base The target obtained, the keys must contain 4 digits in binary representation as a string. For example: For the input - 8 {"0000": 0, "0001": 1, "0010": 2, "0011": 3, "0100": 4, "0101": 5, "0110": 6, "0111": 7} - The dictionary will be accepted

• Do not write the dictionary manually

My code:

def pad_zeros(binary_as_str, length):
    subtraction = length - len(binary_as_str)
    return "0" * subtraction + binary_as_str

 def binary_add_one(binary_as_str):
    binary_string = pad_zeros(binary_as_str, 4)
    for i in range(len(binary_string) - 1, - 1, - 1):
        if binary_string[i] == "0":
            return binary_string[0:i] + "1" + "0" * (len(binary_string) - i - 1)
    return "1" + 0 * binary_string


def number_to_letter():
    str = ""
    for i in range(7):
        str += chr(65 + i)
    return str


def create_dict(destination_base):
    dicti = {}
    key_list = []
    starter = "0000"
    for i in range(destination_base):
        if i < 10:
            key_list.append(starter)
            starter = binary_add_one(starter)
            dicti[key_list[i]] = i
        else:
            key_list.append(starter)
            starter = binary_add_one(starter)
            dicti[key_list[i]] = number_to_letter()[i - 10]
    return dicti


print(create_dict(16))

It works great, but the problem is, I dont know how to do it without putting the list.. Any tip on how to do without using list + int + ascii to do it? without built in functions...

Answer 1

To avoid the literal list, change the definition of value_list to

value_list=[x for x in range(10)]+["{:X}".format(x) for x in range(10,16)]

Or are you not allowed to use range() and str.format() either?