math - Create a base 12 calculator with different limits at diferent digits with python

Question

I want o create a calculator that can add (and multiply, divide, etc) numbers in base 12 and with different limits at the different digits.

Base 12 sequence: [0,1,2,3,4,5,6,7,8,9,"A","B"]

The limits must be:

First digit: limit "B" Second digit: limit 4 Third digit: limit "B"

(The idea would be that it follows the hourly System limits but in base 12 so for example in base 12 there are 50 seconds in a minute)

That means you would count like this:[1,2,3,4,5,6,7,8,9,A,B,10,11,...48,49,4A,4B,100,101,...14B,200,201,...B4B,1000,1001..]

So I made the following code

 import string
digs = string.digits + string.ascii_uppercase


def converter(number):
    #split number in figures
    figures = [int(i,12) for i in str(number)]
    #invert oder of figures (lowest count first)
    figures = figures[::-1]
    result = 0
    #loop over all figures
    for i in range(len(figures)):
        #add the contirbution of the i-th figure
        result += figures[i]*12**i
    return result

def int2base(x):
    if x < 0:
        sign = -1
    elif x == 0:
        return digs[0]
    else:sign = 1

    x *= sign
    digits = []

    while x:
        digits.append(digs[int(x % 12)])
        x = int(x / 12)

    if sign < 0:
        digits.append('-')

    digits.reverse()

    return ''.join(digits)

def calculator (entry1, operation, entry2):
    value1=float(converter(entry1))
    value2=float(converter(entry2))
    if operation == "suma" or "+":
        resultvalue=value1+value2
    else:
        print("operación no encontrada, porfavor ingrese +,-,")
    result=int2base(resultvalue)
    return result


print(calculator(input("Ingrese primer valor"), input("ingrese operación"), input("Ingrese segundo valor")))

The thing is that I dont know how to establish the limits to the different digits If someone could help me I would be extreamly greatful

Answer 1

You can define two converters:

class Base12Convert:
    d = {hex(te)[2:].upper():te for te in range(0,12)}
    d.update({val:key for key,val in d.items()})
    d["-"] = "-"

    @staticmethod
    def text_to_int(text):
        """Converts a base-12 text into an int."""
        if not isinstance(text,str):
            raise ValueError(
                f"Only strings allowed: '{text}' of type '{type(text)}' is invalid")
        t = text.strip().upper()
        if any (x not in Base12Convert.d for x in t):
            raise ValueError(
                f"Only [-0123456789abAB] allowed in string: '{t}' is invalid")            
        if "-" in t.lstrip("-"):
            raise ValueError(f"Sign '-' only allowed in front. '{t}' is invalid")
        # easy way
        return int(t,12)

        # self-calculated way
        # return sum(Base12Convert.d[num]*12**idx for idx,num in enumerate(t[::-1]))

    @staticmethod
    def int_to_text(num):
        """Converts an int into a base-12 string."""

        sign = ""
        if not isinstance(num,int):
            raise ValueError(
                f"Only integer as input allowed: '{num} of type {type(num)}' is invalid")
        if num < 0:
            sign = "-"
            num *= -1

        # get highest possible number
        p = 1
        while p < num:
            p *= 12

        # create string 
        rv = [sign]
        while True:
            p /= 12
            div = num // p
            num -= div*p
            rv.append(Base12Convert.d[div])
            if p == 1:
                break

        return ''.join(rv)

Then you can use them to convert what you want:

text = "14:54:31"  # some time 

# convert each int of the time into base12 - join with | again
base12 = '|'.join(Base12Convert.int_to_text (int(t)) for t in text.split(":"))

# split base12 at |, convert to int, make str from int and join with | again
base10 = '|'.join(map(str,(Base12Convert.text_to_int (b12) for b12 in base12.split("|"))))

# print all 3
print(text,base12,base10,sep="
")

Output:

14:54:31
12|46|27
14|54|31

and enforce whatever restrictions you have on your "digits" using normal ints. You should split up your "digits" of a clock ... 14b (203 in base 10) does not make sense, 1:4b might if you mean 1:59 hours/minutes.