Create A Base 12 Calculator With Different Limits At Diferent Digits With Python

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,'

Solution 1:

You can define two converters:

classBase12Convert:
    d = {hex(te)[2:].upper():te for te inrange(0,12)}
    d.update({val:key for key,val in d.items()})
    d["-"] = "-"    @staticmethoddeftext_to_int(text):
        """Converts a base-12 text into an int."""ifnotisinstance(text,str):
            raise ValueError(
                f"Only strings allowed: '{text}' of type '{type(text)}' is invalid")
        t = text.strip().upper()
        ifany (x notin 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 wayreturnint(t,12)

        # self-calculated way# return sum(Base12Convert.d[num]*12**idx for idx,num in enumerate(t[::-1]))    @staticmethoddefint_to_text(num):
        """Converts an int into a base-12 string."""

        sign = ""ifnotisinstance(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 = 1while p < num:
            p *= 12# create string 
        rv = [sign]
        whileTrue:
            p /= 12
            div = num // p
            num -= div*p
            rv.append(Base12Convert.d[div])
            if p == 1:
                breakreturn''.join(rv)

Then you can use them to convert what you want:

text = "14:54:31"  # sometime 

# converteachintof the timeinto base12 -joinwith| again
base12 ='|'.join(Base12Convert.int_to_text (int(t)) for t in text.split(":"))

# split base12 at|, converttoint, make str fromintandjoinwith| again
base10 ='|'.join(map(str,(Base12Convert.text_to_int (b12) for b12 in base12.split("|"))))

# print all3
print(text,base12,base10,sep="\n")

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.