Convert decimal to ternary(base3) in python

Question

I am trying to make a decimal number ternary in a python function. My idea was to keep dividing until the quotient and remainder were equal, but I can't seem to get that to work. Here's my code:

l = 1


#problem code
def ternary(n):
    e = n/3
    q = n%3
    e= n/3
    q= e%3
    print q

r = input("What number should I convert?: ")
k = bin(r)
v = hex(r)
i = oct(r)
print k+"(Binary)"
print v+"(Hex)"
print i+"(Octals)"
ternary(r)
l+=1
# Variables:
#l,r,k,v,i 
#n,q,e

Answer 1

You can also use the implementation of NumPy: https://numpy.org/doc/stable/reference/generated/numpy.base_repr.html?highlight=base_repr#numpy.base_repr

Though, I agree that a function for ternary exclusively is faster.

import numpy as np

number=100 # decimal
ternary=np.base_repr(number,base=3)
print(ternary)
#10201

Answer 2

My idea was to keep dividing until the quotient and remainder were equal, but I can't seem to get that to work.

Yeah, something like that. Essentially, you want to keep dividing by 3, and collect the remainders. The remainders then make up the final number. In Python, you can use divmod to divide and collect the remainder.

def ternary (n):
    if n == 0:
        return '0'
    nums = []
    while n:
        n, r = divmod(n, 3)
        nums.append(str(r))
    return ''.join(reversed(nums))

Examples:

>>> ternary(0)
'0'
>>> ternary(1)
'1'
>>> ternary(2)
'2'
>>> ternary(3)
'10'
>>> ternary(12)
'110'
>>> ternary(22)
'211'

Answer 3

This can also be done with recursion.

def ternary(n):
    e = n//3
    q = n%3
    if n == 0:
        return '0'
    elif e == 0:
        return str(q)
    else:
        return ternary(e) + str(q)

More generally, you can convert to any base b (where 2<=b<=10) with the following recursive function.

def baseb(n, b):
    e = n//b
    q = n%b
    if n == 0:
        return '0'
    elif e == 0:
        return str(q)
    else:
        return baseb(e, b) + str(q)