Just for fun - the solution as a recursive one-liner:

def tobin(x):
    return tobin(x/2) + [x%2] if x > 1 else [x]

this would be a lot easier no?

conv = str(input('what do you want to convert?'))
if conv == 'binary':
    x = str(input('enter binary : '))
    ansBin = int(x, 2)
    print('decimal for binary ', x, ' is ', ansBin)
elif conv == 'decimal':
    y = int(input('enter decimal : '))
    ansDec = bin(y)[2:]
    print('binary for decimal ', y, ' is ', ansDec)

To convert binary value to decimal you need to do the following:

Take the least significant bit and multiply it by 2^0, then take the next least significant beat and multiply it by 2^1, next one by 2^2 and so on...

Let's say, for example you need to convert a number 1010 to decimal:

You would have 0*2^0 + 1*2^1 + 0*2^2 + 1*2^3 = 0 + 2 + 0 + 8 = 10

So in your python code, you need to:

• read the int that the user inputted (representing the binary value).
• convert that int and convert it to string, so you can break it into list of digits
• make a list of digits from the string you created (a list int python can be created from a string not an int, that's why you need the conversion to string first)
• go trough that list of bits in reverse and multiply every bit by 2^k, k being the counter starting from 0

Here's the code that demonstrates what I just tried to explain:

user_input = int(input("enter a binary value"))

bits = list(str(user_input))

decimal = 0

counter = 0

for i in reversed(bits):
    decimal += 2**counter * int(i)
    counter+=1

print 'The decimal value is: ', decimal

Without the 0b in front:

"{0:b}".format(int_value)

Starting with Python 3.6 you can also use formatted string literal or f-string, --- PEP:

f"{int_value:b}"