Python convert decimal to hex

I have a function here that converts decimal to hex but it prints it in reverse order. How would I fix it?

def ChangeHex(n):
    if (n < 0):
        print(0)
    elif (n<=1):
        print(n)
    else:
        x =(n%16)
        if (x < 10):
            print(x), 
        if (x == 10):
            print("A"),
        if (x == 11):
            print("B"),
        if (x == 12):
            print("C"),
        if (x == 13):
            print("D"),
        if (x == 14):
            print("E"),
        if (x == 15):
            print ("F"),
        ChangeHex( n / 16 )

What about this:

hex(dec).split('x')[-1]

Example:

>>> d = 30
>>> hex(d).split('x')[-1]
'1e'

~Rich

By using -1 in the result of split(), this would work even if split returned a list of 1 element.

This isn't exactly what you asked for but you can use the "hex" function in python:

>>> hex(15)
'0xf'

If you want to code this yourself instead of using the built-in function hex(), you can simply do the recursive call before you print the current digit:

def ChangeHex(n):
    if (n < 0):
        print(0)
    elif (n<=1):
        print n,
    else:
        ChangeHex( n / 16 )
        x =(n%16)
        if (x < 10):
            print(x), 
        if (x == 10):
            print("A"),
        if (x == 11):
            print("B"),
        if (x == 12):
            print("C"),
        if (x == 13):
            print("D"),
        if (x == 14):
            print("E"),
        if (x == 15):
            print ("F"),

I think this solution is elegant:

def toHex(dec):
    digits = "0123456789ABCDEF"
    x = (dec % 16)
    rest = dec // 16
    if (rest == 0):
        return digits[x]
    return toHex(rest) + digits[x]

numbers = [0, 11, 16, 32, 33, 41, 45, 678, 574893]
print [toHex(x) for x in numbers]
print [hex(x) for x in numbers]

This output:

['0', 'B', '10', '20', '21', '29', '2D', '2A6', '8C5AD']
['0x0', '0xb', '0x10', '0x20', '0x21', '0x29', '0x2d', '0x2a6', '0x8c5ad']