Python: Convert list of ints to one number?

Question

I have a list of integers that I would like to convert to one number like:

numList = [1,2,3]
num = magic(numList)

print num, type(num)
>>> 123, <type 'int'>

What is the best way to implement the magic function?

Thanks for your help.

EDIT
I did find this, but it seems like there has to be a better way.

EDIT 2
Let's give some credit to Triptych and cdleary for their great answers! Thanks guys.

Answer 1

# Over-explaining a bit:
def magic(numList):         # [1,2,3]
    s = map(str, numList)   # ['1','2','3']
    s = ''.join(s)          # '123'
    s = int(s)              # 123
    return s


# How I'd probably write it:
def magic(numList):
    s = ''.join(map(str, numList))
    return int(s)


# As a one-liner  
num = int(''.join(map(str,numList)))


# Functionally:
s = reduce(lambda x,y: x+str(y), numList, '')
num = int(s)


# Using some oft-forgotten built-ins:
s = filter(str.isdigit, repr(numList))
num = int(s)

Answer 2

pseudo-code:

int magic(list nums)
{
  int tot = 0

  while (!nums.isEmpty())
  {
    int digit = nums.takeFirst()
    tot *= 10
    tot += digit
  }

  return tot
}

Answer 3

def magic(numbers):
    return int(''.join([ "%d"%x for x in numbers]))

Answer 4

def magic(number):
    return int(''.join(str(i) for i in number))

Answer 5

This seems pretty clean, to me.

def magic( aList, base=10 ):
    n= 0
    for d in aList:
        n = base*n + d
    return n

Answer 6

This method works in 2.x as long as each element in the list is only a single digit. But you shouldn't actually use this. It's horrible.

>>> magic = lambda l:int(`l`[1::3])
>>> magic([3,1,3,3,7])
31337

Answer 7

Using a generator expression:

def magic(numbers):
    digits = ''.join(str(n) for n in numbers)
    return int(digits)