You can use int and set the base to 2 (for binary):

>>> binary = raw_input('enter a number: ')
enter a number: 11001
>>> int(binary, 2)
25
>>>

However, if you cannot use int like that, then you could always do this:

binary = raw_input('enter a number: ')
decimal = 0
for digit in binary:
    decimal = decimal*2 + int(digit)
print decimal

Below is a demonstration:

>>> binary = raw_input('enter a number: ')
enter a number: 11001
>>> decimal = 0
>>> for digit in binary:
...     decimal = decimal*2 + int(digit)
...
>>> print decimal
25
>>>

To expand on my comment: Let's say we want to convert $409$ to binary. We first need to know how many bits we'll need; we need to know the largest power of $2$ that's less than $409$. We'll also need all smaller powers of $2$, so let's write those down (they are all very easy to calculate by hand).

\begin{array}{c|c|c|c|c|c} k & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\\hline 2^k &256 & 128 & 64 & 32 & 16 & 8 & 4 & 2 & 1 \end{array}

So we'll need $9$ bits. To find them, we essentially subtract the largest power of $2$ from our remainders, repeatedly (for example, $409 - 256 = 153$): \begin{align*} 409 &= 256\cdot 1 + 153 \\ 153 &= 128 \cdot 1 + 25 \\ 25 &= 16 \cdot 1 + 9 \\ 9 &= 8 \cdot 1 + 1 \\ 1 &= 1 \cdot 1 \end{align*}

Since we saw $256,\ 128,\ 16,\ 8,$ and $1$ show up in the process, we have $$409_{10} = 110011001_2.$$

Of course, for a number like yours, we would need many more powers of $2$, in general.

But! It just so happens that $2147483647 = 2^{31} - 1$, and its binary representation is literally $$2147483647_{10} = \underbrace{111\ldots 1_2}_{31\text{ times}}$$ (try smaller numbers that are one less than a power of $2$, like $7, 15, 31$, and you'll see the pattern).