(See the Complexity handout for the analysis of Binary Search)

def find_i_binary(L, e):
    '''Returns the index of e in L, or None if e is not in L
    
    Arguments:
    L -- a sorted list of ints
    e -- int
    
    '''
    low = 0
    high = len(L) - 1
    
    while high-low >= 2:
        mid = (low+high)//2
        if L[mid] > e:
            high = mid - 1
        elif L[mid] < e:
            low = mid + 1
        else:
            return mid
            
    if L[low] == e:
        return low
    elif L[high] == e:
        return high
    else:
        return None

So, how does Binary Search compare to our previous attempt (called Linear Search)?

Suppose that one iteration of Binary Search runs $c_1$ ms, and one iteration of Linear Search runs $c_2$ ms. $c_1$ and $c_2$ are quite equal, but are pretty close.

For $n=1,000,000$, Binary Search will take, in the worst case, about $c_1\times \log_2 (1,000,000) = 20c_1$ ms. Linear Search will take $1,000,000 \times c_2$ ms. A big difference!

Of course, that's all assuming that the list is sorted. But if we need to perform multiple look-ups, it's worth it to sort the list the once, and then be able to use the much faster Binary Search.