Binary search algorithm in python

Why not use the bisect module? It should do the job you need---less code for you to maintain and test.

In the case that needle_element > array[mid], you currently pass array[mid:] to the recursive call. But you know that array[mid] is too small, so you can pass array[mid+1:] instead (and adjust the returned index accordingly).

If the needle is larger than all the elements in the array, doing it this way will eventually give you an empty array, and an error will be raised as expected.

Note: Creating a sub-array each time will result in bad performance for large arrays. It's better to pass in the bounds of the array instead.

array[mid:] creates a new sub-copy everytime you call it = slow. Also you use recursion, which in Python is slow, too.

Try this:

def binarysearch(sequence, value):
    lo, hi = 0, len(sequence) - 1
    while lo <= hi:
        mid = (lo + hi) // 2
        if sequence[mid] < value:
            lo = mid + 1
        elif value < sequence[mid]:
            hi = mid - 1
        else:
            return mid
    return None

It would be much better to work with a lower and upper indexes as Lasse V. Karlsen was suggesting in a comment to the question.

This would be the code:

def binary_search(array, target):
    lower = 0
    upper = len(array)
    while lower < upper:   # use < instead of <=
        x = lower + (upper - lower) // 2
        val = array[x]
        if target == val:
            return x
        elif target > val:
            if lower == x:   # these two are the actual lines
                break        # you're looking for
            lower = x
        elif target < val:
            upper = x

• lower < upper will stop once you have reached the smaller number (from the left side)
• if lower == x: break will stop once you've reached the higher number (from the right side)

Example:

>>> binary_search([1,5,8,10], 5)   # return 1
1
>>> binary_search([1,5,8,10], 0)   # return None
>>> binary_search([1,5,8,10], 15)  # return None