Finding the index of an element in a sorted list efficiently

Question

I have a sorted list l (of around 20,000 elements), and would like to find the first element in l that exceeds a given value t_min. Currently, my code is as follows.

def find_index(l):   
    first=next((t for t in l if t>t_min), None) 
    if first==None:
        return None
    else:
        return l.index(first)

To benchmark the code, I used cProfile to run a testing loop, and stripped out the time required to randomly generate lists by comparing the time to a control loop:

import numpy
import cProfile

def test_loop(n):
    for _ in range(n):
        test_l=sorted(numpy.random.random_sample(20000))
        find_index(test_l, 0.5)

def control_loop(n):
    for _ in range(n):
        test_l=sorted(numpy.random.random_sample(20000))

# cProfile.run('test_loop(1000)') takes 10.810 seconds
# cProfile.run('control_loop(1000)') takes 9.650 seconds

Each function call for find_index takes about 1.16 ms. Is there a way to improve the code to make it more efficient, given that we know the list is sorted?

Answer 1

The standard library bisect module is useful for this, and the docs contain an example of exactly this use case.

def find_gt(a, x):
    'Find leftmost value greater than x'
    i = bisect_right(a, x)
    if i != len(a):
        return a[i]
    raise ValueError