In a python list which is sorted, find the closest value to target value and its index in the list

Question

I am trying to get both the closest value and its index in a sorted list in python.

In MATLAB this is possible with:

[closest_val,index] = min(abs(array - target))

I was wondering if there is a similar way this can be implemented in python.

I’ve seen posts which do one or the other, but I haven’t seen both done together.

Link to finding closest value in list post.

Asked By: user94758

||

Source

Answer 1

this is an option:

lst = [
    13.09409,
    12.18347,
    11.33447,
    10.32184,
    9.544922,
    8.813385,
]

target = 11.5

res = min(enumerate(lst), key=lambda x: abs(target - x[1]))
# (2, 11.33447)

enumerate iterates over your list in index, value pairs. the key of the min method tells it to only consider the value.

note that python starts indexing at 0; matlab at 1 as far as i remember. if you want that same behavior:

res = min(enumerate(lst, start=1), key=lambda x: abs(target - x[1]))
# (3, 11.33447)

if the list is big, i strongly suggest you use bisect as suggested in this answer.

Answered By: hiro protagonist

Answer 2

bisect wasn’t used in the linked question because the list was not sorted. Here, we don’t have the same problem, and we can use bisect for the speed it provides:

import bisect

def find_closest_index(a, x):
    i = bisect.bisect_left(a, x)
    if i >= len(a):
        i = len(a) - 1
    elif i and a[i] - x > x - a[i - 1]:
        i = i - 1
    return (i, a[i])

find_closest_index([1, 2, 3, 7, 10, 11], 0)   # => 0, 1
find_closest_index([1, 2, 3, 7, 10, 11], 7)   # => 3, 7
find_closest_index([1, 2, 3, 7, 10, 11], 8)   # => 3, 7
find_closest_index([1, 2, 3, 7, 10, 11], 9)   # => 4, 10
find_closest_index([1, 2, 3, 7, 10, 11], 12)  # => 5, 11

EDIT: In case of descending array:

def bisect_left_rev(a, x, lo=0, hi=None):
    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if a[mid] > x: lo = mid+1
        else: hi = mid
    return lo

def find_closest_index_rev(a, x):
    i = bisect_left_rev(a, x)
    if i >= len(a):
        i = len(a) - 1
    elif i and a[i] - x < x - a[i - 1]:
        i = i - 1
    return (i, a[i])

find_closest_index_rev([11, 10, 7, 3, 2, 1], 0)   # => 5, 1
find_closest_index_rev([11, 10, 7, 3, 2, 1], 7)   # => 2, 7
find_closest_index_rev([11, 10, 7, 3, 2, 1], 8)   # => 2, 7
find_closest_index_rev([11, 10, 7, 3, 2, 1], 9)   # => 1, 10
find_closest_index_rev([11, 10, 7, 3, 2, 1], 12)  # => 0, 11

Answered By: Amadan

In a python list which is sorted, find the closest value to target value and its index in the list

Question:

Answers: