dis.obs dis.obs - 5 months ago 7
Python Question

How do I find the closest n numbers to a given number at x distance from it?

For e.g. I have a list of number like

lst = [1, 2, 3, 3, 4, 4, 5, 6, 7, 8, 9]


And I need 2 numbers at 3 distance from a given number let's say 5,
so the output list should look like this

output_lst = [2, 8]


Here by distance I mean the distance on number-line and not in list index. So 3 numbers, 2 distance from 5 would give

output_lst = [3,3,7]


What I tired doing was to use nsmallest from heapq like this

check_number = 5

output_lst = nsmallest(3, lst, key=lambda x: abs(x - check_number))


But the problem here is that I don't know specify the distance. It will just outputs 3 closest numbers to 5.

[4,4,5]

Answer

You can use a list comprehension for that. See this post for more on list comprehensions.

>>> lst = [1, 2, 3, 3, 4, 4, 5, 6, 7, 8, 9]
>>> given_numer = 5
>>> distance = 3
>>> [i for i in lst if abs(i-given_numer)==distance]
[2, 8]

The logic is quite simple, we just see if the absolute value of the difference between each number and the given number, if so we return the value. Similarly

>>> distance = 2
>>> [i for i in lst if abs(i-given_numer)==distance]
[3, 3, 7]

Let's complicate a bit and try to use filter and closures. The code is:

Just to show that it is an alternative.

def checkdistance(given_number,distance):
    def innerfunc(value):
        return abs(value-given_number)==distance
    return innerfunc


lst = [1, 2, 3, 3, 4, 4, 5, 6, 7, 8, 9]
given_number = 5
distance = 3
checkdistance3from5 = checkdistance(5,3)
list(filter(checkdistance3from5,lst))
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