Finding k closest numbers to a given number

Question

Say I have a list [1,2,3,4,5,6,7]. I want to find the 3 closest numbers to, say, 6.5. Then the returned value would be [5,6,7].

Finding one closest number is not that tricky in python, which can be done using

min(myList, key=lambda x:abs(x-myNumber))

But I am trying not to put a loop around this to find k closest numbers. Is there a pythonic way to achieve the above task?

Asked By: frazman

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Source

Answer 1

You could compute distances, and sort:

[n for d, n in sorted((abs(x-myNumber), x) for x in myList)[:k]]

This does the following:

Create a sequence of tuples (d, x) where d is the distance to your target
Select the first k elements of that list
Extract just the number values from the result, discarding the distance

Answered By: Greg Hewgill

Answer 2

The short answer

The heapq.nsmallest() function will do this neatly and efficiently:

>>> from heapq import nsmallest
>>> s = [1,2,3,4,5,6,7]
>>> nsmallest(3, s, key=lambda x: abs(x - 6.5))
[6, 7, 5]

Essentially this says, "Give me the three input values that have the smallest absolute difference from the number 6.5".

Optimizing for repeated lookups

In the comments, @Phylliida, asked how to optimize for repeated lookups with differing start points. One approach would be to pre-sort the data and then use bisect to locate the center of a small search segment:

from bisect import bisect

def k_nearest(k, center, sorted_data):
    'Return *k* members of *sorted_data* nearest to *center*'
    i = bisect(sorted_data, center)
    segment = sorted_data[max(i-k, 0) : i+k]
    return nsmallest(k, segment, key=lambda x: abs(x - center))

For example:

>>> s.sort()
>>> k_nearest(3, 6.5, s)
[6, 7, 5]
>>> k_nearest(3, 0.5, s)
[1, 2, 3]
>>> k_nearest(3, 4.5, s)    
[4, 5, 3]
>>> k_nearest(3, 5.0, s)
[5, 4, 6]

Answered By: Raymond Hettinger

Answer 3

Both answers were good, and Greg was right, Raymond’s answer is more high level and easier to implement, but I built upon Greg’s answer because it was easier to manipulate to fit my need.

In case anyone is searching for a way to find the n closest values from a list of dicts.

My dict looks like this, where npi is just an identifier that I need along with the value:

mydict = {u'fnpi': u'1982650024',
 u'snpi': {u'npi': u'1932190360', u'value': 2672},
 u'snpis': [{u'npi': u'1831289255', u'value': 20},
  {u'npi': u'1831139799', u'value': 20},
  {u'npi': u'1386686137', u'value': 37},
  {u'npi': u'1457355257', u'value': 45},
  {u'npi': u'1427043645', u'value': 53},
  {u'npi': u'1477548675', u'value': 53},
  {u'npi': u'1851351514', u'value': 57},
  {u'npi': u'1366446171', u'value': 60},
  {u'npi': u'1568460640', u'value': 75},
  {u'npi': u'1326046673', u'value': 109},
  {u'npi': u'1548281124', u'value': 196},
  {u'npi': u'1912989989', u'value': 232},
  {u'npi': u'1336147685', u'value': 284},
  {u'npi': u'1801894142', u'value': 497},
  {u'npi': u'1538182779', u'value': 995},
  {u'npi': u'1932190360', u'value': 2672},
  {u'npi': u'1114020336', u'value': 3264}]}

value = mydict['snpi']['value'] #value i'm working with below
npi = mydict['snpi']['npi'] #npi (identifier) i'm working with below
snpis = mydict['snpis'] #dict i'm working with below

To get an [id, value] list (not just a list of values) , I use this:

[[id,val] for diff, val, id in sorted((abs(x['value']-value), x['value'], x['npi']) for x in snpis)[:6]]

Which produces this:

[[u'1932190360', 2672],
 [u'1114020336', 3264],
 [u'1538182779', 995],
 [u'1801894142', 497],
 [u'1336147685', 284],
 [u'1912989989', 232]]

EDIT

I actually found it pretty easy to manipulate Raymond’s answer too, if you’re dealing with a dict (or list of lists).

from heapq import nsmallest
[[i['npi'], i['value']] for i in nsmallest(6, snpis, key=lambda x: abs(x['value']-value))]

This will produce the same as the above output.

And this

nsmallest(6, snpis, key=lambda x: abs(x['value']-value)) will produce a dict instead.

Answered By: tmthyjames

Answer 4

For those who wants the index instead:

def find_nearest_index(array, value, k):
    array = np.array(array)
    return np.argsort(abs(array - value))[:k]

Example:

find_nearest_index([-3,0,1,2,4,5], 0.2, 4)

# array([1, 2, 3, 0], dtype=int64)
# distance = [3.20 0.20 0.80 1.80 3.80 4.80]

Answered By: Muhammad Yasirroni

Finding k closest numbers to a given number

Question:

Answers:

The short answer

Optimizing for repeated lookups