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I would like to find out the three closest numbers in a vector.
Something like

v = c(10,23,25,26,38,50)
c = findClosest(v,3)
c
23 25 26

I tried with sort(colSums(as.matrix(dist(x))))[1:3], and kind of works, but it selects the three numbers with minimum overall distance not the three closest numbers.

There is already an answer for matlab, but I do not know how to translate it to R:

%finds the index with the minimal difference in A
minDiffInd = find(abs(diff(A))==min(abs(diff(A))));
%extract this index, and it's neighbor index from A
val1 = A(minDiffInd);
val2 = A(minDiffInd+1);

How to find two closest (nearest) values within a vector in MATLAB?

My assumption is that the for the n nearest values, the only thing that matters is the difference between the v[i] - v[i - (n-1)]. That is, finding the minimum of diff(x, lag = n - 1L).

findClosest <- function(x, n) 
 x <- sort(x)
 x[seq.int(which.min(diff(x, lag = n - 1L)), length.out = n)]


findClosest(v, 3L)

[1] 23 25 26

Let's define "nearest numbers" by "numbers with minimal sum of L1 distances". You can achieve what you want by a combination of diff and windowed sum.

You could write a much shorter function but I wrote it step by step to make it easier to follow.

v <- c(10,23,25,26,38,50)

#' Find the n nearest numbers in a vector
#'
#' @param v Numeric vector
#' @param n Number of nearest numbers to extract
#'
#' @details "Nearest numbers" defined as the numbers which minimise the
#' within-group sum of L1 distances.
#' 
findClosest <- function(v, n) 
 # Sort and remove NA
 v <- sort(v, na.last = NA)

 # Compute L1 distances between closest points. We know each point is next to
 # its closest neighbour since we sorted.
 delta <- diff(v)

 # Compute sum of L1 distances on a rolling window with n - 1 elements
 # Why n-1 ? Because we are looking at deltas and 2 deltas ~ 3 elements.
 withingroup_distances <- zoo::rollsum(delta, k = n - 1)

 # Now it's simply finding the group with minimum within-group sum
 # And working out the elements
 group_index <- which.min(withingroup_distances)
 element_indices <- group_index + 0:(n-1)

 v[element_indices]


findClosest(v, 2)
# 25 26
findClosest(v, 3)
# 23 25 26

An idea is to use zoo library to do a rolling operation, i.e.

library(zoo)
m1 <- rollapply(v, 3, by = 1, function(i)c(sum(diff(i)), c(i)))
m1[which.min(m1[, 1]),][-1]
#[1] 23 25 26

Or make it into a function,

findClosest <- function(vec, n) 
 require(zoo)
 vec1 <- sort(vec)
 m1 <- rollapply(vec1, n, by = 1, function(i) c(sum(diff(i)), c(i)))
 return(m1[which.min(m1[, 1]),][-1])


findClosest(v, 3)
#[1] 23 25 26

A base R option, idea being we first sort the vector and subtract every ith element with i + n - 1 element in the sorted vector and select the group which has minimum difference.

closest_n_vectors <- function(v, n) 
 v1 <- sort(v)
 inds <- which.min(sapply(head(seq_along(v1), -(n - 1)), function(x) 
 v1[x + n -1] - v1[x]))
 v1[inds: (inds + n - 1)]


closest_n_vectors(v, 3)
#[1] 23 25 26

closest_n_vectors(c(2, 10, 1, 20, 4, 5, 23), 2)
#[1] 1 2

closest_n_vectors(c(19, 23, 45, 67, 89, 65, 1), 2)
#[1] 65 67

closest_n_vectors(c(19, 23, 45, 67, 89, 65, 1), 3)
#[1] 1 19 23

In case of tie this will return the numbers with smallest value since we are using which.min.

BENCHMARKS

Since we have got quite a few answers, it is worth doing a benchmark of all the solutions till now

set.seed(1234)
x <- sample(100000000, 100000)

identical(findClosest_antoine(x, 3), findClosest_Sotos(x, 3), 
 closest_n_vectors_Ronak(x, 3), findClosest_Cole(x, 3))
#[1] TRUE

microbenchmark::microbenchmark(
 antoine = findClosest_antoine(x, 3),
 Sotos = findClosest_Sotos(x, 3), 
 Ronak = closest_n_vectors_Ronak(x, 3),
 Cole = findClosest_Cole(x, 3),
 times = 10
)



#Unit: milliseconds
# expr min lq mean median uq max neval cld
#antoine 148.751 159.071 163.298 162.581 167.365 181.314 10 b 
# Sotos 1086.098 1349.762 1372.232 1398.211 1453.217 1553.945 10 c
# Ronak 54.248 56.870 78.886 83.129 94.748 100.299 10 a 
# Cole 4.958 5.042 6.202 6.047 7.386 7.915 10 a