Finding the closest value in a vector, for each value in the same vector, excluding the element in question in R?

Question

I am trying to find a way to find the closest value in a vector, from elements in the SAME vector, but excluding the row in question. For example, suppose I have dataframe A with one column (column_1):

column_1
1
5
6
2
3
0
5
2
1
9

I want to add a second column which, for every element in column_1 finds the closest value in THAT SAME vector excluding the row in question. Desired output is below:

column_1    column_2
1           1
5           5
6           5
2           2
3           2
0           1
5           5
2           2
1           1
9           6

I have seen people discuss how to do this where the closest value for each element in a vector (a) is identified from another vector (b) via the following:

which(abs(a-b)==min(a-b))

Does anyone know how to modify the above, or do this in some other way, so that I can look within the same vector and exclude the row in question (example: the third row in column_1 is closest to 5 not 6, since I exclude its own row from the search vector. However, the fourth row in column_1 is closest to 2 since even when excluding the fourth row, there is another 2 value in the 8th row)

Answer 1

You could sort the vector, then check only the number before and number after, and replace the original vector:

y <- sort(x)
z <- c(-Inf, y, Inf)
b <- cbind(head(z, -2), tail(z, -2)) 
x[order(x)] <- b[cbind(seq_along(y), max.col(-abs(b - y)))]
x
[1] 1 5 5 2 2 1 5 2 1 6

---

Notice that this method has the lowest complexity as compared to the above solutions. Ie it is the fastest:

onyambu <- function(x){
  y <- sort(x)
  z <- c(-Inf, y, Inf)
  b <- cbind(head(z, -2), tail(z, -2)) 
  x[order(x)] <- b[cbind(seq_along(y), max.col(-abs(b - y)))]
  x
}

gregor <- function(x){
  dist = outer(x, x, FUN = \(a, b) abs(a - b))
  diag(dist) = Inf
  mins = max.col(-dist, ties.method = 'first')
  x[mins]
}
x <- rnorm(1000)

microbenchmark::microbenchmark(onyambu(x), gregor(x), check = 'equal', times = 1)
Unit: milliseconds
       expr       min        lq      mean    median        uq       max neval
 onyambu(x)    5.7839    5.7839    5.7839    5.7839    5.7839    5.7839     1
  gregor(x) 3021.9425 3021.9425 3021.9425 3021.9425 3021.9425 3021.9425     1

Answer 2

# sample data
x = c(1, 5, 6, 2, 3, 0, 5, 2, 1, 9)

# make a distance matrix and set diagonal to Inf
dist = outer(x, x, FUN = \(a, b) abs(a - b))
diag(dist) = Inf

# find the index of the min value on each row
#   (which is the index of the max negative value
#    so we can use the convenient max.col)
mins = max.col(-dist)

# show the result
y = x[mins]
cbind(x, y)
#       x y
#  [1,] 1 1
#  [2,] 5 5
#  [3,] 6 5
#  [4,] 2 2
#  [5,] 3 2
#  [6,] 0 1
#  [7,] 5 5
#  [8,] 2 2
#  [9,] 1 1
# [10,] 9 6

Answer 3

Here is a way. Compute all distances from each element of the column to all other elements. To set the diagonal element to Inf is a trick to avoid index arithmetic. Then return the index to the minimum distance. This is used to assign the wanted values.

df1 <- 
structure(list(
  column_1 = c(1, 5, 6, 2, 3, 0, 5, 2, 1, 9)), 
  class = "data.frame", row.names = c(NA, -10L))

i <- sapply(seq_along(df1$column_1), \(i) {
  d <- abs(df1$column_1[i] - df1$column_1)
  d[i] <- Inf
  which.min(d)
})
df1$column_2 <- df1$column_1[i]
df1
#>    column_1 column_2
#> 1         1        1
#> 2         5        5
#> 3         6        5
#> 4         2        2
#> 5         3        2
#> 6         0        1
#> 7         5        5
#> 8         2        2
#> 9         1        1
#> 10        9        6

^{Created on 2022-11-14 with reprex v2.0.2}

Answer 4

Using the get.knn function from the FNN package:

x <- c(1, 5, 6, 2, 3, 0, 5, 2, 1, 9)
cbind(x, x[FNN::get.knn(x, 1)$nn.index])
#>       x  
#>  [1,] 1 1
#>  [2,] 5 5
#>  [3,] 6 5
#>  [4,] 2 2
#>  [5,] 3 2
#>  [6,] 0 1
#>  [7,] 5 5
#>  [8,] 2 2
#>  [9,] 1 1
#> [10,] 9 6

Answer 5

We can use dist + max.col to locate the closest neighbor

> x <- c(1, 5, 6, 2, 3, 0, 5, 2, 1, 9)

> cbind(x, y = x[max.col(-`diag<-`(as.matrix(dist(x)), Inf))])
      x y
 [1,] 1 1
 [2,] 5 5
 [3,] 6 5
 [4,] 2 2
 [5,] 3 2
 [6,] 0 1
 [7,] 5 5
 [8,] 2 2
 [9,] 1 1
[10,] 9 6