how to find the columns with the most similar values - r

Question

Just starting to use R and am feeling a bit confused. Suppose I have three columns
data = data.frame(id=c(101, 102, 103),column1=c(2, 4, 9),
column2=c(3, 4, 2), column3=c(5, 15, 7))
How can I create a new column (e.g., colmean) that is the mean of the two columns closest in value? I thought about doing a bunch of ifelse statements, but that seemed unnecessarily messy.
In this case, for instance, colmean=c(2.5, 4, 8).

Answer 1

Borrowing the function findClosest() created here by #Cole, we can do the following,
findClosest <- function(x, n) {
x <- sort(x)
x[seq.int(which.min(diff(x, lag = n - 1L)), length.out = n)]
}
colMeans(apply(data[-1], 1, function(i)findClosest(i, 2)))
#[1] 2.5 4.0 8.0

Answer 2

A vectorized function using the Rfast package:
library(Rfast)
fClosest <- function(m, n) {
m <- colSort(t(m))
matrix(
m[
sequence(
rep(n, ncol(m)),
seq(0, nrow(m)*(ncol(m) - 1), nrow(m)) + colMins(diff(m, lag = n - 1))
)
],
ncol(m), n, TRUE
)
}
m <- matrix(sample(10, 24, 1), 4)
m
#> [,1] [,2] [,3] [,4] [,5] [,6]
#> [1,] 4 2 6 2 5 3
#> [2,] 3 4 7 3 4 7
#> [3,] 4 2 7 6 10 2
#> [4,] 8 1 10 8 2 9
fClosest(m, 3L)
#> [,1] [,2] [,3]
#> [1,] 2 2 3
#> [2,] 3 3 4
#> [3,] 2 2 4
#> [4,] 8 8 9
rowMeans(fClosest(m, 3L))
#> [1] 2.333333 3.333333 2.666667 8.333333

Answer 3

Here is a version with a loop:
data = data.frame(id=c(101, 102, 103),column1=c(2, 4, 9),
column2=c(3, 4, 2), column3=c(5, 15, 7))
data$colmean <- NaN # set up empty column for results
for(i in seq(nrow(data))){
data.i <- data[i,-1] # get ith row
d <- as.matrix(dist(c(data.i))) # get distances between values
diag(d) <- NaN # replace diagonal of distance matrix with NaN
hit <- which.min(d) # identify value of lowest distance
pos <- c(row(d)[hit], col(d)[hit]) # get the position (i.e. the values that are closest)
data$colmean[i] <- mean(unlist(data.i[pos])) # calculate mean
}
data
# id column1 column2 column3 colmean
# 1 101 2 3 5 2.5
# 2 102 4 4 15 4.0
# 3 103 9 2 7 8.0

Answer 4

Here's a self-contained solution, based on the tidyverse, that is independent of the number of columns to be compared.
library(tidyverse)
data %>%
# Add the means of smallest pairwise differences to the input data
bind_cols(
data %>%
# Make the data tidy (and hence independent of the number of "column"s)
pivot_longer(starts_with("column")) %>%
# For each id/row (replace with rowwise() if appropriate)
group_by(id) %>%
group_map(
function(.x, .y) {
# Form a tibble of all pairwise ciombinations of values
as_tibble(t(combn(.x$value, 2))) %>%
# Calculate pairwise differences
mutate(difference = abs(V1 - V2)) %>%
# Find the smallest pairwise difference
arrange(difference) %>%
head(1) %>%
# Calculate the mean of this pair
pivot_longer(starts_with("V")) %>%
summarise(colmean=mean(value))
}
) %>%
# Convert list of values to column
bind_rows()
)
id column1 column2 column3 colmean
1 101 2 3 5 2.5
2 102 4 4 15 4.0
3 103 9 2 7 8.0