R: Euclidian distances between objects in a group - r

Question

I want to create a matrix with similarities based on two identifiers, consider following matrix:
x1 <- c(2,2,2,3,1,2,4,6,4)
y1 <- c(5,4,3,3,4,2,1,6,3)
x2 <- c(8,2,7,3,1,2,2,2,6)
y2 <- c(1,3,3,3,1,2,4,3,8)
x3 <- c(4,4,1,2,4,6,3,2,9)
y3 <- c(1,2,3,3,1,2,4,6,1)
id1 <- c("a","a","a","a","b","b","b","b","b")
id2 <- c(2002,2002,2003,2003,2002,2002,2003,2003,2003)
dat <- data.frame(x1,y1,x2,y2,x3,y3,id1,id2)
For the groups marked by id1 and id2 I want to create the euclidean distance (sqrt((x1a-x1b)^2+(y1a-y1b)^2 + ... + (y3a-y3b)^2)) between the lines in the dataset. In the best case, there would be a new variable that indicates the distances of each line to each other line with the same id1 and id2. Please note that different numbers of members can be in each group as for instance in 2003 in the b-group there are three cases.
Any advice would be great!!!

Answer 1

I think it would be a good idea first to distinguish the lines whose distances you want to calculate. For example, for id1 == b and id2 == 2003 you have 3 lines, and you want to calculate 3 different distances (between each possible pair). So let's first assign each of these a unique id.
f <- function(n) {
# Returns a vector
# 1, 2, 1, 3, ..., 1, n, 2, 3, 2, 4, ..., 2, n, ..., (n-1), n
m <- matrix(ncol = 2, nrow = n * (n-1) / 2)
m[, 1] <- rep(1:(n-1), (n-1):1)
m[, 2] <- unlist(lapply(2:n, function(x) x:n))
as.numeric(t(m))
}
# Alternatively,
# f <- function(n) {
# d <- expand.grid(a = 1:n, b = 1:n)
# d <- d[d$a < d$b, ]
# unlist(d)
# }
# but this is slower
# Using plyr...
library(plyr)
dat <- ddply(dat, .(id1, id2), function(d) {
d <- d[f(nrow(d)), ]
d$id3 <- paste0(d$id1, rep(1:(nrow(d) / 2), each = 2))
d
})
# ...or using base R
dat <- do.call(rbind,
by(dat, list(dat$id1, dat$id2), function(d) {
d <- d[f(nrow(d)), ]
d$id3 <- paste0(d$id1, rep(1:(nrow(d) / 2), each = 2))
d
}))
Now there will only be two lines for each (id3, id2) pair and you can calculate the differences as follows
# Using plyr
result <- ddply(dat, .(id3, id2), function(d) {
d <- d[paste0(rep(c("x", "y"), 3), 1:3)]
d$dist <- sqrt(sum((d[1, ] - d[2, ])^2))
d
})
# Base R
result <- do.call(rbind,
by(dat[paste0(rep(c("x", "y"), 3), 1:3)],
list(dat$id3, dat$id2),
function(d){
d$dist <- sqrt(sum((d[1, ] - d[2, ])^2))
d
}
))
result[c("id3", "id2")] <- dat[c("id3", "id2")]
result
# x1 y2 x3 y1 x2 y3 dist id3 id2
# 1 2 1 4 5 8 1 6.480741 a1 2002
# 2 2 3 4 4 2 2 6.480741 a1 2002
# 5 1 1 4 4 1 1 3.464102 b1 2002
# 6 2 2 6 2 2 2 3.464102 b1 2002
# 3 2 3 1 3 7 3 4.242641 a1 2003
# 4 3 3 2 3 3 3 4.242641 a1 2003
# 7 4 4 3 1 2 4 5.916080 b1 2003
# 8 6 3 2 6 2 6 5.916080 b1 2003
# 7.1 4 4 3 1 2 4 9.000000 b2 2003
# 9 4 8 9 3 6 1 9.000000 b2 2003
# 8.1 6 3 2 6 2 6 11.313708 b3 2003
# 9.1 4 8 9 3 6 1 11.313708 b3 2003

Answer 2

Maybe this could be helpful.
dist(dat[which(dat[,"id1"]=="a" & dat[,"id2"]=="2002"),], method ="euclidean")
dist(dat[which(dat[,"id1"]=="b" & dat[,"id2"]=="2003"),], method ="euclidean")