计算数据帧的每一行与另一个数据帧中的所有其他行之间的欧几里德距离

Question

I need to generate a dataframe with minimum euclidean distance between each row of a dataframe and all other rows of another dataframe.Both my dataframes are large (approx 40,000 rows).This is what I could work out till now.

x<-matrix(c(3,6,3,4,8),nrow=5,ncol=7,byrow = TRUE)     
y<-matrix(c(1,4,4,1,9),nrow=5,ncol=7,byrow = TRUE)


sed.dist<-numeric(5)
for (i in 1:(length(sed.dist))) {
sed.dist[i]<-(sqrt(sum((y[i,1:7] - x[i,1:7])^2)))
}

But this only works when i=j.What I essentially need is to find the minimum euclidean distance by looping over every row one by one ( y[1,1:7],then y[2,1:7] and so on till i= 5 ) of the "y" dataframe with all the rows of the "x"dataframe(x[i,1:7]).Each time it does this,I need it to find the minimum euclidean distance for each computation of row i of the y dataframe and all the rows of the x dataframe and store it in another dataframe.

Answer 1

Try this:

apply(y,1,function(y) min(apply(x,1,function(x,y)dist(rbind(x,y)),y)))
# [1] 5.196152 5.385165 4.898979 4.898979 5.385165

Working from the inside out, we bind a row of x to a row of y and calcualte the distance between them usin the dist(...) function (written in C). We do this for a given row of y with each row of x in turn, using the inner apply(...) , and then find the minimum of the result. Then we do this for each row of y in the outer call to apply(...) .

Answer 2

Expanding on my comment on the question, a pretty fast approach would be the following, although with 40,000 rows you'll have to wait a bit, I guess:

unlist(lapply(seq_len(nrow(y)), function(i) min(sqrt(colSums((y[i, ] - t(x))^2)))))
#[1] 5.196152 5.385165 4.898979 4.898979 5.385165

And a comparing benchmarking:

x = matrix(runif(1e2*5), 1e2)
y = matrix(runif(1e2*5), 1e2)
library(microbenchmark)
alex = function() unlist(lapply(seq_len(nrow(y)), 
                           function(i) min(sqrt(colSums((y[i, ] - t(x))^2)))))
jlhoward = function() apply(y,1,function(y)
                                  min(apply(x,1,function(x,y)dist(rbind(x,y)),y)))
all.equal(alex(), jlhoward())
#[1] TRUE
microbenchmark(alex(), jlhoward(), times = 20)
#Unit: milliseconds
#       expr        min         lq     median         uq        max neval
#     alex()   3.369188   3.479011   3.600354   4.513114   4.789592    20
# jlhoward() 422.198621 431.565643 436.561057 442.643181 602.929742    20