来自两个独立数据帧的距离矩阵

Question

I'd like to create a matrix which contains the euclidean distances of the rows from one data frame versus the rows from another. For example, say I have the following data frames:

a <- c(1,2,3,4,5)
b <- c(5,4,3,2,1)
c <- c(5,4,1,2,3)
df1 <- data.frame(a,b,c)

a2 <- c(2,7,1,2,3)
b2 <- c(7,6,5,4,3)
c2 <- c(1,2,3,4,5)
df2 <- data.frame(a2,b2,c2)

I would like to create a matrix with the distances of each row in df1 versus the rows of df2.

So matrix[2,1] should be the euclidean distance between df1[2,] and df2[1,]. matrix[3,2] the distance between df[3,] and df2[2,], etc.

Does anyone know how this could be achieved?

Answer 1

Perhaps you could use the fields package: the function rdist might do what you want:

rdist : Euclidean distance matrix
Description: Given two sets of locations computes the Euclidean distance matrix among all pairings.

> rdist(df1, df2)
     [,1]     [,2]     [,3]     [,4]     [,5]
[1,] 4.582576 6.782330 2.000000 1.732051 2.828427
[2,] 4.242641 5.744563 1.732051 0.000000 1.732051
[3,] 4.123106 5.099020 3.464102 3.316625 4.000000
[4,] 5.477226 5.000000 4.358899 3.464102 3.316625
[5,] 7.000000 5.477226 5.656854 4.358899 3.464102

Similar is the case with the pdist package

pdist : Distances between Observations for a Partitioned Matrix
Description: Computes the euclidean distance between rows of a matrix X and rows of another matrix Y.

> pdist(df1, df2)
An object of class "pdist"
Slot "dist":
[1] 4.582576 6.782330 2.000000 1.732051 2.828427 4.242640 5.744563 1.732051
[9] 0.000000 1.732051 4.123106 5.099020 3.464102 3.316625 4.000000 5.477226
[17] 5.000000 4.358899 3.464102 3.316625 7.000000 5.477226 5.656854 4.358899
[25] 3.464102
attr(,"Csingle")
[1] TRUE

Slot "n":
[1] 5

Slot "p":
[1] 5

Slot ".S3Class":
[1] "pdist"

#

NOTE: If you're looking for the Euclidean norm between rows, you might want to try:

> rdist(df1, df2)
         [,1]     [,2]     [,3]
[1,] 6.164414 7.745967 0.000000
[2,] 5.099020 4.472136 6.324555
[3,] 4.242641 5.291503 5.656854

This gives:

 > rdist(df1, df2) [,1] [,2] [,3] [1,] 6.164414 7.745967 0.000000 [2,] 5.099020 4.472136 6.324555 [3,] 4.242641 5.291503 5.656854 

Answer 2

This is adapted from my previous answer here .

For general n -dimensional Euclidean distance, we can exploit the equation (not R, but algebra):

square_dist(b,a) = sum_i(b[i]*b[i]) + sum_i(a[i]*a[i]) - 2*inner_prod(b,a)

where the sums are over the dimensions of vectors a and b for i=[1,n] . Here, a and b are one pair of columns from df1 and df2 , respectively. The key here is that this equation can be written as a matrix equation for all pairs in df1 and df2 .

In code:

d <- sqrt(matrix(rowSums(expand.grid(rowSums(df1*df1),rowSums(df2*df2))),
                 nrow=nrow(df1)) - 
          2. * as.matrix(df1) %*% t(as.matrix(df2)))

Notes:

1. The inner rowSums compute sum_i(a[i]*a[i]) and sum_i(b[i]*b[i]) for each a in df1 and b in df2 , respectively.
2. expand.grid then generates all pairs between df1 and df2 .
3. The outer rowSums computes the sum_i(a[i]*a[i]) + sum_i(b[i]*b[i]) for all these pairs.
4. This result is then reshaped into a matrix . Note that the number of rows of this matrix is the number of rows of df1 .
5. Then subtract two times the inner product of all pairs. This inner product can be written as a matrix multiply df1 %*% t(df2) where I left out the coercion to matrix for clarity.
6. Finally, take the square root.

Using this code with your data:

print(d)
##         [,1]     [,2]     [,3]     [,4]     [,5]
##[1,] 4.582576 6.782330 2.000000 1.732051 2.828427
##[2,] 4.242641 5.744563 1.732051 0.000000 1.732051
##[3,] 4.123106 5.099020 3.464102 3.316625 4.000000
##[4,] 5.477226 5.000000 4.358899 3.464102 3.316625
##[5,] 7.000000 5.477226 5.656854 4.358899 3.464102

Note that this code will work for any n > 1 . In your case, n=3 .