[英]Pairwise distance two matrix R
I want to run distance bewteen matrix in R. In this example I use manhattan distance but I would like to apply other formuls. 我想在R中的矩阵之间运行距离。在此示例中,我使用曼哈顿距离,但我想应用其他公式。 My question is , is there one way to apply function to row of a matrix by row of other matrix in R?
我的问题是,是否有一种方法可以将函数逐行应用于R中的其他矩阵?
In this example I have only two variables but I would like to apply with more than 10 vars. 在此示例中,我只有两个变量,但是我想应用10个以上的变量。
Thanks. 谢谢。
set.seed(123)
mat1 <- data.frame(x=sample(1:10000,3),
z=sample(1:10000,3))
mat2 <- data.frame(x=sample(1:100,3),
z=sample(1:1000,3))
dista<-matrix(0,ncol=2,nrow=2)
for (j in 1:nrow(mat1)){
for(i in 1:nrow(mat2)){
dista[i,j]<-sqrt((mat1[i,1]-mat2[j,1]) + (mat1[i,2]-mat2[j,2]))
}
}
dista
You can use the proxy
package for these problems. 您可以使用
proxy
程序包来解决这些问题。 By default, proxy::dist
considers each row of a matrix or data frame as a single "object". 默认情况下,
proxy::dist
将矩阵或数据帧的每一行视为一个“对象”。
library(proxy)
proxy::dist(mat1, mat2, method="Manhattan")
[,1] [,2] [,3]
[1,] 4804 4832 4656
[2,] 3708 3736 3560
[3,] 17407 17435 17259
proxy::dist(mat1, mat2, method="Euclidean")
[,1] [,2] [,3]
[1,] 3397.036 3417.059 3295.962
[2,] 2761.996 2787.495 2708.075
[3,] 12308.674 12328.422 12204.286
Type vignette("overview", "proxy")
in the R console to see which similarities and distances it includes, and check the documentation of proxy::pr_DB
if you would like to add your own functions that can be used with proxy::dist
. 在R控制台中键入
vignette("overview", "proxy")
以查看其包含哪些相似性和距离,并查看proxy::pr_DB
的文档(如果要添加可与proxy::dist
一起使用的函数) proxy::dist
。
Your code is still wrong as the output should be of length nrow(mat1) * nrow(mat2)
which is 9
, and that cannot fit in a 2x2 matrix (which you previously define). 您的代码仍然是错误的,因为输出的长度应为
nrow(mat1) * nrow(mat2)
,其长度为9
,并且不能适合2x2矩阵(您先前定义)。 Also, the i
should run through mat1
and j
through mat2
; 另外,
i
应该穿过mat1
, j
应该穿过mat2
; you have it the other way around. 反之亦然。 Changing
dista[i,j] <-
for a print()
, you'd obtain: 为
print()
更改dista[i,j] <-
,您将获得:
dista<-matrix(0,ncol=2,nrow=2)
for (i in 1:nrow(mat1)){
for(j in 1:nrow(mat2)){
print(sqrt((mat1[i,1]-mat2[j,1]) + (mat1[i,2]-mat2[j,2])))
}
}
[1] 105.8159
[1] 129.5261
[1] 63.52165
[1] 103.257
[1] 127.4441
[1] 59.1608
[1] 105.8253
[1] 129.5338
[1] 63.53739
You can use outer
to limit the calculations to just one vectorised function 您可以使用
outer
将计算限制为仅一个矢量化函数
y = outer(1:nrow(mat1),1:nrow(mat2),paste)
y
[,1] [,2] [,3]
[1,] "1 1" "1 2" "1 3"
[2,] "2 1" "2 2" "2 3"
[3,] "3 1" "3 2" "3 3"
sapply(as.vector(y), function(x){
aux = as.numeric(strsplit(x," ")[[1]])
sqrt((mat1[aux[1],1]-mat2[aux[2],1]) + (mat1[aux[1],2]-mat2[aux[2],2]))})
1 1 2 1 3 1 1 2 2 2 3 2 1 3 2 3 3 3
105.81588 129.52606 63.52165 103.25696 127.44411 59.16080 105.82533 129.53378 63.53739
Here, we first create a y
matrix which contains all the i
and j
combinations, feed it to sapply
and then split it to get i
and j
individually. 在这里,我们首先创建一个包含所有
i
和j
组合的y
矩阵,将其馈给sapply
,然后将其拆分以分别获得i
和j
。
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