R:在矩阵中成对比较所有列

Question

我有一个41行和6列的矩阵.这就是第一部分的样子.

      X13  X15  X17  X19  X21  X23 
 [1,] "7"  "6"  "5"  "8"  "1"  "8" 
 [2,] "7"  "6"  "5"  "8"  "14" "3" 
 [3,] "7"  "6"  "1"  "3"  "12" "3" 
 [4,] "7"  "6"  "1"  "5"  "6"  "14"
 [5,] "2"  "6"  "1"  "5"  "16" "3" 
 [6,] "2"  "3"  "5"  "5"  "2"  "3" 
 [7,] "7"  "5"  "5"  "17" "7"  "3" 
 [8,] "7"  "2"  "5"  "2"  "2"  "14"
 [9,] "2"  "2"  "10" "10" "2"  "3" 
[10,] "2"  "2"  "10" "5"  "2"  "6" 

我的目标是,将所有列彼此进行比较,并查看2列中有多少数字相等.我试着这样做:

s <- sum(matrix[,1]==matrix[,2])

但由于我需要比较所有可能的对,因此无效.把它放在一个循环中会很好,但我不知道如何.

我想以6x6相似度矩阵的形式得到我的结果.像这样的东西:

      X13 X15 X17 X19 X21 X23
 X13   0   0   3   2   2   3
 X15   0   0   9  11   4   6
 X17   3   9   0   5   1   3
 X19   2  11   5   0   9  10
 X21   2   4   1   9   0   9
 X23   3   6   3  10   9   0

如您所见,我希望在将列与iteslf进行比较时将零填充到矩阵中.

由于我是初学R用户,这个任务semms对我来说真的很复杂.我需要将这个比较用于50个矩阵,所以如果你能帮助我,我会很高兴.我将不胜感激任何提示/建议.我的英语也不太好,但我希望我能够很好地解释我的问题.:)

Answer 1

下面是使用一种完全矢量化溶液expand.grid来计算指数和colSums并matrix包裹起来的结果.

#  Some reproducible 6x6 sample data
set.seed(1)
m <- matrix( sample(10,36,repl=TRUE) , ncol = 6 )
#     [,1] [,2] [,3] [,4] [,5] [,6]
#[1,]    3   10    7    4    3    5
#[2,]    4    7    4    8    4    6
#[3,]    6    7    8   10    1    5
#[4,]   10    1    5    3    4    2
#[5,]    3    3    8    7    9    9
#[6,]    9    2   10    2    4    7


#  Vector source for column combinations
n <- seq_len( ncol(m) )

#  Make combinations
id <- expand.grid( n , n )

#  Get result
out <- matrix( colSums( m[ , id[,1] ] == m[ , id[,2] ] ) , ncol = length(n) )
diag(out) <- 0
#    [,1] [,2] [,3] [,4] [,5] [,6]
#[1,]    0    1    1    0    2    0
#[2,]    1    0    0    1    0    0
#[3,]    1    0    0    0    1    0
#[4,]    0    1    0    0    0    0
#[5,]    2    0    1    0    0    1
#[6,]    0    0    0    0    1    0

Answer 2

这是一种非矢量化(但可能更有内存效率)的方法:

# Fancy way.
similarity.matrix<-apply(matrix,2,function(x)colSums(x==matrix))
diag(similarity.matrix)<-0


# More understandable. But verbose.
similarity.matrix<-matrix(nrow=ncol(matrix),ncol=ncol(matrix))
for(col in 1:ncol(matrix)){
  matches<-matrix[,col]==matrix
  match.counts<-colSums(matches)
  match.counts[col]<-0 # Set the same column comparison to zero.
  similarity.matrix[,col]<-match.counts
}