如何根据行名和列名的比较构建方阵答案

【问题标题】：How to build a square matrix based on a comparison of row and column names如何根据行名和列名的比较构建方阵
【发布时间】：2016-11-07 05:31:34
【问题描述】：

我的矩阵中的列和行的名称与组相关，这些组与其他组具有不同的关系。我想创建一个矩阵，其中的值基于行名和列名以及对应关系。

我已经创建了一种低效的方法来执行此操作，这对于小型 3*3 矩阵来说是可以的，但对于大型矩阵来说并不实用。

我的例子数据如下：

tom <- data.frame("w"=c(7,1,2),"x"=c(2,4,4),"y"=c(12,4,8))
row.names(tom) <- colnames(tom)

same <- data.frame("trait"=c("w","x","y"),
                   "group"=c(1,2,2),
                   "own_group_relationship"=c(0.86,0.55,0.55))

diff <- data.frame("trait"=c("w","x","y"),
                    "diff_group_relationship"=c(0.23,0.23,0.23))

我的陈述是这样的：

a1 <- if ( row.names(tom[c(1),]) == colnames(tom[c(1)]) ) {
  merge(data.frame("trait" = row.names(tom[c(1),])),same)[1,3]
} else {
  merge(data.frame("trait" = row.names(tom[c(1),])),diff)[1,2]
}

这目前一次只适用于一个元素。我不得不再重复这段代码 8 次，更改相应的行名以获得 9 个值（a1 到 a9）。

a2 <- if ( row.names(tom[c(1),]) == colnames(tom[c(2)]) ) {merge(data.frame("trait" = row.names(tom[c(1),])),same)[1,3]} else {merge(data.frame("trait" = row.names(tom[c(1),])),diff)[1,2]}   
a3 <- if ( row.names(tom[c(1),]) == colnames(tom[c(3)]) ) {merge(data.frame("trait" = row.names(tom[c(1),])),same)[1,3]} else {merge(data.frame("trait" = row.names(tom[c(1),])),diff)[1,2]}   
a4 <- if ( row.names(tom[c(2),]) == colnames(tom[c(1)]) ) {merge(data.frame("trait" = row.names(tom[c(1),])),same)[1,3]} else {merge(data.frame("trait" = row.names(tom[c(1),])),diff)[1,2]}   
a5 <- if ( row.names(tom[c(2),]) == colnames(tom[c(2)]) ) {merge(data.frame("trait" = row.names(tom[c(1),])),same)[1,3]} else {merge(data.frame("trait" = row.names(tom[c(1),])),diff)[1,2]}   
a6 <- if ( row.names(tom[c(2),]) == colnames(tom[c(3)]) ) {merge(data.frame("trait" = row.names(tom[c(1),])),same)[1,3]} else {merge(data.frame("trait" = row.names(tom[c(1),])),diff)[1,2]}   
a7 <- if ( row.names(tom[c(3),]) == colnames(tom[c(1)]) ) {merge(data.frame("trait" = row.names(tom[c(1),])),same)[1,3]} else {merge(data.frame("trait" = row.names(tom[c(1),])),diff)[1,2]}   
a8 <- if ( row.names(tom[c(3),]) == colnames(tom[c(2)]) ) {merge(data.frame("trait" = row.names(tom[c(1),])),same)[1,3]} else {merge(data.frame("trait" = row.names(tom[c(1),])),diff)[1,2]}   
a9 <- if ( row.names(tom[c(3),]) == colnames(tom[c(3)]) ) {merge(data.frame("trait" = row.names(tom[c(1),])),same)[1,3]} else {merge(data.frame("trait" = row.names(tom[c(1),])),diff)[1,2]}

这 9 个值很容易转换为 3*3 矩阵，但必须有更优雅的解决方案。

vec <- c(a1,a2,a3,a4,a5,a6,a7,a8,a9)
mtrx <- matrix(vec, nrow=3, ncol=3)
mtrx # the resulting matrix of group inter group relationships
     [,1] [,2] [,3]
[1,] 0.86 0.23 0.23
[2,] 0.23 0.86 0.23
[3,] 0.23 0.23 0.86

【问题讨论】：

你不能用vec <- matrix(NA, 3, 3); diag(vec) <- "condition A"; vec[row(vec) != col(vec)] <- "condition B"之类的东西吗？类似于this question

标签： r matrix

【解决方案1】：

这是使用outer 和match 的解决方案：

outer(rownames(tom), colnames(tom),
      FUN=function(x, y) {
                          (x==y) * same$own_group_relationship[match(x, same$trait)] +
                          (x!=y) * diff$diff_group_relationship[match(x, diff$trait)]
      })

     [,1] [,2] [,3]
[1,] 0.86 0.23 0.23
[2,] 0.23 0.55 0.23
[3,] 0.23 0.23 0.55

内部函数生成一个向量，该向量将根据行名和列名是否匹配来提取正确的值。 outer 函数返回正确的位置和尺寸。

【讨论】：