Question

我有22个变量，我想得到相关分数，不是作为相关矩阵，而是在数据框中，通过成对...

我的意思是......不喜欢这个

    v1  v2  v3  v4
v1  1   x   x   x
v2  x   1   x   x
v3  x   x   1   x
v4  x   x   x   1

但是像这样：

var1  var2 cor
v1    v2   x
v1    v3   x
v1    v4   x
v2    v3   x
v2    v4   x
v3    v4   x

我是R的新手，我一直在研究很多，最后我得到了一个代码，真诚地，根本没有效率...我的代码创建了一个巨大的数据框，包含所有可能的组合22变量（4194304 combinatios ... 很多!!! ）...然后代码只分配前211行的相关性，这是仅有2个变量的组合......然后我排除了我不感兴趣的一切。嗯......我得到了我需要的东西。但我确信这是一种非常愚蠢的方式，我想学习一种更好的方法...... 有什么提示吗？

我的代码：

#Getting the variable names from the data frame
av_variables<-variable.names(data.1)

#Creating a huge data frame for all possible combinations
corr_combinations <- as.data.frame(matrix(1,0,length(av_variables)))
for (i in 1:length(av_variables)){
  corr_combinations.i <- t(combn(av_variables,i))
  corr_combinations.new <- as.data.frame(matrix(1,length(corr_combinations.i[,1]),length(av_variables)))
  corr_combinations.new[,1:i] <- corr_combinations.i
  corr_combinations <- rbind(corr_combinations,corr_combinations.new)

#How many combinations for 0:2 variables?
comb_par_var<-choose(20, k=0:2)
##211

#A new column to recieve the values
corr_combinations$cor <- 0


  #Getting the correlations and assigning to the empty column
 for (i in (length(av_variables)+1):(length(av_variables)+ sum(comb_par_var) +1)){
  print(i/length(corr_combinations[,1]))
  corr_combinations$cor[i] <- max(as.dist(abs(cor(data.1[,as.character(corr_combinations[i,which(corr_combinations[i,]!=0&corr_combinations[i,]!=1)])]))))
  # combinations$cor[i] <- max(as.dist(abs(cor(data.0[,as.character(combinations[i,combinations[i,]!=0&combinations[i,]!=1])]))))
  }

#Keeping only the rows with the combinations of 2 variables
corr_combinations[1:(length(av_variables)+ sum(comb_par_var) +2),21]
corr_combinations<-corr_combinations[1:212,]
corr_combinations<-corr_combinations[21:210,]

#Keeping only the columns var1, var2 and cor
corr_combinations<-corr_combinations[,c(1,2,21)]

#Ordering to keep only the pairs with correlation >0.95, 
#which was my purpose the whole time
corr_combinations <- corr_combinations[order(corr_combinations$cor),]
corr_combinations<-corr_combinations[corr_combinations$cor >0.95, ] 
}

Answer 1

您可以一次性计算完整的相关矩阵。那你就需要重塑一下。一个例子，

cr <- cor(mtcars)
# This is to remove redundancy as upper correlation matrix == lower 
cr[upper.tri(cr, diag=TRUE)] <- NA
reshape2::melt(cr, na.rm=TRUE, value.name="cor")

Answer 2

一个基本R替代方法是对与combn一起拉出的行/列名称使用矩阵子集。

# get pairwise combination of variable names
vars <- t(combn(colnames(myMat), 2))

# build data.frame with matrix subsetting
data.frame(vars, myMat[vars])
  X1 X2 myMat.vars.
1 V1 V2   0.8500071
2 V1 V3  -0.2828288
3 V1 V4  -0.2867921
4 V2 V3  -0.2698210
5 V2 V4  -0.2273411
6 V3 V4   0.9962044

您也可以使用setNames在一行中添加列名。

setNames(data.frame(vars, myMat[vars]), c("var1", "var2", "corr"))

数据

set.seed(1234) myMat <- cor(matrix(rnorm(16), 4, dimnames=list(paste0("V", 1:4), paste0("V", 1:4)))) myMat V1 V2 V3 V4 V1 1.0000000 0.8500071 -0.2828288 -0.2867921 V2 0.8500071 1.0000000 -0.2698210 -0.2273411 V3 -0.2828288 -0.2698210 1.0000000 0.9962044 V4 -0.2867921 -0.2273411 0.9962044 1.0000000

Answer 3

您可以使用tidyr重塑相关矩阵。

首先，create a correlation matrix：

> d <- data.frame(x1=rnorm(10),
+                 x2=rnorm(10),
+                 x3=rnorm(10))
> x <- cor(d) # get correlations (returns matrix)
> x
           x1         x2         x3
x1  1.0000000  0.3096685 -0.5358578
x2  0.3096685  1.0000000 -0.7497212
x3 -0.5358578 -0.7497212  1.0000000

然后，使用tidyr重塑：

> y <- as.data.frame(x)
> y$var1 <- row.names(y)
> library(tidyr)
> gather(data = y, key = "var2", value = "correlation", -var1)
  var1 var2 correlation
1   x1   x1   1.0000000
2   x2   x1   0.3096685
3   x3   x1  -0.5358578
4   x1   x2   0.3096685
5   x2   x2   1.0000000
6   x3   x2  -0.7497212
7   x1   x3  -0.5358578
8   x2   x3  -0.7497212
9   x3   x3   1.0000000

组合对的相关性

3 个答案: