如何在R corrplot中返回有效的匹配? [英] How to return significant matches in R corrplot?

问题描述

我想从图1所示的以下结果中返回重要匹配项

I would like to return the significant matches from the following result shown in Fig. 1

library("corrplot")

M <- cor(mtcars)

# http://www.sthda.com/english/wiki/visualize-correlation-matrix-using-correlogram
cor.mtest <- function(mat, ...) {
    mat <- as.matrix(mat)
    n <- ncol(mat)
    p.mat<- matrix(NA, n, n)
    diag(p.mat) <- 0
    for (i in 1:(n - 1)) {
        for (j in (i + 1):n) {
            tmp <- cor.test(mat[, i], mat[, j], ...)
            p.mat[i, j] <- p.mat[j, i] <- tmp$p.value
        }
    }
  colnames(p.mat) <- rownames(p.mat) <- colnames(mat) 
  p.mat
}

N <- length(mtcars) -1
p.mat <- cor.mtest(mtcars)
head(p.mat[, 1:N])
ids <- seq(1,N) 

corrplot(M, type="upper", order="hclust", tl.pos=c("td"), method="circle",  
    tl.cex = 0.5, tl.col = 'black', 
    diag = FALSE, p.mat = p.mat, 
    sig.level = 0.0000005)

图. 1个输出

预期产量

cyl: wt hp
disp: wt hp cyl
...

R:3.3.1
操作系统:Debian 8.5

R: 3.3.1
OS: Debian 8.5

推荐答案

此处涉及多次测试的常见警告.

The usual warnings regarding multiple testing apply here.

我会写一个向量化的cor.test(有一个原因不能从stats包中获得，请参见上文).

I would write a vectorized cor.test (there is a reason that is not available from the stats package, see above).

cor.test.all <- function(DF) {
  #based on code from stats:::cor.test.default
  #see license() for the license 
  #two-sided test for pearson correlation 
  #without adjustment of p-values
  #no Na treatment
  r <- cor(DF)
  df <- nrow(DF) - 2L
  t <-  sqrt(df) * r/sqrt(1 - r^2)
  2 * pmin(pt(t, df), pt(t,  df, lower.tail = FALSE))
}

cor.p <- cor.test.all(mtcars)
diag(cor.p) <- NA
res <- which(cor.p < 0.0000005, arr.ind = TRUE)

split(colnames(cor.p)[res[,2]], rownames(cor.p)[res[,1]])
#$am
#[1] "gear"
#
#$cyl
#[1] "mpg"  "disp" "hp"   "wt"   "vs"  
#
#$disp
#[1] "mpg" "cyl" "hp"  "wt" 
#
#$gear
#[1] "am"
#
#$hp
#[1] "mpg"  "cyl"  "disp"
#
#$mpg
#[1] "cyl"  "disp" "hp"   "wt"  
#
#$vs
#[1] "cyl"
#
#$wt
#[1] "mpg"  "cyl"  "disp"

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