Question

I have many X and Y variables (something like, 500 x 500). The following just small data:

yvars <- data.frame (Yv1 = rnorm(100, 5, 3), Y2 = rnorm (100, 6, 4),
  Yv3 = rnorm (100, 14, 3))
xvars <- data.frame (Xv1 = sample (c(1,0, -1), 100, replace = T),
 X2 = sample (c(1,0, -1), 100, replace = T), 
 Xv3 = sample (c(1,0, -1), 100, replace = T), 
 D = sample (c(1,0, -1), 100, replace = T))

I want to extact p-values and make a matrix like this:

     Yv1    Y2     Yv3   
Xv1
X2
Xv3
D

Here is my attempt to loop the process:

prob = NULL
   anova.pmat <- function (x) {
            mydata <- data.frame(yvar = yvars[, x], xvars)
            for (i in seq(length(xvars))) {
              prob[[i]] <- anova(lm(yvar ~ mydata[, i + 1],
              data = mydata))$`Pr(>F)`[1]
              }
              }
    sapply (yvars,anova.pmat)
    Error in .subset(x, j) : only 0's may be mixed with negative subscripts
What could be the solution ?

Edit:

For the first Y variable:

For first Y variable:

prob <- NULL
 mydata <- data.frame(yvar = yvars[, 1], xvars)
for (i in seq(length(xvars))) {
              prob[[i]] <- anova(lm(yvar ~ mydata[, i + 1],
              data = mydata))$`Pr(>F)`[1]
              }

prob 
[1] 0.4995179 0.4067040 0.4181571 0.6291167

Edit again:

for (j in seq(length (yvars))){
        prob <- NULL
        mydata <- data.frame(yvar = yvars[, j], xvars)
         for (i in seq(length(xvars))) {
                  prob[[i]] <- anova(lm(yvar ~ mydata[, i + 1],
                  data = mydata))$`Pr(>F)`[1]
                  }
}

Gives the same result as above !!!
Was it helpful?

Solution

Here is an approach that uses plyr to loop over the columns of a dataframe (treating it as a list) for each of the xvars and yvars, returning the appropriate p-value, arranging it into a matrix. Adding the row/column names is just extra.

library("plyr")

probs <- laply(xvars, function(x) {
    laply(yvars, function(y) {
        anova(lm(y~x))$`Pr(>F)`[1]
    })
})
rownames(probs) <- names(xvars)
colnames(probs) <- names(yvars)

OTHER TIPS

Here is one solution, which consists in generating all combinations of Y- and X-variables to test (we cannot use combn) and run a linear model in each case:

dfrm <- data.frame(y=gl(ncol(yvars), ncol(xvars), labels=names(yvars)),
                   x=gl(ncol(xvars), 1, labels=names(xvars)), pval=NA)
## little helper function to create formula on the fly
fm <- function(x) as.formula(paste(unlist(x), collapse="~"))
## merge both datasets
full.df <- cbind.data.frame(yvars, xvars)
## apply our LM row-wise
dfrm$pval <- apply(dfrm[,1:2], 1, 
                   function(x) anova(lm(fm(x), full.df))$`Pr(>F)`[1])
## arrange everything in a rectangular matrix of p-values
res <- matrix(dfrm$pval, nc=3, dimnames=list(levels(dfrm$x), levels(dfrm$y)))

Sidenote: With high-dimensional datasets, relying on the QR decomposition to compute the p-value of a linear regression is time-consuming. It is easier to compute the matrix of Pearson linear correlation for each pairwise comparisons, and transform the r statistic into a Fisher-Snedecor F using the relation F = νar2/(1-r2), where degrees of freedom are defined as νa=(n-2)-#{(xi=NA),(yi=NA)} (that is, (n-2) minus the number of pairwise missing values--if there're no missing values, this formula is the usual coefficient R2 in regression).

Licensed under: CC-BY-SA with attribution
Not affiliated with StackOverflow
scroll top