如何将 mult.chart 中的 T^2 hotelling 分解为 output 中的矩阵？

Question

I run mult.chart package for T^2 hotelling. I want to get just "the first point" that is out of control and "it's decomposition" as a matrix.But running these codes get all out of control points and all decomposition in the type of "list". and i can't separate the first point and it's decomposition and save them. what should i do?

    a<- runif(400,0,1)
    b<- matrix(a, nrow=100,ncol=4)
    output <- mult.chart(type="t2", alpha=0.07,b)

I expect the output for example number 70 (out of control point) and the decomposition matrix of that with 15 row and 7 columns (titles: t2 decomp, ucl, p-value,1,2,3, 4), but the actual output is a list with all out of control points and the decomposition of those as a list.

Answer 1

The table you saw, is only printed when your type is t2 and is not stored in output. You can see in the code the table is variable q, and not returned.

I basically took the calculation part out (nasty bit of code) and wrote it into a function:

printOutlier = function(x,output,alpha){
p <- ncol(x)
m <- nrow(x)
x <- array(data.matrix(x), c(m, p, 1))
n <- dim(x)[3]
phase <- 2
x.jk <- matrix(0, m, p)
t2=output$t2
x.jk <- apply(x, 1:2, mean)
Xmv <- output$Xmv
S <- output$covariance
colm <- nrow(x)
ucl = output$ucl

t3 <- which(t2 > ucl)
res = vector("list",length(t3))
for (ii in 1:length(t3)) {
       v = 1
       k = 0
       for (i in 1:p) {
              k <- k + factorial(p)/(factorial(i) * factorial(p -i))
                }
       q <- matrix(0, k, p + 3)
       for (i in 1:p) {
                  a <- t(combn(p, i))
                  for (l in 1:nrow(a)) {
                    for (j in 1:ncol(a)) {
                      q[v, j + 3] <- a[l, j]
                    }
                    v = v + 1
                  }
                }
       for (i in 1:nrow(q)) {
              b <- subset(q[i, 4:ncol(q)], q[i, 4:ncol(q)] > 0)
              di <- length(b)
              if (length(b) > 1) {
              q[i, 1] <- n * t(Xmv[b] - x.jk[t3[ii], ][b]) %*% 
              solve(S[b, b]) %*% (Xmv[b] - x.jk[t3[ii],][b])
                  }
              else (q[i, 1] <- n * (x.jk[t3[ii], ][b] - Xmv[b])^2/S[b, b])
              ifelse(n == 1, ifelse(phase == 1, q[i, 2] <- ((colm - 
                    1)^2)/colm * qbeta(1 - alpha, di/2, (((2 * 
                    (colm - 1)^2)/(3 * colm - 4) - di - 1)/2)), 
                    q[i, 2] <- ((di * (colm + 1) * (colm - 1))/((colm^2) - 
                      colm * di)) * qf(1 - alpha, di, colm - 
                      di)), ifelse(phase == 1, q[i, 2] <- (di * 
                    (colm - 1) * (n - 1))/(colm * n - colm - 
                    di + 1) * qf(1 - alpha, di, colm * n - colm - 
                    di + 1), q[i, 2] <- (di * (colm + 1) * (n - 
                    1))/(colm * n - colm - di + 1) * qf(1 - alpha, 
                    di, colm * n - colm - di + 1)))
                  q[i, 3] <- 1 - pf(q[i, 1], di, colm - 1)
                }
                colnames(q) <- c("t2 decomp", "ucl", "p-value", 1:p)
                names(res)[ii] <- paste(`Decomposition of` = t3[ii])
                res[[ii]] <- round(q, 4)
            }
       return(res)
}

Now if you run your mult.chart again, and use this function, it should give you the table:

library(MSQC)
set.seed(111)
a<- runif(400,0,1)
b<- matrix(a, nrow=100,ncol=4)
output <- mult.chart(type="t2", alpha=0.07,b)

printOutlier(b,output,0.07)