r - 如何在 R 中获取多个 Stuart-Maxwell 检验的 p 值矩阵？问

Question

我有一个包含 1o 个问题(Q1-Q10)的数据集。我想提取成对 Stuart-Maxwell 检验的 P 值，并将它们打印在带有列名标题的矩阵中，以便我可以检测到哪个 P 值指的是哪个成对比较。我从专家那里得到了如何运行测试和提取结果的帮助，我试图用标题制作矩阵，但我失败了。如果有人可以在下面修改我的代码，我将不胜感激。

data <- data.frame(Q1=sample(1:5, 20, replace=T),
               Q2=sample(1:5, 20, replace=T),
               Q3=sample(1:5, 20, replace=T),
               Q4=sample(1:5, 20, replace=T),
               Q5=sample(1:5, 20, replace=T),
               Q6=sample(1:5, 20, replace=T),
               Q7=sample(1:5, 20, replace=T),
               Q8=sample(1:5, 20, replace=T),
               Q9=sample(1:5, 20, replace=T),
               Q10=sample(1:5, 20,replace=T) )  #fake data
Labels<- names(data)
# Matrix to store the result
enter code here
groups <- unique( Labels )
result <- matrix(NA, nc=length(groups), nr=length(groups))
colnames(result) <- rownames(result) <- groups
# Loop
for( g1 in groups ) {
for( g2 in groups ) {
result[ g1, g2 ] <- 
      sapply(labels <-combn(groups, 2, simplify = FALSE), function(i) {


    require(irr)

    xtab <- table(data[,i[1]], data[,i[2]])
    test <- try(stuart.maxwell.mh(xtab))
    ifelse(class(test) == "try-error", NA, test$p)
})



}

}

result

Answer 1

你们很接近。你只是混淆了两种方法。您应该将 sapply 与 combn 或双 for-loop 一起使用。在您的情况下，根据您想要的输出，for-loop 似乎更容易:

require(irr)
data <- data.frame(Q1=sample(1:5, 20, replace=T),
                   Q2=sample(1:5, 20, replace=T),
                   Q3=sample(1:5, 20, replace=T),
                   Q4=sample(1:5, 20, replace=T),
                   Q5=sample(1:5, 20, replace=T),
                   Q6=sample(1:5, 20, replace=T),
                   Q7=sample(1:5, 20, replace=T),
                   Q8=sample(1:5, 20, replace=T),
                   Q9=sample(1:5, 20, replace=T),
                   Q10=sample(1:5, 20,replace=T) )  #fake data

# Loop
labels<- names(data)
groups <- unique(labels)
result <- matrix(NA, nc=length(groups), nr=length(groups))
colnames(result) <- labels
rownames(result) <- labels

for( g1 in groups ) {
 for( g2 in groups ) {

  xtab <- table(data[,g1], data[,g2])
  test <- try(stuart.maxwell.mh(xtab), silent = TRUE)
  pval <- ifelse(class(test) == "try-error", NA, test$p)
  result[g1, g2] <- pval

 }
}

输出:

> result
           Q1         Q2         Q3        Q4        Q5        Q6         Q7         Q8
Q1         NA 0.05881900 0.26298902        NA 0.7790233 0.5177394 0.73754470 0.30610257
Q2  0.0588190         NA 0.58635825        NA 0.1005838 0.3678794 0.17078180 0.05626191
Q3  0.2629890 0.58635825         NA 0.3173105 0.3916252 0.2541583 0.09984288 0.07472086
Q4  0.3173105 0.36787944         NA        NA        NA        NA         NA         NA
Q5  0.7790233 0.10058381 0.39162518 0.2231302        NA 0.3114032 0.14247485 0.17708783
Q6  0.5177394 0.36787944 0.25415830        NA 0.3114032        NA 0.21636999 1.00000000
Q7  0.7375447 0.17078180 0.09984288        NA 0.1424749 0.2163700         NA 0.41099506
Q8  0.3061026 0.05626191 0.07472086 0.3173105 0.1770878 1.0000000 0.41099506         NA
Q9  0.6712714 0.53408090 0.16832466 0.3173105 0.2110881 0.2635971 0.28933534 0.31414685
Q10 0.2220359 0.38332585 0.05941603 1.0000000 0.1652989 0.2578472 0.23889094 0.08826479

#the output is truncated