这就是我所拥有的：
library(knitr)
table_MSE_EB <- matrix(c(0.0053,0.1140,0.1542,0.1677,0.0049,0.1177,0.1570,0.1762,0.0040,0.1211,0.1578,0.1803),4,3)
colnames(table_MSE_EB) <- c("I=10", "I=20", "I=50")
rownames(table_MSE_EB) <- c("d=2", "d=3", "d=5", "d=10")
kable(head(table_MSE_EB), format = "pandoc", caption = "Err for EB")
table_MSE_MLE <- matrix(c(0.034,0.0388,0.0544,0.0847,0.0263,0.0332,0.0445,0.0778,0.0193,0.0263,0.0376,0.0742),4,3)
colnames(table_MSE_MLE) <- c("I=10", "I=20", "I=50")
rownames(table_MSE_MLE) <- c("d=2", "d=3", "d=5", "d=10")
kable(head(table_MSE_MLE), format = "pandoc", caption = "Err for MLE")
我想要的是将上面的两个表组合起来形成下表：
table <- matrix(c(0.0053,0.034,0.1140,0.0388,0.1542,0.0544,0.1677,0.0847,0.0049,0.0263,0.1177,0.0332,0.1570,0.0445,0.1762,0.0778,0.004,0.0193,0.1211,0.0263,0.1578,0.0376,0.1803,0.0742),8,3)
colnames(table) <- c("I=10", "I=20", "I=50")
rownames(table) <- c("d=2(EB)", "d=2(MLE)", "d=3(EB)", "d=3(MLE)", "d=5(EB)","d=5(MLE)", "d=10(EB)", "d=10(MLE)")
kable(head(table), format = "pandoc", caption = "Err for MLE/EB")
另外，我想知道是否有某种方法让R根据I和d定义输出I和d的值，而不是手动输入。
感谢您的帮助。
答案 0 :(得分：3)
如果您有两个向量并且想要获得结果（如评论中所述）
a <- c(5,6,2,4,4,3)
b <- c(10,13,12,12, 15,15)
c(rbind(a,b))
#[1] 5 10 6 13 2 12 4 12 4 15 3 15
关于组合两个tables
row.names(table_MSE_EB) <- paste0(row.names(table_MSE_EB), "(EB)")
row.names(table_MSE_MLE) <- paste0(row.names(table_MSE_MLE), "(MLE)")
tableNew <- rbind(table_MSE_EB, table_MSE_MLE)
tableNew1 <- tableNew[c(matrix(1:nrow(tableNew),nrow=2, byrow=TRUE)),]
或重新排列rows的其他方法是
n <- nrow(table_MSE_EB)
indx <- c(rbind(seq(1,n), seq(n+1, 2*n)))
tableNew1 <- tableNew[indx,]
kable(head(tableNew1), format='pandoc', caption='Err for MLE/EB')
#Table: Err for MLE/EB
# I=10 I=20 I=50
#--------- ------- ------- -------
#d=2(EB) 0.0053 0.0049 0.0040
#d=2(MLE) 0.0340 0.0263 0.0193
#d=3(EB) 0.1140 0.1177 0.1211
#d=3(MLE) 0.0388 0.0332 0.0263
#d=5(EB) 0.1542 0.1570 0.1578
#d=5(MLE) 0.0544 0.0445 0.0376