我试图根据其他三个分组变量生成具有不同参数的正常随机变量。有关详细信息，请参阅下面的示例（假设我有两个明智的分组变量）：
V1 <- rep(c(1,2,3), each = 3, times = 2) # first grouping variable
V2 <- rep(c(1,2,3), times = 6) # 2nd grouping variable
V <- cbind(V1,V2)
显然，我有9种不同的情况{（1,1），（1,2），......（3,3）}。所以我想使用rnorm(n = 1, mean = x, sd = 20)生成V3，其中x因具体情况而异。
我知道我可以使用if，但对于大量数据，它不会是一个不错的选择。我使用以下代码（这里只是示例）：
V3 <-
rnorm(n = sum(V[,1] == 1 & V[,2] == 1), mean = 1, sd = 20)*(V[,1] == 1 & V[,2] == 1)+
rnorm(n = sum(V[,1] == 1 & V[,2] == 2), mean = 2, sd = 20)*(V[,1] == 1 & V[,2] == 2)+
rnorm(n = sum(V[,1] == 1 & V[,2] == 3), mean = 3, sd = 20)*(V[,1] == 1 & V[,2] == 3)+
rnorm(n = sum(V[,1] == 2 & V[,2] == 1), mean = 4, sd = 20)*(V[,1] == 2 & V[,2] == 1)+
rnorm(n = sum(V[,1] == 2 & V[,2] == 2), mean = 5, sd = 20)*(V[,1] == 2 & V[,2] == 2)+
rnorm(n = sum(V[,1] == 2 & V[,2] == 3), mean = 6, sd = 20)*(V[,1] == 2 & V[,2] == 3)+
rnorm(n = sum(V[,1] == 3 & V[,2] == 1), mean = 7, sd = 20)*(V[,1] == 3 & V[,2] == 1)+
rnorm(n = sum(V[,1] == 3 & V[,2] == 2), mean = 8, sd = 20)*(V[,1] == 3 & V[,2] == 2)+
rnorm(n = sum(V[,1] == 3 & V[,2] == 3), mean = 9, sd = 20)*(V[,1] == 3 & V[,2] == 3)
同样，我的数据很多，有很多不同的群体，所以这不是一种有效的方法。我也可以使用for loop，但我试图避免它。我想使用apply，但不确定它是否适用于此。
任何建议表示赞赏。感谢