R迭代数据集生成

Question

我想基于存储在[1:21] num/chr 字符串中的 21 个参数生成 21 个具有两个变量的随机正态分布 data.frames（使用 mvrnorm）。另外，每个 data.frame 应该包含两列，其中的字符串对于 21 个集合中的每一个集合都是相同的（一旦我 rbind 将它们放在一起，这将是有意义的）

假设 AGQ.mu2.mat 存储第一个变量 (stddev = 1) 的平均值，Std.mu2.mat 存储第二个变量 (stddev = 20)， ORIG 和 ID 存储两个字符串（我用一些 LotR 名称替换了数据），并让 corr = .30 和 n 总是 300：

r<-0.30
ORIG<-c("Ang","Ang","Ang","DGu","DGu","DGu", "Har", 
        "Har", "Har", "Isg", "Isg", "Isg", "Mor", "Mor", 
        "Mor", "MMo", "MMo", "MMo", "Rhu", "Rhu", "Rhu")
ID<-c("a","b","c","a","b","c","a","b","c",
         "a","b","c","a","b","c","a","b","c","a","b","c")
AGQ.mu2.mat <- c(-1.06, -1.27, -1.97, -3.27, -3.32, -2.32, -1.05, -0.36, -1.85, -2.30, -2.49, -2.90, -0.80, -0.62, -0.56,  1.22,  1.62,  0.44, -1.43, -1.08, -2.34)
Std.mu2.mat <- c(107, 100, 102, 124, 118, 120, 147, 152, 141, 154, 160, 155, 125, 123, 116, 103, 106, 107, 123, 126, 117)

我想迭代

dt<-data.frame(mvrnorm(n=300, mu=c(AGQ.mu2.mat[1],Std.mu2.mat[1]),
                       Sigma=matrix(c(1, r*20, r*20, (20^2)),
                                  nrow=2), empirical=TRUE))
dt<-rename(dt, "AGQ" = "X1","Std" = "X2")
dt$Herkunft<-ORIG[1]
dt$Bat_ID<-ID[1]

在参数向量的所有 21 个条目上，产生 21 个唯一命名的数据帧。

我试过了

for (i in 1:21)
  {
dt<-data.frame(mvrnorm(n=300, mu=c(AGQ.mu2.mat[i],Std.mu2.mat[i]),
                       Sigma=matrix(c(1, r*20, r*20, (20^2)),
                                  nrow=2), empirical=TRUE))
dt<-rename(dt, "AGQ" = "X1","Std" = "X2")
dt$Herkunft<-ORIG[i]
dt$Bat_ID<-ID[i]
}

和

dt<-data.frame(mvrnorm(n=300, mu=c(rep(AGQ.mu2.mat,each=1),rep(Std.mu2.mat, each=1)),
                       Sigma=matrix(c(1, r*20, r*20, (20^2)),
                                    nrow=2), empirical=TRUE))
dt<-rename(dt, "Aggr" = "X1","Std" = "X2")
dt$Herkunft<-rep(ORIG,each=1)
dt$Bat_ID<-rep(ID,each=1)

无济于事；即使这样，这也会一遍又一遍地生成相同的数据集，而不会创建唯一命名的新数据集。

我非常感谢任何有关如何迭代它的帮助或方法。提前致谢！

Answer 1

您需要使用某种方式来收集结果。我选择使用附加的list。其他解决方案可能是使用rbind。你必须用 nrow/ncol 参数来塑造你的矩阵。

library(MASS)

r<-0.30
ORIG<-c("Ang","Ang","Ang","DGu","DGu","DGu", "Har", 
        "Har", "Har", "Isg", "Isg", "Isg", "Mor", "Mor", 
        "Mor", "MMo", "MMo", "MMo", "Rhu", "Rhu", "Rhu")
ID<-c("a","b","c","a","b","c","a","b","c",
         "a","b","c","a","b","c","a","b","c","a","b","c")

AGQ.mu2.mat <- 1:21
Std.mu2.mat <- 1:21

results <- list()

for (i in 1:21) {
dt<-data.frame(mvrnorm(n=300, mu=c(AGQ.mu2.mat[i],Std.mu2.mat[i]),
                       Sigma=matrix(c(1, r*20, r*20, (20^2)), nrow = 2), empirical=TRUE))
colnames(dt) <- c("AGQ","Std")
dt$Herkunft<-ORIG[i]
dt$Bat_ID<-ID[i]

results[[i]] <- dt
}

加入所有结果

do.call(rbind, results)

验证：

> cor(results[[1]][,1:2]);apply(results[[1]][,1:2], 2, mean);apply(results[[1]][,1:2], 2, sd)
    AGQ Std
AGQ 1.0 0.3
Std 0.3 1.0
AGQ Std 
  1   1 
AGQ Std 
  1  20 

Answer 2

这个使用dplyr 和purrr 的解决方案可能会有所帮助...

library(MASS)
library(dplyr)
library(purrr)

r <-0.30

ORIG<-c("Ang","Ang","Ang","DGu","DGu","DGu", "Har", 
        "Har", "Har", "Isg", "Isg", "Isg", "Mor", "Mor", 
        "Mor", "MMo", "MMo", "MMo", "Rhu", "Rhu", "Rhu")

ID<-c("a","b","c","a","b","c","a","b","c",
      "a","b","c","a","b","c","a","b","c","a","b","c")

AGQ.mu2.mat <- c(-1.06, -1.27, -1.97, -3.27, -3.32, -2.32, -1.05, -0.36, -1.85, -2.30, -2.49, -2.90, -0.80, -0.62, -0.56,  1.22,  1.62,  0.44, -1.43, -1.08, -2.34)
Std.mu2.mat <- c(107, 100, 102, 124, 118, 120, 147, 152, 141, 154, 160, 155, 125, 123, 116, 103, 106, 107, 123, 126, 117)

listofdf <- purrr::map2(AGQ.mu2.mat, 
            Std.mu2.mat,
            ~ data.frame(MASS::mvrnorm(n = 300, 
                                       mu=c(.x, .y),
                                       Sigma = matrix(c(1, r*20, r*20, (20^2)), nrow=2), 
                                       empirical=TRUE))
) %>% 
   purrr::map2(ORIG, ~ mutate(.x, Herkunft = .y)) %>%
   purrr::map2(ID, ~ mutate(.x, Bat_ID = .y))

head(listofdf[[4]])
#>          X1         X2 Herkunft Bat_ID
#> 1 0.4329040  17.616240      DGu      a
#> 2 2.6589457  -9.385099      DGu      a
#> 3 2.3626722   9.815020      DGu      a
#> 4 1.8998816 -24.551505      DGu      a
#> 5 0.2506713   0.792565      DGu      a
#> 6 3.4961960   8.112878      DGu      a


### since you mentioned binding them all together
onebigdf <- bind_rows(listofdf)