替換數據框列表中的列值

Question

我有以下示例數據框列表：

cat <- rnorm(5)
dog <- rnorm(5)
mouse <- rnorm(5)

df1 <- cbind(cat,dog,mouse)
df2 <- cbind(cat,dog,mouse)
df3 <- cbind(cat,dog,mouse)

list.1 <- list(df1 = df1,df2 = df2,df3 = df3)
list.1

$df1
            cat        dog      mouse
[1,] -0.6991598 -0.8630006 -0.7564806
[2,]  0.7645475  1.3571995  0.8939621
[3,]  1.0608070 -0.8455111  0.5198387
[4,] -0.2008916 -0.7971714  0.8477894
[5,] -0.6988800  1.0717351 -1.3684944

$df2
            cat        dog      mouse
[1,] -0.6991598 -0.8630006 -0.7564806
[2,]  0.7645475  1.3571995  0.8939621
[3,]  1.0608070 -0.8455111  0.5198387
[4,] -0.2008916 -0.7971714  0.8477894
[5,] -0.6988800  1.0717351 -1.3684944

$df3
            cat        dog      mouse
[1,] -0.6991598 -0.8630006 -0.7564806
[2,]  0.7645475  1.3571995  0.8939621
[3,]  1.0608070 -0.8455111  0.5198387
[4,] -0.2008916 -0.7971714  0.8477894
[5,] -0.6988800  1.0717351 -1.3684944

我想將每個數據框中的dog列替換為另一個數據框中的相應列。

用變量“ dog”的新值創建一個數據框

new.dog1 <- c(1,1,2,2,3)
new.dog2 <- c(10,10,20,20,30)
new.dog3 <- c(100,100,200,200,300)
new.dogs <- cbind(new.dog1, new.dog2, new.dog3)
new.dogs

     new.dog1 new.dog2 new.dog3
[1,]        1       10      100
[2,]        1       10      100
[3,]        2       20      200
[4,]        2       20      200
[5,]        3       30      300

我要執行的操作的偽代碼（不起作用）：

updated.list <- for(i in list.1) {
  list.1[[i]][,2] <- new.dogs[,i]
  return(list.1)
  }

輸出應如下所示：

> updated.list
$df1
            cat dog      mouse
[1,] -0.6991598   1 -0.7564806
[2,]  0.7645475   1  0.8939621
[3,]  1.0608070   2  0.5198387
[4,] -0.2008916   2  0.8477894
[5,] -0.6988800   3 -1.3684944

$df2
            cat dog      mouse
[1,] -0.6991598  10 -0.7564806
[2,]  0.7645475  10  0.8939621
[3,]  1.0608070  20  0.5198387
[4,] -0.2008916  20  0.8477894
[5,] -0.6988800  30 -1.3684944

$df3
            cat dog      mouse
[1,] -0.6991598 100 -0.7564806
[2,]  0.7645475 100  0.8939621
[3,]  1.0608070 200  0.5198387
[4,] -0.2008916 200  0.8477894
[5,] -0.6988800 300 -1.3684944

在我的for循環中，我認為問題出在new.dogs[,i]位代碼上？ 理想情況下，如果可能的話，我寧願使用lapply或tidyverse解決方案，也不願使用for循環...

Answer 1

與基數R：

updated.list <- mapply(function(old, new, which) {
  old[,which] <- new
  old
}, list.1, data.frame(new.dogs), "dog", SIMPLIFY = FALSE)

Answer 2

如果要使用tidyverse ，則可以保留new.dogs一個列表並清理留下的矩陣混亂cbind() ，然后使用map2()來成對地遍歷兩個列表，如下所示：

library(tidyverse)

# use new.dogs as a list instead
new.dogs <- list(new.dog1, new.dog2, new.dog3)

# cbind() creates matrixes from vectors, not tidy tibbles/dataframes
list.1 <- map(list.1, as.tibble) 

# iterate and replace pairwise (list.1[[i]] <- .; new.dogs[[i]] <- .y)
map2(list.1, new.dogs, ~ mutate(., dog = .y))