如何使用 mutate 替换当前列中的数据？

Question

我想按年份对数据框进行分组并标准化某些列（在本例中为 BioTest、MathExam 和 WritingScore）并用新数据替换旧数据。以下是我的数据框示例：

DF:

Var1   Var2   Year  BioTest   MathExam   WritingScore   Var3  Var 4
 X      X     2016   165        140         10           X     X
 X      X     2017   172        128         11           X     X
 X      X     2018   169        115          8           X     X
 X      X     2016   166        139         10           X     X
 X      X     2017   165        140         12           X     X

我尝试了以下代码的变体：

DF<- DF %>% group_by(Year)%>% mutate(across(BioTest:WritingScore),scale)

DF<- DF %>% group_by(Year)%>% mutate(across(select(BioTest:WritingScore)),scale)

我得到的回报是相同的 DF，没有任何变化。我想要的是：

 DF:

 Var1   Var2   Year  BioTest   MathExam   WritingScore   Var3  Var 4
 X      X     2016   NewData     NewData      NewData      X     X
 X      X     2017   NewData     NewData      NewData      X     X
 X      X     2018   NewData     NewData      NewData      X     X
 X      X     2016   NewData     NewData      NewData      X     X
 X      X     2017   NewData     NewData      NewData      X     X

非常感谢任何帮助。

Answer 1

问题可能是dplyr::mutate 被plyr::mutate 掩盖了。它可以被复制（以及across在没有功能的情况下关闭的事实）

iris %>%
    group_by(Species) %>%
    plyr::mutate(across(where(is.numeric), scale))
# A tibble: 150 x 5
# Groups:   Species [3]
#   Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#          <dbl>       <dbl>        <dbl>       <dbl> <fct>  
# 1          5.1         3.5          1.4         0.2 setosa 
# 2          4.9         3            1.4         0.2 setosa 
# 3          4.7         3.2          1.3         0.2 setosa 
# 4          4.6         3.1          1.5         0.2 setosa 
# 5          5           3.6          1.4         0.2 setosa 
# 6          5.4         3.9          1.7         0.4 setosa 
# 7          4.6         3.4          1.4         0.3 setosa 
# 8          5           3.4          1.5         0.2 setosa 
# 9          4.4         2.9          1.4         0.2 setosa 
#10          4.9         3.1          1.5         0.1 setosa 
# … with 140 more rows

与初始的“iris”数据集相同

现在，检查正确的dplyr::mutate

iris %>% 
   group_by(Species) %>%
   dplyr::mutate(across(where(is.numeric), scale))
# A tibble: 150 x 5
# Groups:   Species [3]
#   Sepal.Length[,1] Sepal.Width[,1] Petal.Length[,1] Petal.Width[,1] Species
#              <dbl>           <dbl>            <dbl>           <dbl> <fct>  
# 1           0.267           0.190            -0.357          -0.436 setosa 
# 2          -0.301          -1.13             -0.357          -0.436 setosa 
# 3          -0.868          -0.601            -0.933          -0.436 setosa 
# 4          -1.15           -0.865             0.219          -0.436 setosa 
# 5          -0.0170          0.454            -0.357          -0.436 setosa 
# 6           1.12            1.25              1.37            1.46  setosa 
# 7          -1.15           -0.0739           -0.357           0.512 setosa 
# 8          -0.0170         -0.0739            0.219          -0.436 setosa 
# 9          -1.72           -1.39             -0.357          -0.436 setosa 
#10          -0.301          -0.865             0.219          -1.39  setosa 
# … with 140 more rows

---

因此，在 OP 的代码中，我们只需要使用 dplyr::mutate 或重新启动一个仅加载了 dplyr 的新 R 会话

DF %>% 
   group_by(Year)%>% 
   dplyr::mutate(across(BioTest:WritingScore, scale))

scale 返回带有一些属性的matrix。如果我们只需要numeric vector 部分，我们可以使用as.vector 或as.numeric

DF %>% 
   group_by(Year)%>% 
   dplyr::mutate(across(BioTest:WritingScore, ~ as.numeric(scale(.)))

注意：across 中不需要 select

Answer 2

也许试试这个。问题出在您的across() 声明中。函数必须在里面：

library(dplyr)
#Code
DF %>%
  group_by(Year) %>%
  mutate(across(BioTest:WritingScore,~scale(.)[,1]))

输出：

# A tibble: 5 x 9
# Groups:   Year [3]
  Var1  Var2   Year BioTest[,1] MathExam[,1] WritingScore[,1] Var3  Var   X4   
  <chr> <chr> <int>       <dbl>        <dbl>            <dbl> <chr> <chr> <lgl>
1 X     X      2016      -0.707        0.707          NaN     X     X     NA   
2 X     X      2017       0.707       -0.707           -0.707 X     X     NA   
3 X     X      2018     NaN          NaN              NaN     X     X     NA   
4 X     X      2016       0.707       -0.707          NaN     X     X     NA   
5 X     X      2017      -0.707        0.707            0.707 X     X     NA   

使用的一些数据：

#Data
DF <- structure(list(Var1 = c("X", "X", "X", "X", "X"), Var2 = c("X", 
"X", "X", "X", "X"), Year = c(2016L, 2017L, 2018L, 2016L, 2017L
), BioTest = c(165L, 172L, 169L, 166L, 165L), MathExam = c(140L, 
128L, 115L, 139L, 140L), WritingScore = c(10L, 11L, 8L, 10L, 
12L), Var3 = c("X", "X", "X", "X", "X"), Var = c("X", "X", "X", 
"X", "X"), X4 = c(NA, NA, NA, NA, NA)), class = "data.frame", row.names = c(NA, 
-5L))