R 添加一个新级别的变量，并为其分配每个id的其他变量的平均值

我想在数据框中添加一个变量的级别，并将其指定为同一变量中其他级别的平均值。我不知道怎么做。因此，我想在变量trt中添加级别“base”。这里我想得到变量“pointA”、“pointB”和“pointC”的trt“OA”和“OB”的平均值

ID51base973
##>62A2996
##>7 2 OA 6 3 4
##>82B6410
##>9 2 OB 8 2 4
##>102基地984
##>11 3 A 7 1 8
##>12 3 OA 3 10 2
##>133B246
##>14 3 OB 2 9
##>15 3基地36 8
df[5,3]82B6.0410.0
##>9 2 OB 8.0 2.0 4.0
##>10 2基本7.0 2.5 4.0
##>11 3 A 7.0 1.0 8.0
##>12 3 OA 3.0 10.0 2.0
##>133B2.046.0
##>14 3 OB 2.0 2.0 9.0
##>15 3基础2.5 6.0 5.5

我们可以通过'id'进行分组，然后在

处使用
mutate_替换
将'trt'为'base'的位置替换为值子集的
平均值
，其中'trt'为'OA/OB'

library(dplyr)
df %>% 
   group_by(id) %>%
   mutate_at(vars(starts_with('point')),
        ~ replace(., trt == 'base',  mean(.[trt %in% c("OA", "OB")])))
# Groups:   id [3]
#      id trt   pointA pointB pointC
#   <int> <chr>  <dbl>  <dbl>  <dbl>
# 1     1 A        3      3      5  
# 2     1 OA      10      6      4  
# 3     1 B        9      9      7  
# 4     1 OB      10      5      6  
# 5     1 base    10      5.5    5  
# 6     2 A        2      9      6  
# 7     2 OA       6      3      4  
# 8     2 B        6      4     10  
# 9     2 OB       8      2      4  
#10     2 base     7      2.5    4  
#11     3 A        7      1      8  
#12     3 OA       3     10      2  
#13     3 B        2      4      6  
#14     3 OB       2      2      9  
#15     3 base     2.5    6      5.5


或者在
base R
中使用
split/unsplit

unsplit(lapply(split(df, df$id), function(x) {
     x[x$trt == 'base', 3:5] <- colMeans(subset(x, 
                   trt %in% c("OA", "OB"), select = 3:5))
    x}),
   df$id)

unsplit（lappy（split）（df，df$id），函数（x）{
x[x$trt=='base'，3:5]对于基本R解决方案：

像以前一样重新创建
df
，但首先设置种子：

set.seed(1)
id <- rep(1:3,each=5)
trt <- rep(c("A","OA", "B", "OB","base"),3)
pointA <- sample(1:10,15, replace=TRUE)
pointB<- sample(1:10,15, replace=TRUE)
pointC<- sample(1:10,15, replace=TRUE)
df <- data.frame(id,trt,pointA, pointB,pointC)

然后在这些行上循环，将第3列到第5列中的值替换为
trt
“OA”
和
“OB”
对应个体的值的均值：

df <- structure(list(id = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 
3L, 3L, 3L, 3L, 3L), trt = c("A", "OA", "B", "OB", "base", "A", 
"OA", "B", "OB", "base", "A", "OA", "B", "OB", "base"), pointA = c(3L, 
10L, 9L, 10L, 9L, 2L, 6L, 6L, 8L, 9L, 7L, 3L, 2L, 2L, 3L), pointB = c(3L, 
6L, 9L, 5L, 7L, 9L, 3L, 4L, 2L, 8L, 1L, 10L, 4L, 2L, 6L), pointC = c(5L, 
4L, 7L, 6L, 3L, 6L, 4L, 10L, 4L, 4L, 8L, 2L, 6L, 9L, 8L)), 
class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", 
"14", "15"))

set.seed(1)
id <- rep(1:3,each=5)
trt <- rep(c("A","OA", "B", "OB","base"),3)
pointA <- sample(1:10,15, replace=TRUE)
pointB<- sample(1:10,15, replace=TRUE)
pointC<- sample(1:10,15, replace=TRUE)
df <- data.frame(id,trt,pointA, pointB,pointC)

base_row_ind_vec <- which(df$trt == "base")

for(i in base_row_ind_vec){
  id_curr <- df[i,"id"]
  df[i,3:5] <- lapply(df[df$id == id_curr & df$trt %in% c("OA", "OB"), 3:5], mean)
}

   id  trt pointA pointB pointC
1   1    A    9.0    9.0   10.0
2   1   OA    4.0    5.0    6.0
3   1    B    7.0    5.0    4.0
4   1   OB    1.0    9.0    4.0
5   1 base    2.5    7.0    5.0
6   2    A    7.0    5.0    9.0
7   2   OA    2.0    5.0    7.0
8   2    B    3.0    2.0    6.0
9   2   OB    1.0   10.0    9.0
10  2 base    1.5    7.5    8.0
11  3    A    5.0    1.0    9.0
12  3   OA   10.0    4.0    7.0
13  3    B    6.0    3.0    8.0
14  3   OB   10.0    6.0    6.0
15  3 base   10.0    5.0    6.5