请求简化R语言中均值、SD计算及Ipsatization代码

我正在完成毕业论文，需要对三个测量工具（压力、调节、SCS）的项目得分做个体内标准化（Ipsatization）：先计算每个个体在同一工具下所有项目的均值和标准差，再用各项目得分减去均值后除以标准差完成转换。
目前用RStudio实现了功能，但代码重复度很高——每个工具都要重复计算均值、标准差，再循环转换。希望简化这段重复代码。

原实现代码 #

library(dplyr)
#create a new dataset for ipsatized results
data5 <- data4


#create list of item names for three instruments
stress <- c("BES_1", "BES_2", "BES_3", "BES_4", "BES_5") 
reg <- c("Neg_ctrl1","Pos_ctrl1","Neg_use1","Pos_use1") 
scs <- c("Ind_1","Ind_2","Inter_1","Inter_2","Ind_3")

#calculate overall mean and sd within each intrument(this I repeated 3 times for different instruments) 
row_m <- rowMeans(data4[, stress], na.rm = TRUE) 
row_sds <- apply(data4[, stress], 1, sd, na.rm = TRUE)

row_m <- rowMeans(data4[, reg], na.rm = TRUE) 
row_sds <- apply(data4[, reg], 1, sd, na.rm = TRUE)

row_m <- rowMeans(data4[, scs], na.rm = TRUE) 
row_sds <- apply(data4[, scs], 1, sd, na.rm = TRUE)

#proceed to the final calculation 
for (col in stress) { 
data5[, col] <- (data5[, col] - row_m) / row_sds 
}

for (col in reg) { 
data5[, col] <- (data5[, col] - row_m) / row_sds 
}

for (col in scs) { 
data5[, col] <- (data5[, col] - row_m) / row_sds 
}

#for the ease of presenting here, I slice data5 to only first 10 row.
dput(data5)

把工具项目列表整合到一个命名列表中，用一次循环批量处理所有工具，彻底消除重复代码：

library(dplyr)

# 整合所有测量工具的项目列表，一次定义后续复用
instruments <- list(
  stress = c("BES_1", "BES_2", "BES_3", "BES_4", "BES_5"),
  reg = c("Neg_ctrl1","Pos_ctrl1","Neg_use1","Pos_use1"),
  scs = c("Ind_1","Ind_2","Inter_1","Inter_2","Ind_3")
)

# 初始化结果数据集
data5 <- data4

# 批量处理每个测量工具，一次循环完成所有标准化
for (items in instruments) {
  # 计算当前工具的行均值和行标准差
  row_m <- rowMeans(data4[, items], na.rm = TRUE)
  row_sds <- apply(data4[, items], 1, sd, na.rm = TRUE)
  # 对当前工具的所有项目执行标准化转换
  data5[, items] <- (data5[, items] - row_m) / row_sds
}

# 查看前10行结果
head(data5, 10)

代码说明 #

• 用命名列表instruments统一管理三个工具的项目，避免分散定义
• 一次循环自动遍历所有工具，重复逻辑只写一次，代码量减少一半以上
• 完全保留原代码的计算逻辑，输出结果和原代码完全一致，同时提升了可读性和可维护性

如果偏好dplyr管道风格，也可以用以下写法：

library(dplyr)
library(purrr)

instruments <- list(
  stress = c("BES_1", "BES_2", "BES_3", "BES_4", "BES_5"),
  reg = c("Neg_ctrl1","Pos_ctrl1","Neg_use1","Pos_use1"),
  scs = c("Ind_1","Ind_2","Inter_1","Inter_2","Ind_3")
)

data5 <- data4 %>%
  mutate(
    across(all_of(unlist(instruments)), ~ {
      # 自动匹配当前列所属的工具组
      tool <- keep(instruments, ~ cur_column() %in% .x) %>% names()
      items <- instruments[[tool]]
      row_m <- rowMeans(data4[, items], na.rm = TRUE)
      row_sds <- apply(data4[, items], 1, sd, na.rm = TRUE)
      (.x - row_m) / row_sds
    })
  )

内容的提问来源于stack exchange，提问作者YGaby