如何一步将df_old的变量标签批量赋值给df_new对应变量？

我有一个带变量标签的dataframe（df_old），想把这些标签提取出来赋值给另一个dataframe（df_new）的对应变量。尝试在循环里执行以下代码但没生效：

attr(df_new[ , j], "label") <- attr(df_old[ , j], "label") 

后来用下面的方法绕开了问题：

Tmp <- data.frame(df_old[ , j ])
names(Tmp) <- "Var"
attr(df_new[ , j], "label") <- attr(Tmp$Var, "label") 

请问有没有一步到位的实现方法？

补充示例代码 #

df_old <- structure(list(A1_1 = structure(c(1, 3, 2, 1, 6, 1, 5, 7, 1, 
                                  1), label = "Use Facebook", format.spss = "F1.0", display_width = 1L, labels = c(`Every day` = 1, 
                                                                                                                                                                   `A few times a week` = 2, `A few times a month` = 3, `Once a month` = 4, 
                                                                                                                                                                   `Every few months` = 5, `Less often` = 6, `I never do this activity` = 7
                                  ), class = c("haven_labelled", "vctrs_vctr", "double")), A1_2 = structure(c(1, 
                                                                                                              4, 1, 7, 7, 1, 1, 7, 1, 6), label = "Use Instagram", format.spss = "F1.0", display_width = 1L, labels = c(`Every day` = 1, 
                                                                                                                                                                                                                                                                        `A few times a week` = 2, `A few times a month` = 3, `Once a month` = 4, 
                                                                                                                                                                                                                                                                        `Every few months` = 5, `Less often` = 6, `I never do this activity` = 7
                                                                                                              ), class = c("haven_labelled", "vctrs_vctr", "double")), A1_3 = structure(c(1, 
                                                                                                                                                                                          1, 2, 7, 7, 6, 1, 7, 2, 1), label = "Use TikTok", format.spss = "F1.0", display_width = 1L, labels = c(`Every day` = 1, 
                                                                                                                                                                                                                                                                                                                                                 `A few times a week` = 2, `A few times a month` = 3, `Once a month` = 4, 
                                                                                                                                                                                                                                                                                                                                                 `Every few months` = 5, `Less often` = 6, `I never do this activity` = 7
                                                                                                                                                                                          ), class = c("haven_labelled", "vctrs_vctr", "double"))), row.names = c(NA, 
                                                                                                                                                                                                                                                                  -10L), class = c("tbl_df", "tbl", "data.frame"), label = "TvTPQ2_0.sav")

df_new <- data.frame(matrix(0, ncol = ncol(df_old), nrow = nrow(df_old)  )) 
names(df_new) <- paste0(names(df_old),"_new")

方法1：基础循环+直接引用列 #

先提取df_old的所有变量标签，再通过循环直接给df_new的列赋值，注意用[[j]]而非[,j]来避免复制副本的问题：

# 提取df_old所有变量的标签
labels_old <- sapply(df_old, attr, "label")

# 循环赋值给df_new的对应列
for(j in seq_along(df_new)) {
  attr(df_new[[j]], "label") <- labels_old[j]
}

方法2：用haven包的set_label函数（更直观） #

如果使用haven包处理带标签的数据，可以直接用set_label函数批量设置：

library(haven)

labels_old <- sapply(df_old, attr, "label")
for(j in seq_along(df_new)) {
  df_new[[j]] <- set_label(df_new[[j]], labels_old[j])
}

方法3：purrr批量处理（简洁版） #

借助purrr包的map2函数可以一行完成批量赋值：

library(purrr)
library(dplyr)

labels_old <- sapply(df_old, attr, "label")
df_new <- map2(df_new, labels_old, function(col, lbl) {
  attr(col, "label") <- lbl
  col
}) %>% bind_cols()

为什么原代码不生效？ #

原代码用df_new[,j]提取列时，返回的是向量的副本，修改副本的属性不会影响原dataframe中的列。而用df_new[[j]]是直接引用原dataframe中的列，修改属性会直接生效。

内容的提问来源于stack exchange，提问作者Shaun