我正在处理包含值标签的调查数据。Haven 包允许导入具有值标签属性的数据。有时这些价值标签需要以常规方式进行编辑。
我在这里给出的示例非常简单,但我正在寻找一种可以应用于大型 data.frames 中的类似问题的解决方案。
d <- dput(structure(list(var1 = structure(c(1, 2, NA, NA, 3, NA, 1, 1), labels = structure(c(1,
2, 3, 8, 9), .Names = c("Protection of environment should be given priority",
"Economic growth should be given priority", "[DON'T READ] Both equally",
"[DON'T READ] Don't Know", "[DON'T READ] Refused")), class = "labelled")), .Names = "var1", row.names = c(NA,
-8L), class = c("tbl_df", "tbl", "data.frame")))
d$var1
<Labelled double>
[1] 1 2 NA NA 3 NA 1 1
Labels:
value label
1 Protection of environment should be given priority
2 Economic growth should be given priority
3 [DON'T READ] Both equally
8 [DON'T READ] Don't Know
9 [DON'T READ] Refused
如果值标签以“[DON'T READ]”开头,我想从标签的开头删除“[DON'T READ]”并在末尾添加“(VOL)”。因此,“[DON'T READ] Both 同样”现在将变为“Both equal (VOL)”。
当然,使用 Haven 的相关标签包中的函数编辑这个单独的变量很简单。但我想将此解决方案应用于 data.frame 中的所有变量。
library(labelled)
val_labels(d$var1) <- c("Protection of environment should be given priority" = 1,
"Economic growth should be given priority" = 2,
"Both equally (VOL)" = 3,
"Don't Know (VOL)" = 8,
"Refused (VOL)" = 9)
如何以可以应用于 data.frame 中的每个变量的方式直接实现上面函数的结果?
无论 具体值如何,该解决方案都必须有效。(在这种情况下,需要更改的是值 3,8 和 9,但不一定是这种情况)。