R中是否有一个函数可以同时改变几个相似的因子水平？

Question

我在堆栈溢出时的第二个问题，欢迎所有提示:)

对于临床研究，我必须重新编码许多二分基线特征，其中包含“是”和“否”的多种变体。

目前我正在一个一个地重新编码这些变量，但是它需要很多行代码并且所有不同变量之间的变化非常相似。在未知或 NA 的情况下，我想重新编码为 0。

例子

library(dplyr)

A <- c("Yes", "y", "no", "n", "UK")
B <- c("yes", "Yes", "y", "no", "no")
C <- c("Y", "y", "n", "no", "uk")

#attempt 1 was to recode all variables one by one

A <- recode(A,  "Yes" = "yes", "y" = "yes", "n" = "no", "UK" = "no")
B <- recode (B, "Yes" = "yes", "y" = "yes")
C <- recode(C, "Y" = "yes", "y" = "yes", "n" = "no", "uk" = "no")

#attempt 2 was to use a list option on all vectors.

levels(A) <- list("yes"=c("Likely", "y", "Y", "Yes", "yes"), "no" = c("", "No", "UK", "no", "N", "n"))

我想知道是否有一种方法可以在包含所有 A、B、C 的 list/vector 上执行此列表选项？或者也许有另一种方法可以更轻松、更有效地重新编码这些变量？

任何帮助都会很棒 :)

Answer 1

如果向量长度相同，您可以将它们放入数据框中，或者如果它们长度不同，则将它们放入列表中，然后使用 lapply 对所有向量应用相同的函数。您可以使用 forcats::fct_collapse 将多个级别合并为一个级别。

list_vec <- list(A, B, C)

list_vec <- lapply(list_vec, function(x) forcats::fct_collapse(x, 
            "yes"=c("Likely", "y", "Y", "Yes", "yes"), 
            "no" = c("", "No", "UK", "no", "N", "n", "uk")))

Answer 2

您可以从向量中使用 grepl 到 select yes 或 no。

c("0","no","yes")[1 + grepl("^no?", A, TRUE) + 2*grepl("^ye?s?", A, TRUE)]
#[1] "yes" "yes" "no"  "no"  "0"

要对许多向量进行此操作，您可以使用如下循环：

for(x in c("A","B","C")) {
  assign(x, c("0","no","yes")[1 + grepl("^no?", get(x), TRUE) +
                              2*grepl("^ye?s?", get(x), TRUE)])
}
A
#[1] "yes" "yes" "no"  "no"  "0"  
B
#[1] "yes" "yes" "yes" "no"  "no" 
C
#[1] "yes" "yes" "no"  "no"  "0"

R中是否有一个函数可以同时改变几个相似的因子水平？

Is there a function in R to change several similar factor levels at once?

r

levels

recode