【发布时间】:2020-10-20 09:00:41
【问题描述】:
我的第二个问题是堆栈溢出,所以欢迎所有提示:)
对于临床研究,我必须重新编码许多二分法基线特征,其中包含“是”和“否”的几种变体。
目前我正在对这些变量一一重新编码,但它需要很多行代码,并且所有不同变量之间的变化非常相似。如果未知或不适用,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 的列表/向量上执行此列表选项?或者也许还有另一种方法可以让我更轻松、更高效地重新编码这些变量?
任何帮助都会很棒:)
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