我有一个看起来像这样的数据框:
df1 <- data.frame(V1=rnorm(n = 100, mean=0, sd=1),
Edu=sample(x = c(-999,12,13,14,16,1), size = 100,
replace = T, prob = c(0.05,0.2,.2,0.2,0.2,0.15)))
我想转换变量
Edu
到有序因子变量。我可以使用以下代码将其转换为字符变量:lutedu <- c('-999' = NA, '12' = "High School", '13' = "Associate's",
'14' = "Associate's", '16' = "Bachelor's",
'18' = "Master's, Graduate/professional", '21' = "PhD")
df1$Edu <- lutedu[as.character(df1$Edu)]
从那里我可以使用
ordered()
将字符变量转换为有序因子:df1$Edu <-
ordered(
x = df1$Edu, levels = c(
"High School", "Associate's", "Bachelor's",
"Master's, Graduate/professional", "PhD"
)
)
有没有更好的方法来做到这一点?
最佳答案
而不是使用命名向量重新编码然后调用 ordered
,您可以调用 ordered
为自己省一步并同时使用 levels
和 labels
论据:
ordered(edu, levels=c(-999, 12, 13, 14, 16, 1),
labels=c("NA", "High School", "Associate's", "Bachelor's",
"Master's/Graduate", "PhD"))
# [1] High School Master's/Graduate Master's/Graduate Bachelor's Associate's
# [6] Master's/Graduate High School Master's/Graduate High School PhD
# ...
数据 :
set.seed(144)
edu <- sample(x = c(-999,12,13,14,16,1), size = 100,
replace = T, prob = c(0.05,0.2,.2,0.2,0.2,0.15))
关于r - 将数字变量转换为有序因子的最佳方法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32183314/