给出一个向量v <- seq(0,23)
。
我想将它转换为具有以下级别的因子:
0 = All numbers from 0 to 4
1 = All numbers from 5 to 8
2 = Numbers 9, 11, 23
3 = Others
请指导我该怎么做?
我做过factor(v, levels = as.character(seq(0,23)))
。然后
levels(v[levels(v) %in% as.character(seq(0,4))] <- "0"
levels(v[levels(v) %in% as.character(seq(5,8))] <- "1"
levels(v[levels(v) %in% as.character(c(9,11,23)] <- "2"
答案 0 :(得分:1)
几个选项
v <- 0:23
v <- factor(v)
levels(v) <-
list("0 - 4" = 0:4,
"5 - 8" = 5:8,
"9, 11, 23" = c(9, 11, 23),
"Others" = levels(v)[!levels(v) %in% c(0:9, 11, 23)])
forcats
包(tidyverse范例)library(magrittr)
library(forcats)
v <- 0:23 %>%
factor() %>%
fct_collapse("0 - 4" = as.character(0:4),
"5 - 8" = as.character(5:8),
"9, 11, 23" = as.character(c("9", "11", "23"))) %>%
fct_other(keep = c("0 - 4", "5 - 8", "9, 11, 23"),
other_level = "Others")
答案 1 :(得分:1)
不同的Base R解决方案
v <- 0:23
v <- factor(ifelse(v <= 4, "0",
ifelse(v > 4 & v <= 8, "1",
ifelse(v %in% c(9,11,23), "2", "3"))))