Question

我想根据字符串的结尾创建一个新列Trial。例如，以 2 结尾的字符（例如A3-H2或A9-H2）将被视为试用 2 ，而那些不以数字为结尾的字符A3-H或A9-H将被视为试用版1.这应该是一个简单的ifelse语句，但我不知道如何根据字符串的结尾来执行此操作。

它会离开：

      Plant     Trtmt    
1:     SC       A3-H
2:     SC       A3-H2
3:     SC       A9-H
4:     SC       A9-H2

对此：

      Plant     Trtmt    Trial
1:     SC       A3-H       1
2:     SC       A3-H2      2
3:     SC       A9-H       1
4:     SC       A9-H2      2

真实数据：

dput(stack.df)
structure(list(Plant = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("SC", 
"W"), class = "factor"), Trtmt = c("A3-H", "A3-H", "A3-H", "A3-H", 
"A3-H", "A9-H", "A9-H", "A9-H", "A9-H", "A9-H", "A3-H2", "A3-H2", 
"A3-H2", "A3-H2", "A3-H2", "A9-H2", "A9-H2", "A9-H2", "A9-H2", 
"A9-H2")), .Names = c("Plant", "Trtmt"), row.names = c(6L, 7L, 
8L, 9L, 10L, 16L, 17L, 18L, 19L, 20L, 66L, 67L, 68L, 69L, 70L, 
76L, 77L, 78L, 79L, 80L), class = "data.frame")

Answer 1

library(tidyverse)

stack.df <- stack.df %>% 
  mutate(Trial = ifelse(grepl("2$", Trtmt), 2, 1))

Answer 2

library(dplyr)
library(stringr)
stack.df %>% mutate(Trial = ifelse(str_sub(Trtmt,-1)=="2", 2, 1))

Answer 3

以下是str_extract

的一个选项

library(stringr)
library(data.table)
setDT(stack.df)[, Trial := pmax(as.numeric(str_extract(Trtmt, "\\d+$")), 1, na.rm = TRUE)]

Answer 4

您可以使用substr(stack.df$Trtmt,nchar(stack.df$Trtmt)-1,nchar(stack.df$Trtmt))

获取最后一个字符

就像你说的那样，从那里开始很简单： - ）

根据R中字符串的结尾创建新列

4 个答案: