R 从数据框中的列中提取前两个字符

Question

我有一个包含多个的数据集，我想characters从该sr列中提取前两个。最后，这些字符将存储在一个新列中。

基本上，我想要一个新列permit_type，其中包含srieAP和SP的前两个字符值MP。

我怎样才能做到这一点？

样本数据

structure(list(date_received = c("11/30/2021  ", "11/30/2021  ", 
"11/30/2021  ", "11/30/2021  ", "11/30/2021  ", "11/17/2021  ", 
"12/3/2021  ", "12/3/2021  ", "12/13/2021  "), date_approved = c("11/30/2021", 
"11/30/2021", "11/30/2021", "11/30/2021", "11/30/2021", "11/17/2021", 
"12/3/2021", "12/3/2021", "12/3/2021"), sr = c("AP-21-080", "SP-21-081", 
"AP-21-082", "SP-21-083", "MP-21-084", "AP-21-085", "AP-21-086", 
"MP-21-087", "SP-21-088"), permit = c("AP1766856 Classroom C", 
"AP1766858 Classroom A", "AP1766862 Landscape Area", "AP1766864 Classroom B", 
"AO1766867", "06-SE-2420566", "06-E-2425187", "", "06-SM-2424110"
)), row.names = c(NA, -9L), class = c("tbl_df", "tbl", "data.frame"
))

方法一

library(tidyverse)
df$permit_type= df%>% str_split_fixed(df$sr, "-", 2)

# Error
Error in str_split_fixed(., df$sr, "-", 2) : 
  unused argument (2)

方法二

df$permit_type = df%>% str_extract(sr, "^.{2}")

# Error
Error in str_extract(., sr, "^.{2}") : unused argument ("^.{2}")

方法三

df = df %>%  mutate(permit_type = str_extract_all(sr, "\\b[a-z]{2}")) 

# Returns permit_type with `Character(0)` values

Answer 1

我刚刚意识到，实际上，您可以简单地使用substr()函数来提取前两个字符。按照 akrun 的正确答案，例如，

df$permit_type<- substr(df$sr, 1, 2)

如果这是有道理的话。