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，我想从 sr 列中提取前两个 characters。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，它具有 sr 的前两个字符值，即 AP、SP 和 MP .

我该怎么做？

示例数据

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

对于最后一个选项，它应该是大写字符 ([A-Z]) 而不是小写字符 ([a-z])，因为输入 'sr' 列仅显示大写。此外，当模式多次出现时使用 str_extract_all，它 returns 和 list（默认情况下为 simplify = FALSE）。在这里，示例显示了一次出现，因此 str_extract 会更有用，因为它 returns a vector

library(dplyr)
library(stringr)
df %>% 
   mutate(permit_type = str_extract(sr, "\b[A-Z]{2}"))
# A tibble: 9 × 5
  date_received  date_approved sr        permit                     permit_type
  <chr>          <chr>         <chr>     <chr>                      <chr>      
1 "11/30/2021  " 11/30/2021    AP-21-080 "AP1766856 Classroom C"    AP         
2 "11/30/2021  " 11/30/2021    SP-21-081 "AP1766858 Classroom A"    SP         
3 "11/30/2021  " 11/30/2021    AP-21-082 "AP1766862 Landscape Area" AP         
4 "11/30/2021  " 11/30/2021    SP-21-083 "AP1766864 Classroom B"    SP         
5 "11/30/2021  " 11/30/2021    MP-21-084 "AO1766867"                MP         
6 "11/17/2021  " 11/17/2021    AP-21-085 "06-SE-2420566"            AP         
7 "12/3/2021  "  12/3/2021     AP-21-086 "06-E-2425187"             AP         
8 "12/3/2021  "  12/3/2021     MP-21-087 ""                         MP         
9 "12/13/2021  " 12/3/2021     SP-21-088 "06-SM-2424110"            SP

通过 str_split_fixed 直接应用于数据，我们可以将调用包装在 {}

内

df%>% 
   {str_split_fixed(.$sr, "-", 2)[,1]} 
[1] "AP" "SP" "AP" "SP" "MP" "AP" "AP" "MP" "SP"

第二种情况类似

df%>% 
  {str_extract(.$sr, "^.{2}")}
[1] "AP" "SP" "AP" "SP" "MP" "AP" "AP" "MP" "SP"

Answer 2

在 Base R 中，您可以使用：

transform(df, permit_type = substr(sr,1,2))

  date_received date_approved        sr                   permit permit_type
1  11/30/2021      11/30/2021 AP-21-080    AP1766856 Classroom C          AP
2  11/30/2021      11/30/2021 SP-21-081    AP1766858 Classroom A          SP
3  11/30/2021      11/30/2021 AP-21-082 AP1766862 Landscape Area          AP
4  11/30/2021      11/30/2021 SP-21-083    AP1766864 Classroom B          SP
5  11/30/2021      11/30/2021 MP-21-084                AO1766867          MP
6  11/17/2021      11/17/2021 AP-21-085            06-SE-2420566          AP
7   12/3/2021       12/3/2021 AP-21-086             06-E-2425187          AP
8   12/3/2021       12/3/2021 MP-21-087                                   MP
9  12/13/2021       12/3/2021 SP-21-088            06-SM-2424110          SP

R从数据框中的列中提取前两个字符

R Extract first two characters from a column in a dataframe

r

stringr