如何使用从 R 中现有列中提取的名称将列添加到 data.frame？

Question

我有 DF data.frame。我想添加另一个 column (i.e., call it station_no)，它将 extract number 在 Variables column 的 underscore 之后 number。

library(lubridate)
library(tidyverse)

set.seed(123)

DF <- data.frame(Date = seq(as.Date("1979-01-01"), to = as.Date("1979-12-31"), by = "day"),
                 Grid_2 = runif(365,1,10), Grid_20 = runif(365,5,15)) %>% 
      pivot_longer(-Date, names_to = "Variables", values_to = "Values")

期望输出：

DF_out <- data.frame(Date = c("1979-01-01","1979-01-01"),Variables = c("Grid_2","Grid_20"), 
                     Values = c(0.95,1.3),    Station_no = c(2,20))

Answer 1

简单的选项是 parse_number 其中 returns 数值转换值

library(dplyr)
DF %>% 
   mutate(Station_no  = readr::parse_number(Variables))

或使用 str_extract（如果我们想按模式进行）

library(stringr)
DF %>%
   mutate(Station_no  = str_extract(Variables, "(?<=_)\d+"))

或使用base R

DF$Station_no <-  trimws(DF$Variables, whitespace = '\D+')

Answer 2

一个base R解决方案是：

#Code
DF$Station_no <- sub("^[^_]*_", "", DF$Variables)

输出（一些行）：

# A tibble: 730 x 4
   Date       Variables Values Station_no
   <date>     <chr>      <dbl> <chr>     
 1 1979-01-01 Grid_2      3.59 2         
 2 1979-01-01 Grid_20    12.8  20        
 3 1979-01-02 Grid_2      8.09 2         
 4 1979-01-02 Grid_20     6.93 20        
 5 1979-01-03 Grid_2      4.68 2         
 6 1979-01-03 Grid_20     5.18 20        
 7 1979-01-04 Grid_2      8.95 2         
 8 1979-01-04 Grid_20     9.07 20        
 9 1979-01-05 Grid_2      9.46 2         
10 1979-01-05 Grid_20     9.83 20        
# ... with 720 more rows

如何使用从 R 中现有列中提取的名称将列添加到 data.frame？

How add a column to a data.frame with name extracted from an existing column in R?

r

columnname

dataframe

lubridate

tidyverse