在R時間序列資料框中，如何基于正則運算式進行分離和分類

您好，我有一個時間序列資料框，其中包含一個產品串列及其不同的稅率，我需要將其分為兩類：百分比數字（AV）和文本（沒有百分比數字（SPEC）的所有其他內容，由第一個分隔字符向量中的加號：

#note there are many more years
product <- c("01","02")
yr1<-c("0%","11.5%   190 GBP/100kg")
yr2<-c("0%","15%   190 GBP/100kg   MAX 8.5%/100kg")
yearnum =2

sched <- data.frame(product,yr1,yr2)

#where yearnum is the number of years
schedule<-c(paste0("yr",1:yearnum))
#categorize av and specific DUTY rates
for(j in 1:yearnum){
  for(i in schedule){
  sched <- sched %>% separate(i, c(paste0("av.yr",j), paste0("spec.yr",j)), " \\  ", remove=F, extra = "merge")}}

我試圖將它們分成下面的結果，但我的 for 回圈公式有問題。有人可以幫忙嗎？

#and the output should be
product <- c("01","02")
yr1<-c("0%","11.5%   190 GBP/100kg")
yr2<-c("0%","15%   190 GBP/100kg   MAX 8.5%/100kg") 
av.yr1<- c("0%","11.5%")
av.yr2 <-c("0%","15%")
spec.yr1 <-c("","190 GBP/100kg")
spec.yr2 <-c("","190 GBP/100kg   MAX 8.5%/100kg")

sched<-data.frame(product,yr1,yr2,av.yr1,av.yr2,spec.yr1,spec.yr2)

uj5u.com熱心網友回復：

如果您有很多年，我認為最好的做法是將您的資料轉換為長格式，使用separate或mutate與正則運算式一起使用，然后將 pivot_back 轉換為寬格式。

pivot_longer(sched, -product) %>%
  separate(value,into=c("av","spec"),sep = " [ ] ",extra = "merge") %>% 
  pivot_wider(names_from=name,values_from=av:spec,names_sep = ".")

輸出：

  product av.yr1 av.yr2 spec.yr1      spec.yr2                      
  <chr>   <chr>  <chr>  <chr>         <chr>                         
1 01      0%     0%     NA            NA                            
2 02      11.5%  15%    190 GBP/100kg 190 GBP/100kg   MAX 8.5%/100kg

這是一個使用 的選項mutate，它也保留了原始列：

pivot_longer(sched, -product, values_to = "yr", names_prefix = "yr") %>%
mutate(av.yr = str_extract(yr,"^\\d*[.]?\\d*%"),
       spec.yr = str_remove(yr, "^\\d*[.]?\\d*%( [ ] )?")) %>% 
pivot_wider(names_from=name, values_from=yr:spec.yr, names_sep = "")

輸出

  product yr1                   yr2                                  av.yr1 av.yr2 spec.yr1        spec.yr2                        
  <chr>   <chr>                 <chr>                                <chr>  <chr>  <chr>           <chr>                           
1 01      0%                    0%                                   0%     0%     ""              ""                              
2 02      11.5%   190 GBP/100kg 15%   190 GBP/100kg   MAX 8.5%/100kg 11.5%  15%    "190 GBP/100kg" "190 GBP/100kg   MAX 8.5%/100kg"

uj5u.com熱心網友回復：

您只需要遍歷一個索引：

library(tidyr)
#note there are many more years
product <- c("01","02")
yr1<-c("0%","11.5%   190 GBP/100kg")
yr2<-c("0%","15%   190 GBP/100kg   MAX 8.5%/100kg")
yearnum =2

sched <- data.frame(product,yr1,yr2)

#where yearnum is the number of years
schedule<-c(paste0("yr",1:yearnum))
#categorize av and specific DUTY rates
for(j in 1:yearnum){
  i <- schedule[j]
  sched <- sched %>% separate(i, c(paste0("av.yr",j), paste0("spec.yr",j)), 
                              " \\  ", remove=F, extra = "merge", fill = "right")
}
sched
#>   product                   yr1 av.yr1      spec.yr1
#> 1      01                    0%     0%          <NA>
#> 2      02 11.5%   190 GBP/100kg  11.5% 190 GBP/100kg
#>                                    yr2 av.yr2                       spec.yr2
#> 1                                   0%     0%                           <NA>
#> 2 15%   190 GBP/100kg   MAX 8.5%/100kg    15% 190 GBP/100kg   MAX 8.5%/100kg

^{由reprex 包于 2022-05-25 創建(v2.0.1)}

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