r - 通过将列保存在列表中来跨列应用 grepl？

Question

我有一个格式如下所示的数据框:

                  String  Keyword                           
1 Apples bananas mangoes   mangoes                    
2 Apples bananas mangoes   bananas                    
3 Apples bananas mangoes   peach   
.....  

它是一个数据框(50000 多行)。我目前正在批量手动使用ifelse语句。

data$Result<- ifelse(grepl("apples",data$String,ignore.case = TRUE)==TRUE,"apples",  
              ifelse(grepl("bananas",data$String,ignore.case = TRUE)==TRUE,"bananas",
               ifelse(grepl("mangoes",data$String,ignore.case = TRUE)==TRUE,"mangoes","unavailable")))


                String    Keyword Result
Apples bananas mangoes    mangoes mangoes  
Apples bananas mangoes    bananas bananas  
Apples bananas mangoes    peach   unavailable

有没有办法，我可以将字符串和关键字存储在一个列表中，然后对整个列表应用 grepl？

Answer 1

这是一个结合了data.table 和stringi 包的简单高效的解决方案:

library(data.table)
library(stringi)
setDT(df)[stri_detect_fixed(String, Keyword, case_insensitive = TRUE), result := Keyword]
#                    String Keyword  result
# 1: Apples bananas mangoes mangoes mangoes
# 2: Apples bananas mangoes bananas bananas
# 3: Apples bananas mangoes   peach      NA

---

或者，data.table-only 版本:

library(data.table)
setDT(df)[, result := Keyword[grep(Keyword, String, ignore.case = TRUE)], by = .(Keyword, String)]

---

基准

这是针对 mapply 答案的 5e5 数据集的基准测试。 (for 循环答案还没有运行完):

set.seed(123)
df1 <- data.frame(String = rep('Apples bananas mangoes', 5e5),
                  Keyword = sample(c("mangoes", "bananas", "peach"), 5e5, replace = TRUE))


system.time(df1$result2 <- ifelse(mapply(grepl,df1$Keyword, df1$String, ignore.case = TRUE), as.character(df1$Keyword), "Unavailable"))
# user  system elapsed 
# 40.78    0.02   41.12 
system.time(setDT(df1)[stri_detect_fixed(String, Keyword, case_insensitive = TRUE), result3 := Keyword])
# user  system elapsed 
# 0.52    0.01    0.53