在 R 中识别两个字符串中的模式

Question

我想评估 ColA 是否包含比 ColB 新的字符串。但是，我对某些类型的字符串不感兴趣，例如 oil。我想要一个指标变量如下：

  ColA                           ColB                       Ind
--------------------------      ------------------------   -----
 coconut+grape+pine              grape+coconut              TRUE
 orange+apple+grape+pine         grape+coconut              TRUE
 grape+pine                      grape+oil                  TRUE
 oil+grape                       grape+apple                FALSE
 grape                           grape+oil                  FALSE
 grape+pine                      grape+orange+pine          FALSE

对 R 有什么建议吗？

非常感谢！

Answer 1

由于我们需要拆分字符串，我们将从strsplit开始，

strsplit(dat$ColA, '+', fixed = TRUE)
# [[1]]
# [1] "coconut" "grape"   "pine"   
# [[2]]
# [1] "orange" "apple"  "grape"  "pine"  
# [[3]]
# [1] "grape" "pine" 
# [[4]]
# [1] "oil"   "grape"
# [[5]]
# [1] "grape"
# [[6]]
# [1] "grape" "pine" 

从这里，我们想确定ColA 中的内容是什么，而不是ColB 中的内容。我将使用Map 在每个集合上运行setdiff（ColA 的[[1]] 和ColB 的[[1]] 等）。

Map(setdiff, strsplit(dat$ColA, '+', fixed = TRUE), strsplit(dat$ColB, '+', fixed = TRUE))
# [[1]]
# [1] "pine"
# [[2]]
# [1] "orange" "apple"  "pine"  
# [[3]]
# [1] "pine"
# [[4]]
# [1] "oil"
# [[5]]
# character(0)
# [[6]]
# character(0)

要确定哪个有“新词”，我们可以使用lengths(.) > 0检查非零长度：

lengths(Map(setdiff, strsplit(dat$ColA, '+', fixed = TRUE), strsplit(dat$ColB, '+', fixed = TRUE))) > 0
# [1]  TRUE  TRUE  TRUE  TRUE FALSE FALSE

但由于您不关心 oil，我们也需要删除它。

lapply(Map(setdiff, strsplit(dat$ColA, '+', fixed = TRUE), strsplit(dat$ColB, '+', fixed = TRUE)), setdiff, "oil")
# [[1]]
# [1] "pine"
# [[2]]
# [1] "orange" "apple"  "pine"  
# [[3]]
# [1] "pine"
# [[4]]
# character(0)
# [[5]]
# character(0)
# [[6]]
# character(0)
lengths(lapply(Map(setdiff, strsplit(dat$ColA, '+', fixed = TRUE), strsplit(dat$ColB, '+', fixed = TRUE)),
               setdiff, "oil")) > 0
# [1]  TRUE  TRUE  TRUE FALSE FALSE FALSE

---

@akrun 提出了一个 tidyverse 变体：

library(dplyr)
library(purrr)   # map2_lgl
library(stringr) # str_extract_all
dat %>%
  mutate(
    new = map2_lgl(
      str_extract_all(ColB, "\\w+"), str_extract_all(ColA, "\\w+"),
      ~ !all(setdiff(.y, "oil") %in% .x)
    )
  )
#                      ColA              ColB   Ind   new
# 1      coconut+grape+pine     grape+coconut  TRUE  TRUE
# 2 orange+apple+grape+pine     grape+coconut  TRUE  TRUE
# 3              grape+pine         grape+oil  TRUE  TRUE
# 4               oil+grape       grape+apple FALSE FALSE
# 5                   grape         grape+oil FALSE FALSE
# 6              grape+pine grape+orange+pine FALSE FALSE

---

数据

dat <- structure(list(ColA = c("coconut+grape+pine", "orange+apple+grape+pine", "grape+pine", "oil+grape", "grape", "grape+pine"), ColB = c("grape+coconut", "grape+coconut", "grape+oil", "grape+apple", "grape+oil", "grape+orange+pine"), Ind = c(TRUE, TRUE, TRUE, FALSE, FALSE, FALSE)), class = "data.frame", row.names = c(NA, -6L))

Answer 2

这是一个类似于r2evans's 的解决方案，它在do.call 的帮助下只调用一次strsplit。

rid <- function(x) x[!x %in% z]  ## helper FUN to get rid of the oil

z <- "oil"
L <- sapply(unname(dat), strsplit, "\\+")
dat$ind <- sapply(1:nrow(L), function(x) length(do.call(setdiff, rev(Map(rid, L[x,]))))) > 0
dat
#                  V1                      V2   ind
# 1     grape+coconut      coconut+grape+pine  TRUE
# 2     grape+coconut orange+apple+grape+pine  TRUE
# 3         grape+oil              grape+pine  TRUE
# 4       grape+apple               oil+grape FALSE
# 5         grape+oil                   grape FALSE
# 6 grape+orange+pine              grape+pine FALSE

---

数据：

dat <- structure(list(V1 = c("grape+coconut", "grape+coconut", "grape+oil", 
"grape+apple", "grape+oil", "grape+orange+pine"), V2 = c("coconut+grape+pine", 
"orange+apple+grape+pine", "grape+pine", "oil+grape", "grape", 
"grape+pine")), row.names = c(NA, -6L), class = "data.frame")