2

我想在 和 之间重复字符串的某些部分，]作为由前面;分隔的元素的数量within 。因此，所需的输出将是. 任何开始的提示。谢谢;[][A1, AB11; A2, AB22] I1, C1[A1, AB11] I1, C1; [A2, AB22] I1, C1

df1 <-
  data.frame(
   String = c(
    "[A1, AB11; A2, AB22] I1, C1; [A3, AB33] I3, C1"
  , "[A4, AB44] I4, C4; [A5, AB55; A6, AB66; A7, AB77] I7, C7"
  )
  )
df1

                                                    String
1           [A1, AB11; A2, AB22] I1, C1; [A3, AB33] I3, C1
2 [A4, AB44] I4, C4; [A5, AB55; A6, AB66; A7, AB77] I7, C7


df2 <-
  data.frame(
   String = c(
    "[A1, AB11] I1, C1; [A2, AB22] I1, C1; [A3, AB33] I3, C1"
  , "[A4, AB44] I4, C4; [A5, AB55] I7, C7;[A6, AB66] I7, C7; [A7, AB77] I7, C7"
  )
  )

df2

                                                                     String
1                   [A1, AB11] I1, C1; [A2, AB22] I1, C1; [A3, AB33] I3, C1
2 [A4, AB44] I4, C4; [A5, AB55] I7, C7;[A6, AB66] I7, C7; [A7, AB77] I7, C7
4

3 回答 3

2

我过去曾尝试过类似的东西，并认为调整 usingglue和ungluepackages 可能会很有趣。

首字母strsplit以分号分隔，忽略括号之间的分号。

对于每一行，这unglue将分离出括号之间重复的内容，以及括号外附加的内容。

library(glue)
library(unglue)
library(purrr)

my_fun <- function(inside, outside) {
  glue("[{inside}] {outside}")
}

sapply(strsplit(df1$String, '\\[[^]]*\\](*SKIP)(*F)|;\\s', perl = T), function(x) {
  ud <- unglue_data(x, patterns = "[{Inside}] {Outside}")
  ud_in <- map(ud[['Inside']], strsplit, split = "; ")
  ud_map <- map(seq_along(ud[['Inside']]), function(y) {
    map2(unlist(ud_in[y]), ud[['Outside']][y], my_fun)
  })
  paste(unlist(ud_map), collapse = '; ')
})

输出

[1] "[A1, AB11] I1, C1; [A2, AB22] I1, C1; [A3, AB33] I3, C1"                   
[2] "[A4, AB44] I4, C4; [A5, AB55] I7, C7; [A6, AB66] I7, C7; [A7, AB77] I7, C7"
于 2020-11-24T21:59:47.720 回答
2

这是一个基本的 R 解决方案：

sapply(strsplit(paste0(df1$String, ";"), "\\[|\\]"), function(x) {
  for(i in seq_along(x))
  {
    if(i %% 2 == 0) {
      x[i] <- paste0("[", gsub(";", paste0("]", x[i + 1], " ["), x[i]), "]")
    }
  }
  paste(x, collapse = "")
})
#> [1] "[A1, AB11] I1, C1;  [ A2, AB22] I1, C1; [A3, AB33] I3, C1;"                   
#> [2] "[A4, AB44] I4, C4; [A5, AB55] I7, C7; [ A6, AB66] I7, C7; [ A7, AB77] I7, C7;"
于 2020-11-24T19:14:00.540 回答
1

不是最整洁的解决方案，但它使用的是 stringr

str_split(df1$String, ";(?= *\\[)") %>%
  map(str_match, "\\[(.+?)\\] (.+)") %>%
   map( ~ paste(unlist(map2(paste0(str_split(.x[,2], "; ?")), .x[,3], ~ paste0("[", .x,"] ",.y ))), collapse="; "))

更好的解决方案：

as_tibble(df1) %>%
  mutate(splits=str_split(String, "; *(?=\\[)")) %>%
   unnest_longer(col=splits) %>%
    mutate(splits=map(str_split(splits,"\\[|\\] ?"), str_split, "; ?"))  %>%
     unnest_wider(splits) %>%
      mutate(val=map2(...2, ...3, ~ paste0("[", .x ,"] ", .y, collapse="; ") )) %>%
       group_by(String) %>%
        summarise(val=paste0(val, collapse="; "))
# A tibble: 2 x 2
  String                             val                                        
  <fct>                              <chr>
1 [A1, AB11; A2, AB22] I1, C1; [A3,… [A1, AB11] I1, C1; [A2, AB22] I1, C1; [A3, AB33] I3, C1
2 [A4, AB44] I4, C4; [A5, AB55; A6,… [A4, AB44] I4, C4; [A5, AB55] I7, C7; [A6, AB66] I7, C7; [A7, AB77] I7, C7
于 2020-11-24T19:11:32.537 回答