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r - 字符串按另一列的值拆分

问题描述

嗨我有这个数据框（DF1）

structure(list(Value = list("Peter", "John", c("Patric", "Harry")),Text = c("Hello Peter How are you","Is it John? Yes It is John, Harry","Hello Patric, how are you. Well, Harry thank you."))  , class = "data.frame", row.names = c(NA, -3L)) 

             Value                                              Text
1            Peter                           Hello Peter How are you
2             John                 Is it John? Yes It is John, Harry
3 c(Patric, Harry) Hello Patric, how are you. Well, Harry thank you.

我想通过 Value 中的名称拆分 Text 中的句子以获得这个

             Value                                              Text   Split
1            Peter                           Hello Peter How are you  c("Hello", "Peter How are you")
2             John                 Is it John? Yes It is John, Harry  c("Is it", "John? Yes It is John, Harry")
3 c(Patric, Harry) Hello Patric, how are you. Well, Harry thank you   c("Hello", "Patric, how are you. Well,", "Harry thank you")

我试过这个

DF1 %>% mutate(Split = strsplit(as.character(Text),as.character(Value)))

但它不起作用

标签： rstringdplyrstrsplit

解决方案

数据

假设这是真实的结构：

df <- structure(list(Value = list("Peter", "John", c("Patric", "Harry")),
                     Text = c("Hello Peter How are you","Is it John? Yes It is John, Harry","Hello Patric, how are you. Well, Harry thank you.")),
                class = "data.frame", row.names = c(NA, -3L))

第一个解决方案：双重for循环

您可以使用双 for 循环来解决您的问题。这可能是一个更易读的解决方案并且更容易调试。

library(stringr)

Split <- list()

for(i in seq_len(nrow(df))){
 
 text  <- df$Text[i]
 value <- df$Value[[i]]
 
 for(j in seq_along(value)){
  
  text2 <- str_split(text[length(text)], paste0("(?<=.)(?=", value[[j]], ")"), n = 2)[[1]]
  text <- c(text[-length(text)], text2)
  
 }
 
 Split[[i]] <- text
 
}

df$Split <- Split

如果您打印df，它看起来就像您有一个唯一的字符串，但事实并非如此。

df$Split
#> [[1]]
#> [1] "Hello "            "Peter How are you"
#> 
#> [[2]]
#> [1] "Is it "                      "John? Yes It is John, Harry"
#> 
#> [[3]]
#> [1] "Hello "                      "Patric, how are you. Well, " "Harry thank you."           
#>

第二种解决方案：tidyverse 和递归 fn

由于在您最初尝试使用dplyr函数时，您也可以使用递归函数以这种方式编写它。此解决方案不使用 for 循环。

library(stringr)
library(purrr)
library(dplyr)

str_split_recursive <- function(string, pattern){
 
 string <- str_split(string[length(string)], paste0("(?<=.)(?=", pattern[1], ")"), n = 2)[[1]]
 pattern <- pattern[-1]
 if(length(pattern) > 0) string <- c(string[-length(string)], str_split_recursive(string, pattern))
 string
 
}

df <- df %>% 
 mutate(Split = map2(Text, Value, str_split_recursive))

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