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如何使用R在另一个字符串向量中提取字符串向量的外观?

stringi stringr regex r

2

Ben G · 技术社区 · 8 年前

我有一个这样的弦向量:

strings <- tibble(string = c("apple, orange, plum, tomato",
                             "plum, beat, pear, cactus",
                             "centipede, toothpick, pear, fruit"))

我有一个水果载体:

fruits <- tibble(fruit =c("apple", "orange", "plum", "pear"))

我想要的是一个data.frame/tibble和原始的 strings 数据框,包含原始列中所有水果的第二个列表或字符列。像这样的东西。

strings <- tibble(string = c("apple, orange, plum, tomato",
                             "plum, beat, pear, cactus",
                             "centipede, toothpick, pear, fruit"),
                   match = c("apple, orange, plum",
                             "plum, pear",
                             "pear")
                  )

我试过了 str_extract(strings, fruits) 我得到了一个清单,上面所有东西都是空白的,还有警告:

Warning message:
In stri_detect_regex(string, pattern, opts_regex = opts(pattern)):
longer object length is not a multiple of shorter object length

我试过了 str_extract_all(strings, paste0(fruits, collapse = "|")) 我得到和得到同样的警告信息。

我看过这个 Find matches of a vector of strings in another vector of strings ,但这似乎没有帮助。

任何帮助都将不胜感激。

3 回复 | 直到 8 年前

1

2

Kara Woo 8 年前

这里有一个选择。首先我们将 string 列成单独的字符串(现在 "apple, orange, plum, tomato" 都是一个字符串)。然后我们将字符串列表与 fruits$fruit 列并将匹配值的列表存储在新的 fruits 列。

library("tidyverse")
strings <- tibble(
  string = c(
    "apple, orange, plum, tomato",
    "plum, beat, pear, cactus",
    "centipede, toothpick, pear, fruit"
  )
)

fruits <- tibble(fruit =c("apple", "orange", "plum", "pear"))

strings %>%
  mutate(str2 = str_split(string, ", ")) %>%
  rowwise() %>%
  mutate(fruits = list(intersect(str2, fruits$fruit)))
#> Source: local data frame [3 x 3]
#> Groups: <by row>
#> 
#> # A tibble: 3 x 3
#>   string                            str2      fruits   
#>   <chr>                             <list>    <list>   
#> 1 apple, orange, plum, tomato       <chr [4]> <chr [3]>
#> 2 plum, beat, pear, cactus          <chr [4]> <chr [2]>
#> 3 centipede, toothpick, pear, fruit <chr [4]> <chr [1]>

于2018-08-07由 reprex package (第0.2.0版)。

2

user5176207user5176207 8 年前

下面是一个使用purr的例子

strings <- tibble(string = c("apple, orange, plum, tomato",
                         "plum, beat, pear, cactus",
                         "centipede, toothpick, pear, fruit"))

fruits <- tibble(fruit =c("apple", "orange", "plum", "pear"))

extract_if_exists <- function(string_to_parse, pattern){
  extraction <- stringi::stri_extract_all_regex(string_to_parse, pattern)
  extraction <- unlist(extraction[!(is.na(extraction))])
  return(extraction)
}

strings %>%
  mutate(matches = map(string, extract_if_exists, fruits$fruit)) %>%
  mutate(matches = map(string, str_c, collapse=", ")) %>%
  unnest

3

1

s_baldur 8 年前

下面是base-R解决方案:

strings[["match"]] <- 
  sapply(
    strsplit(strings[["string"]], ", "), 
    function(x) {
      paste(x[x %in% fruits[["fruit"]]], collapse = ", ")
    }
  )

导致:

  string                            match              
  <chr>                             <chr>              
1 apple, orange, plum, tomato       apple, orange, plum
2 plum, beat, pear, cactus          plum, pear         
3 centipede, toothpick, pear, fruit pear