使用 purrr::map 在 for 循環中重寫向量

Question

如何使用purrr::map()返回示例的 for 循環的結果：

vct_string <- c("old ccar","new carr", "house", "oold house")

df_correction <- data.frame(
  pattern  = c("ccar", "carr", "oold"),
  replacement = c("car", "car", "old"),
  stringsAsFactors = FALSE
)

for(i in 1:nrow(df_correction)){
  vct_string <- pmap(df_correction, gsub, x = vct_string)[[i]]
}

> vct_string
[1] "old car"   "new car"   "house"     "old house"

Answer 1

實際上，您不需要任何ReduceMap函數。 只需使用str_replace_all其矢量化

library(stringr)
str_replace_all(vct_string, set_names(df_correction$replacement, df_correction$pattern))

[1] "old car"   "new car"   "house"     "old house"

Answer 2

你必須遞歸地修改你的向量，所以在我看來，這是使用 reduce 系列 function 的經典案例。 所以這樣做，你必須將你的向量傳遞給purrr::reduce的.init參數以獲得所需的 output

purrr::reduce(seq(nrow(df_correction)), .init = vct_string, ~ gsub(df_correction$pattern[.y], df_correction$replacement[.y], .x))

#> [1] "old car"   "new car"   "house"     "old house"

這甚至會在給定向量的元素中進行多次替換。 看到這個

#modified example
vct_string <- c("old ccar","new carr", "house", "oold carr")

purrr::reduce(seq(nrow(df_correction)), .init = vct_string, ~ gsub(df_correction$pattern[.y], df_correction$replacement[.y], .x))
[1] "old car" "new car" "house"   "old car"

Answer 3

首先寫一個 function 用於替換

word_correct <- function(string) {
  df_correction <- data.frame(
    pattern  = c("old ccar", " new carr", "oold house", "house"), # changed from OP
    replacement = c("car", "car", "old", "house"),
    stringsAsFactors = FALSE
  )
  df_correction[ which(df_correction$pattern == string), "replacement"]
}

# Testing
word_correct("oold")
word_correct("ccar")

然后你可以將 function 作為參數傳遞給purrr::map

map_chr(vct_string, word_correct) # using map_chr to return a vector instead of a list which is what map returns

由於您使用映射表來替換單個單詞，因此您實際上可以在第二個 function 中使用map來獲得您想要的結果。

vct_string <- c("old ccar","new carr", "house", "oold house")

single_word_correct <- function(string) {
  
  df_correction <- data.frame(
    pattern  = c("ccar", "carr", "oold"),
    replacement = c("car", "car", "old"),
    stringsAsFactors = FALSE
  )
  if(string %in% df_correction$pattern){
    df_correction[ which(df_correction$pattern == string), "replacement"]
  } else {
    string
  }
}
multi_word_correct <- function(string){
  strings <- strsplit(string, " ")[[1]]
  paste(map_chr(strings, single_word_correct), collapse = " ")
}

map_chr(vct_string, multi_word_correct)

Answer 4

下面是使用 base::Reduce 的方法：

Reduce(function(x, y) {
  gsub(df_correction[y, 1], df_correction[y, 2], x)
}, init = vct_string, 1:nrow(df_correction))

[1] "old car"   "new car"   "house"     "old house"