R:多个连续 gsub 的函数式方法

Question

我一直在纠结这个问题，尝试了很多种类的 map , Reduce并且到目前为止还没有成功。

我正在寻找一种功能性、优雅的方法来替换 gsub 的序列如

text_example <- c(
    "I'm sure dogs are the best", 
    "I won't, I can't think otherwise", 
    "We'll be happy to discuss about dogs",
    "cant do it today tho"
)

text_example %>%
    gsub(pattern = "'ll", replacement = " will") %>%
    gsub(pattern = "can'?t", replacement = "can not") %>%
    gsub(pattern = "won'?t", replacement = "will not") %>%
    gsub(pattern = "n't", replacement = " not") %>%
    gsub(pattern = "'m", replacement = " am") %>%
    gsub(pattern = "'s", replacement = " is") %>%
    gsub(pattern = "dog", replacement = "cat") %>%

变成某种形式

text_example %>% 
    ???(dict$pattern, dict$replacement, gsub())

其中，为了可重现的示例， dict可以是一个data.frame，如

dict <- structure(
    list(
      pattern = c("'ll", "can'?t", "won'?t", "n't", "'m", "'s", "dog"), 
      replacement = c(" will", "can not", "will not", " not", " am", " is", "cat")
    ), 
    row.names = c(NA, -7L), 
    class = "data.frame"
) 

(我知道执行的替换在语言上可能不正确，但这不是现在的问题)

当然，残酷的

for(i in seq(nrow(dict))) {

    text_example <- gsub(dict$pattern[i], dict$replacement[i], text_example)

}

会工作，并且 我知道有许多库可以通过某些特定功能解决此问题。但我想了解如何以简单、实用的方式处理递归和此类问题 ，尽可能接近基础 R。我爱我的 lambdas!

预先感谢您的帮助。

Answer 1

您可以使用 mapply对于并行应用效果:

mapply(dict$pattern, dict$replacement, function(pttrn, rep) gsub(pttrn, rep, text_example))

(您可能想使用 SIMPLIFY=FALSE )