删除方括号、括号和/或大括号内的文本

Question

我需要一个函数来提取任何类型的括号ie ()，[]，{}和中间的信息。我创建了它，并让它做我想做的事情，但我得到了一个恼人的警告，我真的不知道它是什么意思。我希望恼人的警告消失，要么修复我的代码错误，要么隐藏警告。我用suppressWarnings()尝试了一下，但它不起作用，因为我不认为我正确地使用了它。

此函数使用regmatches，需要R版本2.14或更高版本

下面是下面的函数和一个重现警告的示例。谢谢你的帮助。

################
# THE FUNCTION #
################
bracketXtract <- function(text, bracket = "all", include.bracket = TRUE) {

    bracketExtract <- if (include.bracket == FALSE) {
        function(Text, bracket) {
                  switch(bracket, 
                        square = lapply(Text, function(j) gsub("[\\[\\]]", "", 
                                 regmatches(j, gregexpr("\\[.*?\\]", j))[[1]], 
                                 perl = TRUE)), 
                        round =  lapply(Text, function(j) gsub("[\\(\\)]", "", 
                                 regmatches(j, gregexpr("\\(.*?\\)", j))[[1]])), 
                        curly =  lapply(Text, function(j) gsub("[\\{\\}]", "", 
                                 regmatches(j, gregexpr("\\{.*?\\}", j))[[1]])), 
                        all =    { P1 <- lapply(Text, function(j) gsub("[\\[\\]]", "", 
                                         regmatches(j, gregexpr("\\[.*?\\]", j))[[1]], 
                                         perl = TRUE))
                                   P2 <- lapply(Text, function(j) gsub("[\\(\\)]", "", 
                                         regmatches(j, gregexpr("\\(.*?\\)", j))[[1]]))
                                   P3 <- lapply(Text, function(j) gsub("[\\{\\}]", "", 
                                         regmatches(j, gregexpr("\\{.*?\\}", j))[[1]]))
                    apply(cbind(P1, P2, P3), 1, function(x) rbind(as.vector(unlist(x))))
                })
        }
    } else {
        function(Text, bracket) {
                  switch(bracket, 
                         square = lapply(Text, function(j) regmatches(j, 
                                  gregexpr("\\[.*?\\]", j))[[1]]), 
                         round =  lapply(Text, function(j) regmatches(j, 
                                  gregexpr("\\(.*?\\)", j))[[1]]), 
                         curly =  lapply(Text, function(j) regmatches(j, 
                                  gregexpr("\\{.*?\\}", j))[[1]]), 
                         all =    { P1 <- lapply(Text, function(j) regmatches(j, 
                                          gregexpr("\\[.*?\\]", j))[[1]])
                                    P2 <- lapply(Text, function(j) regmatches(j, 
                                          gregexpr("\\(.*?\\)", j))[[1]])
                                    P3 <- lapply(Text, function(j) regmatches(j, 
                                          gregexpr("\\{.*?\\}", j))[[1]])
                apply(cbind(P1, P2, P3), 1, function(x) rbind(as.vector(unlist(x))))
            })
        }
    }
    if (length(text) == 1) {
        unlist(lapply(text, function(x) bracketExtract(Text = text,
            bracket = bracket)))
    } else {
        sapply(text, function(x) bracketExtract(Text = text, 
            bracket = bracket))
    }
} 

##################
# TESTING IT OUT #
##################
j <- "What kind of cheese isn't your cheese? {wonder} Nacho cheese! [groan] (Laugh)"                                                          
bracketXtract(j, 'round')
bracketXtract(j, 'round', include.bracket = FALSE)

examp2<-data.frame(var1=1:4)                                                                                                                               
examp2$text<-as.character(c("I love chicken [unintelligible]!", "Me too! (laughter) It's so good.[interupting]", 
             "Yep it's awesome {reading}.", "Agreed."))

#=================================#
# HERE"S WHERE THE WARNINGS COME: #
#=================================#                                                                                                                                                            
examp2$text2<-bracketXtract(examp2$text, 'round')                                                                                                  
   examp2
examp2$text2<-bracketXtract(examp2$text, 'all')                                                                                                  
   examp2

Answer 1

也许这个函数更简单一些？或者至少更紧凑。

bracketXtract <-
    function(txt, br = c("(", "[", "{", "all"), with=FALSE)
{
    br <- match.arg(br)
    left <-        # what pattern are we looking for on the left?
        if ("all" == br) "\\(|\\{|\\["
        else sprintf("\\%s", br)
    map <-         # what's the corresponding pattern on the right?
        c(`\\(`="\\)", `\\[`="\\]", `\\{`="\\}",
          `\\(|\\{|\\[`="\\)|\\}|\\]")
    fmt <-         # create the appropriate regular expression
        if (with) "(%s).*?(%s)"
        else "(?<=%s).*?(?=%s)"
    re <- sprintf(fmt, left, map[left])
    regmatches(txt, gregexpr(re, txt, perl=TRUE))    # do it!
}

不需要lapply；正则表达式函数是以这种方式矢量化的。使用嵌套的圆括号会失败；如果这很重要，那么正则表达式可能不是一个好的解决方案。现在我们开始行动了：

> txt <- c("I love chicken [unintelligible]!",
+          "Me too! (laughter) It's so good.[interupting]",
+          "Yep it's awesome {reading}.",
+          "Agreed.")
> bracketXtract(txt, "all")
[[1]]
[1] "unintelligible"

[[2]]
[1] "laughter"    "interupting"

[[3]]
[1] "reading"

[[4]]
character(0)

这可以毫不费力地安装到data.frame中。

> examp2 <- data.frame(var1=1:4)
> examp2$text <- c("I love chicken [unintelligible]!",
+                  "Me too! (laughter) It's so good.[interupting]",
+                  "Yep it's awesome {reading}.", "Agreed.")
> examp2$text2<-bracketXtract(examp2$text, 'all')
> examp2
  var1                                          text                 text2
1    1              I love chicken [unintelligible]!        unintelligible
2    2 Me too! (laughter) It's so good.[interupting] laughter, interupting
3    3                   Yep it's awesome {reading}.               reading
4    4                                       Agreed.                      

您看到的警告与试图将矩阵插入数据框有关。我认为答案是“不要那样做”。

> df = data.frame(x=1:2)
> df$y = matrix(list(), 2, 2)
> df
  x    y
1 1 NULL
2 2 NULL
Warning message:
In format.data.frame(x, digits = digits, na.encode = FALSE) :
  corrupt data frame: columns will be truncated or padded with NAs

Answer 2

我的想法是制作6个(隐式矢量化)助手函数，但我将研究Martin的代码，因为他在这方面比我强得多：

rm.curlybkt.no <-function(x) gsub("(\\{).*(\\})", "\\1\\2", x, perl=TRUE)
rm.rndbkt.no <-  function(x) gsub("(\\().*(\\))", "\\1\\2", x, perl=TRUE)
rm.sqrbkt.no <-  function(x) gsub("(\\[).*(\\])", "\\1\\2", x, perl=TRUE)

rm.rndbkt.in <- function(x) gsub("\\(.*\\)", "", x)
rm.curlybkt.in <- function(x) gsub("\\{.*\\}", "", x)
rm.sqrbkt.in   <- function(x) gsub("\\[.*\\]", "", x)

Answer 3

假设括号不是嵌套的，并且我们有以下测试数据：

x <- c("a (bb) [ccc]{d}e", "x[a]y")

然后在gsubfn中使用strapply，我们得到了这个两行的解决方案，它首先将所有的圆括号和方括号转换为大括号，然后进行处理：

library(gsubfn)    

xx <- chartr("[]()", "{}{}", x)
s <- strapply(xx, "{([^}]*)}", c)

以上结果如下所示：

> s
[[1]]
[1] "bb"  "ccc" "d"  

[[2]]
[1] "a"