我需要一个提取任何类型括号的函数，即（），[]，{}以及它们之间的信息。我创造它并让它做我想要的但是我得到一个恼人的警告，我不知道它意味着什么。我希望通过修复代码的错误或抑制警告来消除恼人的警告。我尝试使用suppressWarnings（），但它不起作用，因为我认为我没有正确使用它。
此功能使用regmatches并需要R版本2.14或更高版本
以下是下面的功能以及重现警告的示例。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$text, 'all')
examp2
答案 0 :(得分：7)
也许这个功能更直接一点？或者至少更紧凑。
bracketXtract <-
function(txt, br = c("(", "[", "{", "all"), with=FALSE)
{
br <- match.arg(br)
left <- # what pattern are we looking for on the left?
if ("all" == br) "\\(|\\{|\\["
else sprintf("\\%s", br)
map <- # what's the corresponding pattern on the right?
c(`\\(`="\\)", `\\[`="\\]", `\\{`="\\}",
`\\(|\\{|\\[`="\\)|\\}|\\]")
fmt <- # create the appropriate regular expression
if (with) "(%s).*?(%s)"
else "(?<=%s).*?(?=%s)"
re <- sprintf(fmt, left, map[left])
regmatches(txt, gregexpr(re, txt, perl=TRUE)) # do it!
}
无需lapply;正则表达式函数以这种方式进行矢量化。这会因嵌套括号而失败;如果这很重要，正则表达式可能不是一个好的解决方案。我们在这里：
> txt <- c("I love chicken [unintelligible]!",
+ "Me too! (laughter) It's so good.[interupting]",
+ "Yep it's awesome {reading}.",
+ "Agreed.")
> bracketXtract(txt, "all")
[[1]]
[1] "unintelligible"
[[2]]
[1] "laughter" "interupting"
[[3]]
[1] "reading"
[[4]]
character(0)
这适用于data.frame。
> examp2 <- data.frame(var1=1:4)
> examp2$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.
您看到的警告与尝试将矩阵粘贴到数据框中有关。我认为答案是“不要那样做”。
> df = data.frame(x=1:2)
> df$y = matrix(list(), 2, 2)
> df
x y
1 1 NULL
2 2 NULL
Warning message:
In format.data.frame(x, digits = digits, na.encode = FALSE) :
corrupt data frame: columns will be truncated or padded with NAs
答案 1 :(得分：4)
我的想法是制作6个（隐式矢量化）辅助函数，但我会研究Martin的代码，因为他比我更好：
rm.curlybkt.no <-function(x) gsub("(\\{).*(\\})", "\\1\\2", x, perl=TRUE)
rm.rndbkt.no <- function(x) gsub("(\\().*(\\))", "\\1\\2", x, perl=TRUE)
rm.sqrbkt.no <- function(x) gsub("(\\[).*(\\])", "\\1\\2", x, perl=TRUE)
rm.rndbkt.in <- function(x) gsub("\\(.*\\)", "", x)
rm.curlybkt.in <- function(x) gsub("\\{.*\\}", "", x)
rm.sqrbkt.in <- function(x) gsub("\\[.*\\]", "", x)
答案 2 :(得分：3)
假设括号没有嵌套，我们有这个测试数据：
x <- c("a (bb) [ccc]{d}e", "x[a]y")
然后在gsubfn中使用strapply我们有这个两行解决方案，首先将所有括号和方括号翻译成括号，然后处理：
library(gsubfn)
xx <- chartr("[]()", "{}{}", x)
s <- strapply(xx, "{([^}]*)}", c)
以上结果如下：
> s
[[1]]
[1] "bb" "ccc" "d"
[[2]]
[1] "a"
答案 3 :(得分：0)
给它一个机会。我更喜欢stringr套餐！ ：）
bracketXtract <- function(string, bracket = "all", include.bracket = TRUE){
# Load stringr package
require(stringr)
# Regular expressions for your brackets
rgx = list(square = "\\[\\w*\\]", curly = "\\{\\w*\\}", round = "\\(\\w*\\)")
rgx['all'] = sprintf('(%s)|(%s)|(%s)', rgx$square, rgx$curly, rgx$round)
# Ensure you have the correct bracket name
stopifnot(bracket %in% names(rgx))
# Find your matches
matches = str_extract_all(string, pattern = rgx[[bracket]])[[1]]
# Remove brackets from results if needed
if(!include.bracket)
matches = sapply(matches, function(m) substr(m, 2, nchar(m)-1))
unname(matches)
}
j <- "What kind of cheese isn't your cheese? {wonder} Nacho cheese! [groan] (Laugh)"
bracketXtract(j)
# [1] "{wonder}" "[groan]" "(Laugh)"
bracketXtract(j, bracket = "square")
# [1] "[groan]"
bracketXtract(j, include.bracket = F)
# [1] "wonder" "groan" "Laugh"