我希望能够仅在给定的定界符集之外使用 grepl() 和 gsub()，例如我希望能够忽略之间的文本引号。
这是我想要的输出:
grepl2("banana", "'banana' banana \"banana\"", escaped =c('""', "''"))
#> [1] TRUE
grepl2("banana", "'banana' apple \"banana\"", escaped =c('""', "''"))
#> [1] FALSE
grepl2("banana", "{banana} banana {banana}", escaped = "{}")
#> [1] TRUE
grepl2("banana", "{banana} apple {banana}", escaped = "{}")
#> [1] FALSE
gsub2("banana", "potatoe", "'banana' banana \"banana\"")
#> [1] "'banana' potatoe \"banana\""
gsub2("banana", "potatoe", "'banana' apple \"banana\"")
#> [1] "'banana' apple \"banana\""
gsub2("banana", "potatoe", "{banana} banana {banana}", escaped = "{}")
#> [1] "{banana} potatoe {banana}"
gsub2("banana", "potatoe", "{banana} apple {banana}", escaped = "{}")
#> [1] "{banana} apple {banana}"
真实案例可能会以不同的数量和顺序引用子字符串。
我已经编写了以下适用于这些情况的函数，但它们很笨重并且 gsub2() 根本不健壮，因为它暂时用占位符替换了分隔的内容，这些占位符可能是受后续操作影响。
regex_escape <-
function(string,n = 1) {
for(i in seq_len(n)){
string <- gsub("([][{}().+*^$|\\?])", "\\\\\\1", string)
}
string
}
grepl2 <-
function(pattern, x, ignore.case = FALSE, perl = FALSE, fixed = FALSE,
useBytes = FALSE, escaped =c('""', "''")){
escaped <- strsplit(escaped,"")
# TODO check that "escaped" delimiters are balanced and don't cross each other
for(i in 1:length(escaped)){
close <- regex_escape(escaped[[i]][[2]])
open <- regex_escape(escaped[[i]][[1]])
pattern_i <- sprintf("%s.*?%s", open, close)
x <- gsub(pattern_i,"",x)
}
grepl(pattern, x, ignore.case, perl, fixed, useBytes)
}
gsub2 <- function(pattern, replacement, x, ignore.case = FALSE, perl = FALSE,
fixed = FALSE, useBytes = FALSE, escaped =c('""', "''")){
escaped <- strsplit(escaped,"")
# TODO check that "escaped" delimiters are balanced and don't cross each other
matches <- character()
for(i in 1:length(escaped)){
close <- regex_escape(escaped[[i]][[2]])
open <- regex_escape(escaped[[i]][[1]])
pattern_i <- sprintf("%s.*?%s", open, close)
ind <- gregexpr(pattern_i,x)
matches_i <- regmatches(x, ind)[[1]]
regmatches(x, ind)[[1]] <- paste0("((",length(matches) + seq_along(matches_i),"))")
matches <- c(matches, matches_i)
}
x <- gsub(pattern, replacement, x, ignore.case, perl, fixed, useBytes)
for(i in seq_along(matches)){
pattern <- sprintf("\\(\\(%s\\)\\)", i)
x <- gsub(pattern, matches[[i]], x)
}
x
}
是否有使用正则表达式而不使用占位符的解决方案？请注意，我当前的函数支持多对定界符，但我会对仅支持一对定界符的解决方案感到满意，并且不会尝试匹配简单引号之间的子字符串。
强加不同的分隔符也是可以接受的，例如 { 和 } 而不是 2 " 或 2 ' 如果有帮助。
我也可以强加 perl = TRUE
最佳答案
您可以使用 start/end_escape 参数来提供匹配定界符的 LHS 和 RHS，例如 { 和 } 而无需在错误的地方(} 作为 LHS 分隔符)
perl = TRUE 允许环视断言。这些评估其中陈述的有效性，没有在模式中捕获它们。 This post很好地涵盖了它们。
你会在 perl = FALSE 中得到一个错误，因为 TRE ，R 的默认正则表达式引擎，不支持它们。
gsub3 <- function(pattern, replacement, x, escape = NULL, start_escape = NULL, end_escape = NULL) {
if (!is.null(escape) || !is.null(start_escape))
left_escape <- paste0("(?<![", paste0(escape, paste0(start_escape, collapse = ""), collapse = ""), "])")
if (!is.null(escape) || !is.null(end_escape))
right_escape <- paste0("(?![", paste0(escape, paste0(end_escape, collapse = ""), collapse = ""), "])")
patt <- paste0(left_escape, "(", pattern, ")", right_escape)
gsub(patt, replacement, x, perl = TRUE)
}
gsub3("banana", "potatoe", "'banana' banana \"banana\"", escape = "'\"")
#> [1] "'banana' potatoe \"banana\""
gsub3("banana", "potatoe", "'banana' apple \"banana\"", escape = '"\'')
#> [1] "'banana' apple \"banana\""
gsub3("banana", "potatoe", "{banana} banana {banana}", escape = "{}")
#> [1] "{banana} potatoe {banana}"
gsub3("banana", "potatoe", "{banana} apple {banana}", escape = "{}")
#> [1] "{banana} apple {banana}"
下面是 grepl3 - 注意这不需要 perl = TRUE 因为我们不关心模式捕获什么，只关心它匹配。
grepl3 <- function(pattern, x, escape = "'", start_escape = NULL, end_escape = NULL) {
if (!is.null(escape) || !is.null(start_escape))
start_escape <- paste0("[^", paste0(escape, paste0(start_escape, collapse = ""), collapse = ""), "]")
if (!is.null(escape) || !is.null(end_escape))
end_escape <- paste0("[^", paste0(escape, paste0(end_escape, collapse = ""), collapse = ""), "]")
patt <- paste0(start_escape, pattern, end_escape)
grepl(patt, x)
}
grepl3("banana", "'banana' banana \"banana\"", escape =c('"', "'"))
#> [1] TRUE
grepl3("banana", "'banana' apple \"banana\"", escape =c('""', "''"))
#> [1] FALSE
grepl3("banana", "{banana} banana {banana}", escape = "{}")
#> [1] TRUE
grepl3("banana", "{banana} apple {banana}", escape = "{}")
#> [1] FALSE
编辑:
这应该可以解决 gsub 而不会出现 Andrew 提到的问题，只要您可以使用一组成对的运算符。我认为您可以修改它以允许使用多个定界符。感谢这个有趣的问题，在 regmatches 中找到了一个新的 gem!
gsub4 <-
function(pattern,
replacement,
x,
left_escape = "{",
right_escape = "}") {
# `regmatches()` takes a character vector and
# output of `gregexpr` and friends and returns
# the matching (or unmatching, as here) substrings
string_pieces <-
regmatches(x,
gregexpr(
paste0(
"\\Q", # Begin quote, regex will treat everything after as fixed.
left_escape,
"\\E(?>[^", # \\ ends quotes.
left_escape,
right_escape,
"]|(?R))*", # Recurses, allowing nested escape characters
"\\Q",
right_escape,
"\\E",
collapse = ""
),
x,
perl = TRUE
), invert =NA) # even indices match pattern (so are escaped),
# odd indices we want to perform replacement on.
for (k in seq_along(string_pieces)) {
n_pieces <- length(string_pieces[[k]])
# Due to the structure of regmatches(invert = NA), we know that it will always
# return unmatched strings at odd values, padding with "" as needed.
to_replace <- seq(from = 1, to = n_pieces, by = 2)
string_pieces[[k]][to_replace] <- gsub(pattern, replacement, string_pieces[[k]][to_replace])
}
sapply(string_pieces, paste0, collapse = "")
}
gsub4('banana', 'apples', "{banana's} potatoes {banana} banana", left_escape = "{", right_escape = "}")
#> [1] "{banana's} potatoes {banana} apples"
gsub4('banana', 'apples', "{banana's} potatoes {banana} banana", left_escape = "{", right_escape = "}")
#> [1] "{banana's} potatoes {banana} apples"
gsub4('banana', 'apples', "banana's potatoes", left_escape = "{", right_escape = "}")
#> [1] "apples's potatoes"
gsub4('banana', 'apples', "{banana's} potatoes", left_escape = "{", right_escape = "}")
#> [1] "{banana's} potatoes"
关于r - 扩展 gsub 和 grepl 以忽略给定分隔符之间的子字符串，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/58775471/