[英]Replacing multiple patterns with manipulated pattern
我有一个想要转换的文本字符串
text = "end back@drive@o correct back@drive@adjust@cats@do to tok"
至
"end back@drive drive@o correct back@drive drive@adjust adjust@cats cats@do to tok"
相反,我通常要替换
"a@b@c" with "a@b b@c"
"a@b@c@d" with "a@b b@c c@d"
等等。 我在下面的尝试中使用了stringr
包。
patterns = unlist(str_extract_all(text, "([[:alnum:]]+@){2,}[[:alnum:]]+"))
replacements = strsplit(patterns, "@")
replacements = lapply(replacements, function(y) {
pretuples = y[-length(y)]
posttuples = y[-1]
paste(paste0(pretuples, "@", posttuples), collapse = " ")
})
replacements = do.call(c, replacements)
str_replace_all(text, pattern = patterns, replacement = replacements)
我不认为str_replace_all
是我最后要寻找的函数,当然(合理地)返回
[1] "end back@drive drive@o correct back@drive@adjust to tok"
[2] "end back@drive@o correct back@drive drive@adjust adjust@cats cats@do to tok"
谁能帮我解决这个问题?
非常感谢。
编辑:到目前为止,响应一直非常有用,但是我正在解析一个很大的文件,并不真正知道此a@b@c@d...
模式将被链接多少次。 是否有一种更通用的解决方案,该解决方案在模式的长度上不依赖于硬编码(如上所述)?
> gsub(x = text, pattern = '@(.*?)@', replacement = '@\\1 \\1@')
[1] "end back@drive drive@o correct back@drive drive@adjust to tok"
您需要提供更多有关预期会遇到的情况的示例,但是解决方案将与上述相同。
作为回应,您可能需要运行gsub(x = text, pattern = '@([[:alnum:]]{1,})@', replacement = '@\\\\1 \\\\1@')
链gsub(x = text, pattern = '@([[:alnum:]]{1,})@', replacement = '@\\\\1 \\\\1@')
,直到它保持不变。 同样,如果没有更多的测试用例,就无法确定。
我会用gsub
:
> text = "end back@drive@o correct back@drive@adjust to tok"
> gsub(pattern = "([[:alpha:]]+)@([[:alpha:]]+)@([[:alpha:]]+)", replacement = "\\1@\\2 \\2@\\3", x = text)
[1] "end back@drive drive@o correct back@drive drive@adjust to tok"
尝试
pat <- "(\\s|\\b)[^@]+\\s(*SKIP)(*FAIL)|(?<=@)([^@]*)(?=@)"
repl <- "\\2 \\2"
gsub(pat, repl, text, perl=TRUE)
#[1] "end back@drive drive@o correct back@drive drive@adjust adjust@cats cats@do to tok"
对于“ str1”
gsub(pat, repl, str1, perl=TRUE)
#[1] "a@b b@c" "a@b b@c c@d"
#[3] "a@b b@c c@d d@e e@f f@g g@h"
text <- "end back@drive@o correct back@drive@adjust@cats@do to tok"
str1 <- c("a@b@c", "a@b@c@d", "a@b@c@d@e@f@g@h")
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