我有一个文本字符串,我想从
转换 text = "end back@drive@o correct back@drive@adjust@cats@do to tok"
到
"end back@drive drive@o correct back@drive drive@adjust adjust@cats cats@do to tok"
相反,一般来说,我想替换
"a@b@c" with "a@b b@c"
"a@b@c@d" with "a@b b@c c@d"
等等。我在下面尝试使用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
是我在最后寻找的功能,当然它(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"
有人可以帮我解决这个问题吗?
非常感谢。
编辑:到目前为止,回复非常有用,但它是一个我正在解析的大文件,并且不知道这个a@b@c@d...
模式将链接多少次。是否有一个更通用的解决方案,不依赖于模式长度的硬编码(正如我上面尝试过的那样)?
答案 0 :(得分:3)
> gsub(x = text, pattern = '@(.*?)@', replacement = '@\\1 \\1@')
[1] "end back@drive drive@o correct back@drive drive@adjust to tok"
您需要提供更多关于您希望遇到的案例的例子,但解决方案将与上述方向相同。
在回应评论时 - 您可能需要运行链条
您的文字gsub(x = text, pattern = '@([[:alnum:]]{1,})@', replacement = '@\\1 \\1@')
,直到它没有变化。同样,没有更多的测试用例,我们无法确定。
答案 1 :(得分:2)
我会使用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"
答案 2 :(得分:1)
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&#39; str1&#39;
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")