我有一个要从中转换的文本字符串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"
谁能帮我解决这个问题?
非常感谢。
编辑:到目前为止,响应已经非常有用,但是我正在解析这是一个大文件,并且实际上不知道此
a@b@c@d...
模式将被链接多少次。是否有更通用的解决方案,在模式的长度上不依赖于硬编码(如上所述)? 最佳答案
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"
数据
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")