Question

如何在R中执行stringsplit（）？当没有由破折号分隔的名字时，停止分裂。保持结果中给出的右侧子字符串。

a <- c("tim/tom meyer XY900 123kncjd", "sepp/max/peter moser VK123 456xyz")

# result: 
c("tim meyer XY900 123kncjd", "tom meyer XY900 123kncjd", "sepp moser VK123 456xyz", "max moser VK123 456xyz", "peter moser VK123 456xyz")

Answer 1

这是使用一些不同基本字符串函数的一种可能性。

## get the lengths of the output for each first name
len <- lengths(gregexpr("/", sub(" .*", "", a), fixed = TRUE)) + 1L
## extract all the first names 
## using the fact that they all end at the first space character
fn <- scan(text = a, sep = "/", what = "", comment.char = " ")
## paste them together
paste0(fn, rep(regmatches(a, regexpr(" .*", a)), len))
# [1] "tim meyer XY900 123kncjd" "tom meyer XY900 123kncjd"
# [3] "sepp moser VK123 456xyz"  "max moser VK123 456xyz"  
# [5] "peter moser VK123 456xyz"

添加：这是第二种可能性，使用少量代码。可能会快一点。

s <- strsplit(a, "\\/|( .*)")
paste0(unlist(s), rep(regmatches(a, regexpr(" .*", a)), lengths(s)))
# [1] "tim meyer XY900 123kncjd" "tom meyer XY900 123kncjd"
# [3] "sepp moser VK123 456xyz"  "max moser VK123 456xyz"  
# [5] "peter moser VK123 456xyz"

Answer 2

我会这样做（使用stringi）：

library("stringi")

a <- c("tim/tom meyer XY900 123kncjd", "sepp/max/peter moser VK123 456xyz")

stri_split_fixed(stri_match_first_regex(a, "(.+?)[ ]")[,2], "/") -> start
stri_match_first_regex(a, "[ ](.+)")[,2] -> end


for(i in 1:length(end)){
    start[[i]] <- paste(start[[i]], end[i])
}

unlist(start)

## [1] "tim meyer XY900 123kncjd" "tom meyer XY900 123kncjd" "sepp moser VK123 456xyz" 
## [4] "max moser VK123 456xyz"   "peter moser VK123 456xyz"

Answer 3

为什么不再采用一种方法来展示R解决方案的方法有很多种。按/符号拆分字符串。将名字与字符串的其余部分分开。然后与paste结合使用。有趣的问题顺便说一句：

unlist(sapply(strsplit(a, "/"), function(x) {len <- length(x)
  last <- gsub("^(\\w+).*", "\\1", x[len])
  fill <- gsub("^\\w+ ", "", x[len])
  paste(c(x[-len], last), fill)}))
# [1] "tim meyer XY900 123kncjd" "tom meyer XY900 123kncjd" "sepp moser VK123 456xyz" 
# [4] "max moser VK123 456xyz"   "peter moser VK123 456xyz"

Answer 4

这是一种方法：

a <- c('tim/tom meyer XY900 123kncjd','sepp/max/peter moser VK123 456xyz');
do.call(c,lapply(strsplit(a,' '),function(w) apply(expand.grid(strsplit(w,'/')),1,paste,collapse=' ')));
## [1] "tim meyer XY900 123kncjd" "tom meyer XY900 123kncjd" "sepp moser VK123 456xyz"  "max moser VK123 456xyz"   "peter moser VK123 456xyz"

这个解决方案的一个优点是它对每个字符串中的所有单词执行拆分和重组，而不仅仅是第一个单词，正确返回所有单词变体的完整笛卡尔积：

a <- c('a/b/c d/e/f g/h/i','j/k/l m/n/o p/q/r');
do.call(c,lapply(strsplit(a,' '),function(w) apply(expand.grid(strsplit(w,'/')),1,paste,collapse=' ')));
## [1] "a d g" "b d g" "c d g" "a e g" "b e g" "c e g" "a f g" "b f g" "c f g" "a d h" "b d h" "c d h" "a e h" "b e h" "c e h" "a f h" "b f h" "c f h" "a d i" "b d i" "c d i" "a e i" "b e i" "c e i" "a f i" "b f i" "c f i" "j m p" "k m p" "l m p" "j n p" "k n p" "l n p" "j o p" "k o p" "l o p" "j m q" "k m q" "l m q" "j n q" "k n q" "l n q" "j o q" "k o q" "l o q" "j m r" "k m r" "l m r" "j n r" "k n r" "l n r" "j o r" "k o r" "l o r"

R拆分字符串并保持匹配右边的子字符串？

4 个答案: