R分割字符串並保持子串右邊的匹配？

如何在R中執行這stringsplit（）？當沒有破折號分隔的名字時停止分割。保留結果中給出的右側子字符串。R分割字符串並保持子串右邊的匹配？

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")

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2016-02-05 Kay

下面是一個使用幾個不同的基本字符串函數的一種可能性。

## 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"

增加：這是第二個可能性，使用少的代碼。可能會更快一點。

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"

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2016-02-05 15:10:14

完美！ ..no循環和基函數正是我以後;） – Kay

好的解決方案理查德 –

第二個解決方案是金錢1加 –

我會那樣做（與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"

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2016-02-05 15:01:48 Marta

這裏有一個辦法：

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"

這種解決方案的優點是，它執行對每個字符串中的所有單詞拆分和重組，而不僅僅是第一個字，正確返回整個笛卡爾所有單詞變體的產品：

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"

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2016-02-05 15:26:50 bgoldst

爲什麼不再有一種方法來顯示R方案有很多方法。按/符號拆分字符串。將名字與字符串的其餘部分分開。然後結合paste。有趣的問題btw：

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"

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2016-02-05 15:35:25

R分割字符串並保持子串右邊的匹配？

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