gsub返回表達式的所有匹配，而不僅僅是最後一個匹配

Question

我正在尋找一個gsub字符串，它將返回表達式的所有匹配項，而不僅僅是最後一個匹配項。即：

data <- list("a sentence with citation (Ref. 12) and another (Ref. 13)", "single (Ref. 14)") 
gsub(".*(Ref. (\\d+)).*", "\\1", data) 

返回

[1] "Ref. 13" "Ref. 14" 

所以我已經失去了參考。 12.

Answer 1

可以使用strapply功能從gsubfn包做到這一點：

library(gsubfn) 

data <- list("a sentence with citation (Ref. 12) and another (Ref. 13)", "single (Ref. 14)") 
unlist(strapply(data,"(Ref. (\\d+))")) 

Answer 2

這裏是一個函數 - 實質上是一個gregexpr()的包裝 - 它將捕獲單個字符串中的多個引用。

extractMatches <- function(data, pattern) { 
    start <- gregexpr(pattern, data)[[1]] 
    stop <- start + attr(start, "match.length") - 1 
    if(-1 %in% start) { 
     "" ## **Note** you could return NULL if there are no matches 
    } else { 
     mapply(substr, start, stop, MoreArgs = list(x = data)) 
    } 
}  

data <- list("a sentence with citation (Ref. 12), (Ref. 13), and then (Ref. 14)", 
      "another sentence without reference") 
pat <- "Ref. (\\d+)" 

res <- lapply(data, extractMatches, pattern = pat) 
res 
# [[1]] 
# [1] "Ref. 12" "Ref. 13" "Ref. 14" 
# 
# [[2]] 
# [1] "" 

（**注**：如果您返回NULL，而不是""時，有一個字符串的參考，那麼你可以後期處理與do.call("c", res)結果得到只包含匹配引用單個矢量）。

Answer 3

如何

sapply(data,stringr::str_extract_all,pattern="Ref. (\\d+))") 

？

Answer 4

我之前（http://thebiobucket.blogspot.com/2012/03/how-to-extract-citation-from-body-of.html）有一個非常類似的問題以及與此想出了（其實非常接近本的）解決方案：

require(stringr) 
unlist(str_extract_all(unlist(data), pattern = "\\(.*?\\)")) 

捐贈：

[1] "(Ref. 12)" "(Ref. 13)" "(Ref. 14)"