Question

我导入了一些没有列名的数据，所以现在我有超过一百万行和一列（而不是5列）。

每一行的格式如下：

x <- "2012-10-19T16:59:01-07:00 192.101.136.140 <190>Oct 19 2012 23:59:01: %FWSM-6-305011: Built dynamic tcp translation from Inside:10.2.45.62/56455 to outside:192.101.136.224/9874"

strsplit( x , split = c(" ", " ", "%", " "))

得到了

[[1]]
 [1] "2012-10-19T16:59:01-07:00"    "192.101.136.140"             
 [3] "<190>Oct"                     "19"                          
 [5] "2012"                         "23:59:01:"                   
 [7] "%FWSM-6-305011:"              "Built"                       
 [9] "dynamic"                      "tcp"                         
[11] "translation"                  "from"                        
[13] "Inside:10.2.45.62/56455"      "to"                          
[15] "outside:192.101.136.224/9874"

我知道它与回收分裂参数有关，但我似乎无法想象如何得到它我想要的：

    [[1]]
     [1] "2012-10-19T16:59:01-07:00"    "192.101.136.140"             
     [3] "<190>Oct 19 2012 23:59:01     "%FWSM-6-305011
     [5] Built dynamic tcp translation from Inside:10.2.45.62/56455 to outside:192.101.136.224/9874"

每一行都有一个不同的消息作为第五个元素，但在第四个元素后，我只想将其余的字符串保持在一起。

任何帮助都将不胜感激。

Answer 1

您可以使用paste和collapse参数组合从第五个元素开始的每个元素。

A <- strsplit( x = "2012-10-19T16:59:01-07:00 192.101.136.140 <190>Oct 19 2012 23:59:01: %FWSM-6-305011: Built dynamic tcp translation from Inside:10.2.45.62/56455 to outside:192.101.136.224/9874", split = c(" ", " ", "%", " "))

c(A[[1]][1:4], paste(A[[1]][5:length(A[[1]])], collapse=" "))

正如@DWin指出的那样，split = c(" ", " ", "%", " ")未按顺序使用 - 换句话说，它与split = c(" ", "%")相同

Answer 2

我想在这里你不需要使用strsplit。我使用read.table使用text参数读取行。然后使用paste汇总列。由于您有很多行，因此最好在data.table中进行列聚合。

dt <- read.table(text=x)
library(data.table)
DT <- as.data.table(dt)
DT[ , c('V3','V8') := list(paste(V3,V4,V5),
      V8=paste(V8,V9,V10,V11,V12,V13,V14,V15))]
DT[,paste0('V',c(1:3,6:7,8)),with=FALSE]

                        V1              V2               V3        V6              V7
1: 2012-10-19T16:59:01-07:00 192.101.136.140 <190>Oct 19 2012 23:59:01: %FWSM-6-305011:
                                                                                           V8
1: Built dynamic tcp translation from Inside:10.2.45.62/56455 to outside:192.101.136.224/9874

Answer 3

我认为这是一个以您认为strsplit运行的方式工作的函数：

split.seq<-function(x,delimiters) {
  break.point<-regexpr(delimiters[1], x)
  first<-mapply(substring,x,1,break.point-1,USE.NAMES=FALSE)
  second<-mapply(substring,x,break.point+1,nchar(x),USE.NAMES=FALSE)
  if (length(delimiters)==1)  return(lapply(1:length(first),function(x) c(first[x],second[x])))
  else mapply(function(x,y) c(x,y),first, split.seq(second, delimiters[-1]) ,USE.NAMES=FALSE, SIMPLIFY=FALSE)
}

split.seq(x,delimiters)

测试：

x<-rep(x,2)             
delimiters=c(" ", " ", "%", " ")
split.seq(x,delimiters)

[[1]]
[1] "2012-10-19T16:59:01-07:00"                                                                 
[2] "192.101.136.140"                                                                           
[3] "<190>Oct 19 2012 23:59:01: "                                                               
[4] "FWSM-6-305011:"                                                                            
[5] "Built dynamic tcp translation from Inside:10.2.45.62/56455 to outside:192.101.136.224/9874"

[[2]]
[1] "2012-10-19T16:59:01-07:00"                                                                 
[2] "192.101.136.140"                                                                           
[3] "<190>Oct 19 2012 23:59:01: "                                                               
[4] "FWSM-6-305011:"                                                                            
[5] "Built dynamic tcp translation from Inside:10.2.45.62/56455 to outside:192.101.136.224/9874"

在R中拆分字符串，使用不同的拆分参数元素

3 个答案: