拆分字符串而不丢失字符

Question

我希望将字符串拆分为某个字符,同时将该字符保留在第二个结果字符串中.我可以实现几乎所有所需的操作,除了我丢失了我指定的字符strsplit,我猜这个字符称为分隔符.

有没有办法要求strsplit保留分隔符？或者我必须使用某种正则表达式吗？谢谢你的任何建议.这似乎是一个非常基本的问题.对不起,如果它是重复的.我更喜欢使用底座R.

这是一个显示我到目前为止的例子:

my.table <- read.table(text = '
                                                            model npar     AICc 
 AA(~region+state+county+city)BB(~region+state+county+city)CC(~1)   17 11111.11
         AA(~region+state+county)BB(~region+state+county)CC(~123)   14 22222.22
                        AA(~region+state)BB(~region+state)CC(~33)   13 33333.33
                                  AA(~region)BB(~region)CC(~4321)    6 44444.44
', header = TRUE, stringsAsFactors = FALSE)

desired.result <- read.table(text = '
                                                      model        CC npar     AICc
 AA(~region+state+county+city)BB(~region+state+county+city)    CC(~1)   17 11111.11
           AA(~region+state+county)BB(~region+state+county)  CC(~123)   14 22222.22
                         AA(~region+state)BB(~region+state)   CC(~33)   13 33333.33
                                     AA(~region)BB(~region) CC(~4321)    6 44444.44
', header = TRUE, stringsAsFactors = FALSE)

split.model  <- strsplit(my.table$model, 'CC\\(')

split.models <- matrix(unlist(split.model), ncol=2, byrow=TRUE, dimnames = list(NULL, c("model", "CC")))

desires.result2 <- data.frame(split.models, my.table[,2:ncol(my.table)])
desires.result2

#                                                       model     CC npar     AICc
# 1 AA(~region+state+county+city)BB(~region+state+county+city)    ~1)   17 11111.11
# 2           AA(~region+state+county)BB(~region+state+county)  ~123)   14 22222.22
# 3                         AA(~region+state)BB(~region+state)   ~33)   13 33333.33
# 4                                     AA(~region)BB(~region) ~4321)    6 44444.44

Answer 1

基本思想是使用look-around正则表达式中的操作strsplit来获得所需的结果.然而,它比那些具有strsplit积极前瞻性的技术有点棘手.阅读@ JoshO'Brien的这篇优秀文章,以获得解释.

pattern <- "(?<=\\))(?=CC)"
strsplit(my.table$model, pattern, perl=TRUE)
# [[1]]
# [1] "AA(~region+state+county+city)BB(~region+state+county+city)"
# [2] "CC(~1)"                                                    

# [[2]]
# [1] "AA(~region+state+county)BB(~region+state+county)"
# [2] "CC(~123)"                                        

# [[3]]
# [1] "AA(~region+state)BB(~region+state)" "CC(~33)"                           

# [[4]]
# [1] "AA(~region)BB(~region)" "CC(~4321)"             

当然,我完成了任务do.call(rbind, ...)并将cbind决赛desired.output交给你.