r - 拆分字符串并替换索引

Question

我有这样的规则

rule <- "X[,14]<=-0.00479499999999999 & X[,14]<=-0.00490000000000002 & X[,14]>-0.00595499999999993 & X[,14]<=-0.00593499999999991 & X[,96]>0.000395000000000034 & X[,96]<=0.000859999999999972"

1. 我需要按索引拆分规则

   rule1 <- "X[,14]<=-0.0047 & X[,14]<=-0.0049 & X[,14]>-0.0059 & X[,14]<=-0.0059 rule2
   <- X[,96] >0.0003 & X[,96]<=0.0008"

2. 在这种情况下，将索引替换为从 1 到 2 的新索引

14,96 至 1,2

rule1 <- "X[,1]<=-0.0047 & X[,1]<=-0.0049 &
          X[,1]>-0.0059 & X[,1]<=-0.0059  

rule2 <-  X[,2]>0.0003 & X[,2]<=0.0008"

Answer 1

我们将'rule'分割成'，然后用to&提取数字部分，以数字为一组，然后是元素，最后用from替换索引parse_numbersplitpasteliststr_replace_allstringr

library(stringr)
v1 <- strsplit(rule, "\\s*\\&\\s*")[[1]]

v2 <- trimws(sapply(split(v1, readr::parse_number(v1)), function(x)
          sub("\\s*\\&\\s*$", "", paste(x, collapse=" & "))))
str_replace_all(v2, setNames(as.character(seq_along(v2)), names(v2)))

-输出

#[1] "X[,1]<=-0.00479499999999999 & X[,1]<=-0.00490000000000002 & X[,1]>-0.00595499999999993 & X[,1]<=-0.00593499999999991"
#[2] "X[,2]>0.000395000000000034 & X[,2]<=0.000859999999999972"