Question

我一方面有一个字符元素F的载体。

另一方面，一组函数在GF中生成两个子列表的分段版本。输出SA生成包含所需段的列表，但SA的索引具有以下格式。

[[1]]
[[1]]$`1` 
[[1]]$`2`    

[[2]]
[[2]]$`1`....   

F  <-c("a","b","c","d","e","f","g","h","i","a","b","c","d","e","f","g","h","i","j","k","l") 

GF  <- list(c(1,2,3,4,5,6,7,8,9) , c(2,3,4,5,6,3,2,1,3,4,5,2)) 
RET <- seq(length ( GF )) # SEQUENCE : NUMBER OF SUB-LISTS
LS  <- c( 3 , 2 ) # LENGTH OF EACH ADDITIONAL SUB-SUB SEGMENT WITHIN TWO LISTS
fun <- function (x) split ( GF[[x]] , ceiling (seq_along (GF[[x]])/LS[[x]]))
SA  <- (lapply (  RET  ,  fun )) # OUTPUT 
relist(unlist( F ), SA )

我想使用SA的scheleton来重新连接F，但是在输出SA中获取的索引格式是[[和S并且希望它全部[[。

Answer 1

是的，你可以在分割功能中使用unname

F  <-c("a","b","c","d","e","f","g","h","i","a","b","c","d","e","f","g","h","i","j","k","l") 

GF  <- list(c(1,2,3,4,5,6,7,8,9) , c(2,3,4,5,6,3,2,1,3,4,5,2)) 
RET <- seq(length ( GF )) # SEQUENCE : NUMBER OF SUB-LISTS
LS  <- c( 3 , 2 ) # LENGTH OF EACH ADDITIONAL SUB-SUB SEGMENT WITHIN TWO LISTS
fun <- function (x) unname(split ( GF[[x]] , ceiling (seq_along (GF[[x]])/LS[[x]])))
SA  <- (lapply (  RET  ,  fun )) # OUTPUT 
relist(unlist( F ), SA )

可以将分割函数的列表索引产生索引[[]] + S转换为全部[[

1 个答案: