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R 列表的索引是否可以在拆分后生成索引[]和#x2B；让我们转换成所有[[_R - Fatal编程技术网
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  • R 列表的索引是否可以在拆分后生成索引[]和#x2B；让我们转换成所有[[

R 列表的索引是否可以在拆分后生成索引[]和#x2B；让我们转换成所有[[

r

R 列表的索引是否可以在拆分后生成索引[]和#x2B；让我们转换成所有[[,r,R,一方面，我有一个字符元素F的向量 另一方面，生成GF中两个子列表的分段版本的一组函数。输出SA生成包含所需分段的列表，但SA的索引采用以下格式 [[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,

一方面，我有一个字符元素F的向量

另一方面，生成GF中两个子列表的分段版本的一组函数。输出SA生成包含所需分段的列表，但SA的索引采用以下格式

[[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 )  

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

F是的，您可以在split函数中使用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 )  



F所有的方法在我写问题的同时，我尝试了一些替代方法，发现上面的答案很好。