我有一个由n个子串组成的字符串。它看起来像这样:
string <- c("A_AA", "A_BB", "A_BB_AAA", "B_AA", "B_BB", "B_CC")
此字符串中的每个子组件都通过“_”与任何其他子组件分开。这里,第一级包括值“A”和“B”,第二级“AA”,“BB”和“CC”,第三级“AAA”。可以进行更深入的嵌套,解决方案应该扩展到这些情况。嵌套不一定是平衡的,例如“A”只有两个孩子,而“B”有三个孩子,但它也有一个“B”没有的孙子。
基本上,我想在某个R对象中重新创建此字符串中的嵌套结构,最好是列表。因此,嵌套列表结构如下所示:
list("A" = list("AA", "BB" = list("AAA")),
"B" = list("AA", "BB", "CC"))
> $A
$A[[1]]
[1] "AA"
$A$BB
$A$BB[[1]]
[1] "CCC"
$B
$B[[1]]
[1] "AA"
$B[[2]]
[1] "BB"
$B[[3]]
[1] "CC"
对此有任何帮助表示赞赏
答案 0 :(得分:2)
你可以把它变成一个没有太多大惊小怪的矩阵......
string <- c("A_AA", "A_BB", "A_BB_AAA", "B_AA", "B_BB", "B_CC")
splitted<-strsplit(string,"_")
cols<-max(lengths(splitted))
mat<-do.call(rbind,lapply(splitted, "length<-", cols))
答案 1 :(得分:1)
不是那么直接,也不是最漂亮的代码,但它应该完成它的工作并返回一个列表:
string <- c("A_AA", "A_BB", "A_BB_AAA", "B_AA", "B_BB", "B_CC")
# loop through each element of the string "str_el"
list_els <- lapply(string, function(str_el) {
# split the string into parts
els <- strsplit(str_el, "_")[[1]]
# loop backwards through the elements
for (i in length(els):1){
# the last element gives the value
if (i == length(els)){
# assign the value to a list and rename the list
res <- list(els[[i]])
names(res) <- els[[i - 1]]
} else {
# if its not the last element (value) assign the list res to another list
# with the name of that element
if (i != 1) {
res <- list(res)
names(res) <- els[[i - 1]]
}
}
}
return(res)
})
# combine the lists into one list
res_list <- mapply(c, list_els, SIMPLIFY = F)
res_list
# [[1]]
# [[1]]$A
# [1] "AA"
#
#
# [[2]]
# [[2]]$A
# [1] "BB"
#
#
# [[3]]
# [[3]]$A
# [[3]]$A$BB
# [1] "AAA"
#
#
#
# [[4]]
# [[4]]$B
# [1] "AA"
#
#
# [[5]]
# [[5]]$B
# [1] "BB"
#
#
# [[6]]
# [[6]]$B
# [1] "CC"
这会给你你想要的吗?
答案 2 :(得分:0)
我找到了这种方法。很奇怪,但似乎可以正常工作
my_relist <- function(x){
y=list()
#This first loop creates the skeleton of the list
for (name in x){
split=strsplit(name,'_',fixed=TRUE)[[1]]
char='y'
l=length(split)
for (i in 1:(l-1)){
char=paste(char,'$',split[i],sep="")
}
char2=paste(char,'= list()',sep="")
#Example of char2: "y$A$BB=list()"
eval(parse(text=char2))
#Evaluates the expression inside char2
}
#The second loop fills the list with the last element
for (name in x){
split=strsplit(name,'_',fixed=TRUE)[[1]]
char='y'
l=length(split)
for (i in 1:(l-1)){
char=paste(char,'$',split[i],sep="")
}
char3=paste(char,'=c(',char,',split[l])')
#Example of char3: "y$A = c(y$A,"BB")"
eval(parse(text=char3))
}
return(y)
}
这是结果:
example <- c("A_AA_AAA", "A_BB", "A_BB_AAA", "B_AA", "B_BB", "B_CC")
my_relist(example)
#$A
#$BB
#1.'AAA'
#[[2]]
#'AA'
#[[3]]
#'BB'
#$B
#1.'AA'
#2.'BB'
#3.'CC'