从R中的嵌套列表中提取内容

Question

我将客户数据存储在R中的嵌套列表中，其方式与此相同：

myinventedlist <- list("LOLETE" = list("Name" = "LOLETE",
                                "location" = "Huelva",
                                "Employees" = "22",
                                "SM" = "eJeK1",
                                "Groups" = list("ABUELOs" = list("PICHI" = list("fab_name" = "Pichi (ES)", "fab_id" = "2323423ES", "fab_tarif" = "6A"),
                                                                 "PACHA" = list("fab_name" = "Pacha (AG)", "fab_id" = "1231212AG", "fab_tarif" = "6A"),
                                                                 "POCHO" = list("fab_name" = "Pocho (ED)", "fab_id" = "2132192ED", "fab_tarif" = "6A")),
                                             "PRIMOts" = list("MONGO" = list("fab_name" = "MONGO (LB)", "fab_id" = "21332238LB", "fab_tarif" = "6A"),
                                                              "MINGO" = list("fab_name" = "MINGO (NT)", "fab_id" = "22231220NT", "fab_tarif" = "6B"),
                                                              "MUNGO" = list("fab_name" = "MUNGO (CQ)", "fab_id" = "23215001CQ", "fab_tarif" = "6B")))),
                       "GUPERA" =  list("Name" = "GUPERA",
                                          "location" = "Madrid",
                                          "Employees" = "113",
                                          "SM" = "1xa3P",
                                          "Groups" = list("ABUELOs" = list("YYTER" = list("fab_name" = "YYTER (MM)", "fab_id" = "2323423MM", "fab_tarif" = "6A"),
                                                                           "LOLE" = list("fab_name" = "LOLE (NN)", "fab_id" = "1231212NN", "fab_tarif" = "6A"),
                                                                           "PEEE" = list("fab_name" = "PEE (EE)", "fab_id" = "2132192EE", "fab_tarif" = "6A")))))

我想从给定名称的cliente中提取一个带有所有“ fab_id”的向量（在本例中为“ LOLETE”或“ GUPERA”）。

我可以从某个客户端访问所需的内容，即所有“ fab_id”，但这是一种可怕的方式：

cliente <- "LOLETE"
firstindex <- which(names(myinventedlist) == eval(cliente))
secondindex <- which(names(myinventedlist[[firstindex]]) == "Groups")
sapply(myinventedlist[[firstindex]][[secondindex]][[1]], "[[", "fab_id")
sapply(myinventedlist[[firstindex]][[secondindex]][[2]], "[[", "fab_id")

这使：

      PICHI       PACHA       POCHO 
"2323423ES" "1231212AG" "2132192ED" 

       MONGO        MINGO        MUNGO 
"21332238LB" "22231220NT" "23215001CQ

我希望给定客户我可以恢复所有“ fab_id”，而不必考虑它们所属的“组”。客户端作为字符串传递。

换句话说，我希望能够获得列表中某个标题（例如“ fab_name”）下标记的所有元素值，尽管它们可能会包含在嵌套列表（例如“ Groups”）中。

我想尝试一下，问是否在此类列表中存储将在项目中递归使用的数据，将客户端命名为“ CLIENT01”，然后在列表中添加字段“ clien_name”，或者是否将其命名可以直接使用客户端名称来命名列表。典型的做法是什么？

欢迎使用这种方式在R中使用列表的任何良好链接。

提前致谢！

Answer 1

unlist，然后按名称子集grepl：

res <- unlist(myinventedlist[[ cliente ]])
res[ grepl("fab_id", names(res)) ]
# Groups.ABUELOs.PICHI.fab_id Groups.ABUELOs.PACHA.fab_id Groups.ABUELOs.POCHO.fab_id Groups.PRIMOts.MONGO.fab_id 
# "2323423ES"                 "1231212AG"                 "2132192ED"                "21332238LB" 
# Groups.PRIMOts.MINGO.fab_id Groups.PRIMOts.MUNGO.fab_id 
# "22231220NT"                "23215001CQ"