Question

我的客户数据存储在R中的嵌套列表中，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")))))

我想用所有＆＃34; fab_id＆＃34;提取一个向量。来自客户的名字（在这种情况下＆＃34; LOLETE＆＃34;或＆＃34; GUPERA＆＃34;）。

我可以访问所需的内容，即所有＆＃34; fab_id＆＃34;来自某个客户，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")

给出了：

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

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

我希望客户能够恢复所有＆＃34; fab_id＆＃34;无视＆＃34; Group＆＃34;他们属于。客户端作为String传递。

换句话说，我希望能够获得列表中某些标题下标记的所有元素值（例如＆＃34; fab_name＆＃34;），尽管它们可能包含在嵌套列表中（喜欢＆＃34;群组＆＃34;）。

我想冒险并询问是否在这种列表中存储将在项目中使用的数据，最好将客户命名为＆＃34; CLIENT01＆＃34;然后添加字段＆＃34; clien_name＆＃34;在列表中，或者如果可以直接使用客户端的名称命名列表。什么是典型的方式？

欢迎在这种意义上与R中的列表一起使用的任何良好链接。

提前致谢！

Answer 1

unlist，然后按名称grepl进行分组：

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"

从R中的嵌套列表中提取内容

1 个答案: