Question

我有许多列表列表，在每个列表中我需要提取的变量以稍微不同的方式嵌套。有没有一种简单的方法来搜索变量并将其提取出来？

示例列表

list1 <- list(AccountingSupplierParty = list(Party = list(PartyName = "Company Incorporated", PartyType = "The worst party")), DataSet = "Data Set 1")
list2 <- list(SupplierParty = list(Party = list(PartyName = "Company A/S", PartyType = "The best party")), DataSet = "Data Set 2")

我想提取＆＃34; PartyName＆＃34;。在巨大的数据集中学习变量的所有组合并不是那么有效，如下图所示：

Company1 <- list1$AccountingSupplierParty$Party$PartyName
Company2 <- list2$SupplierParty$Party$PartyName

我想要的输出是：

"Company Incorporated"
"Company A/S"

Answer 1

您可以取消列出每个列表，然后清除所有未在PartyName中结束的列表。

list1 <- list(AccountingSupplierParty = list(Party = list(PartyName = "Company Incorporated", PartyType = "The worst party")), DataSet = "Data Set 1")
list2 <- list(SupplierParty = list(Party = list(PartyName = "Company A/S", PartyType = "The best party")), DataSet = "Data Set 2")

c1 <- unlist(list1)
c1 <- c1[grepl("PartyName$", names(c1))]

AccountingSupplierParty.Party.PartyName 
                 "Company Incorporated"

c2 <- unlist(list2)
c2 <- c2[grepl("PartyName$", names(c2))]
c2

SupplierParty.Party.PartyName 
                "Company A/S"

Answer 2

您可以尝试使用rlist库。
https://cran.r-project.org/web/packages/rlist/rlist.pdf

library(rlist)
list.flatten(list1)[1]
list.flatten(list2)[1]

$AccountingSupplierParty.Party.PartyName
[1] "Company Incorporated"
$SupplierParty.Party.PartyName
[1] "Company A/S"

Answer 3

你可以试试这个

dfs <- data.frame(lapply(list1, data.frame, stringsAsFactors = FALSE))
df1[ , grepl( "PartyName" , names( df1 ) ) ]

注意我使用了data.frame两次才能使stringAsFactors工作。这会给你

[1] "Company Incorporated"

希望有所帮助，翁贝托

R：从列表列表中搜索和提取变量

3 个答案: