Question

我有一个列表，我想将其转换为数据表数据如下：

"Customer"
"Steve"
"AddressLine"
"2041"
"Total"
"10"
"MailItemInfo"
"None"
"Customer"
"Mike"
"AddressLine"
"2043"
"Total"
"20"
"MailItemInfo"
"Yes"
"Customer"
"Jenn"
"AddressLine"
"1132"
"Total"
"24"
"MailItemInfo"
"Yes"

此模式最终会重复5个不同的地址。我想把所有其他行放入一个新列，例如：

"Customer"  "AddressLine"  "BatchTotal"  "MailItemInfo" 
"Steve"  "2041"  "10"  "None"
"Mike"  "2043"  "20"  "Yes"
"Jenn"  "1132" "24" "Yes"

我希望每当这些新的“列”继续出现在数据列表中时，它们会自动填充到新列中。任何可以提供解决方案的人都会真正提供帮助！

Answer 1

l <- list("Customer",
          "Steve",
          "AddressLine",
          "2041",
          "Total",
          "10",
          "MailItemInfo",
          "None",
          "Customer",
          "Mike",
          "AddressLine",
          "2043",
          "Total",
          "20",
          "MailItemInfo",
          "Yes",
          "Customer",
          "Jenn",
          "AddressLine",
          "1132",
          "Total",
          "24",
          "MailItemInfo",
          "Yes")

如果结构没有变化，这可行：

data.frame(matrix(unlist(l), ncol=8, byrow=TRUE))[c(2,4,6,8)]

     X2   X4 X6   X8
1 Steve 2041 10 None
2  Mike 2043 20  Yes
3  Jenn 1132 24  Yes

但是如果您想要在显示列时添加列，则可能需要使用已融合的data.frame。实际上iy = t更容易用data.table执行此操作。

首先，您必须确定新行的开头是什么，例如“客户”：

l <- list("Customer",
      "Steve",
      "AddressLine",
      "2041",
      "Total",
      "10",
      "MailItemInfo",
      "None",
      "Customer",
      "Mike",
      "AddressLine",
      "2043",
      "Total",
      "20",
      "MailItemInfo",
      "Yes",
      "Customer",
      "Jenn",
      "AddressLine",
      "1132",
      "Total",
      "24",
      "MailItemInfo",
      "Yes",
      "NewColumn",
      "xxx")

library(data.table)
dt <- data.table(matrix(unlist(l), ncol=2, byrow=TRUE)) # melted data.table
dt[V1=='Customer', id:=.I] # add id attribute to each "Customer"
dt[, id := id[1], by = cumsum(!is.na(id))] # set the id to following attributes
dcast(dt,id~V1, value.var="V2", fill=NA) # reverse the melted to large data.table

   id AddressLine Customer MailItemInfo NewColumn Total
1:  1        2041    Steve         None        NA    10
2:  2        2043     Mike          Yes        NA    20
3:  3        1132     Jenn          Yes       xxx    24

Answer 2

假设上面的列表名为：“mylist”。试试这个：

titleseq<-seq(1, 8, by =2)
titles<-droplevels(mylist[titleseq,1])

nameseq<-seq(2, 40, by=8)
names<-droplevels(mylist[nameseq,1])
addres<-droplevels(mylist[(nameseq+2),1])
tot<-droplevels(mylist[(nameseq+4),1])
mailitem<-droplevels(mylist[(nameseq+6),1])

df<-data.frame(names, addres, tot, mail item)
names(df)<-titles

这是非常粗野的强迫，但是应该这样做。

行中的R列

2 个答案: