如何从R中的现有列值动态创建新列

Question

i have data frame called df,how to create new column from existing list column data frame.

my data frame.

Policy             Item

Checked           list(Processed = "Valid", Gmail = "yy@gmail", Information = list(list(Descrption = "T1, R1", VID = "YUY")))

Sample            list(Processed = "Valid", Gmail = "tt@gmail", Information = list(list(Descrption = "D3, Y3", VID = "RT")))

Processed         list(Processed = "Valid", Gmail = "pp@gmail", Information = list(list(Descrption = "Y2, LE", VID = "UIU")))

my expected data frame.

Policy          Processed    Gmail        Descrption  VID

Checked           Valid      yy@gmail       "T1,R1"  "YUY"

Sample            Valid      tt@gmail       "D3,Y3"  "RT"

Processed         Valid      pp@gmail       "Y2,LE"  "UIU"

i'm using below code to get my expected dataframe .

na_if_null <- function(x) if (is.null(x)) NA else x

new_cols <- lapply(
  Filter(is.list, df),
  function(list_col) {
    names_ <- setNames(nm = unique(do.call(c, lapply(list_col, names))))
    lapply(names_, function(name) sapply(list_col, function(x) 
      trimws(na_if_null(as.list(x)[[name]]))))
  }
)

res <- do.call(
  data.frame,
  c(
    list(df, check.names = FALSE, stringsAsFactors = FALSE),
    do.call(c, new_cols)
  )
)

But i'm getting below Data frame.please help me to done my post.

Policy       Item                                                                                                          Item.Processed    Item.Gmail     Item.Information

Checked      list(Processed = "Valid", Gmail = "yy@gmail", Information = list(list(Descrption = "T1, R1", VID = "YUY")))    Processed        yy@gmail      list(Descrption = "T1, R1", VID = "YUY")

Sample       list(Processed = "Valid", Gmail = "tt@gmail", Information = list(list(Descrption = "D3, Y3", VID = "RT")))     Processed        tt@gmail      list(Descrption = "D3, Y3", VID = "RT")  

Processed    list(Processed = "Valid", Gmail = "pp@gmail", Information = list(list(Descrption = "Y2, LE", VID = "UIU")))    Processed        pp@gmail      list(Descrption = "Y2, LE", VID = "UIU")

dput

    structure(list(Policy = c("Checked", "Sample", "Processed"), Item = list(
    structure(list(Processed = "Valid", Gmail = "yy@gmail", Information = list(
        structure(list(Descrption = "T1, R1", VID = "YUY"), .Names = c("Descrption", 
        "VID"), class = "data.frame", row.names = 1L))), .Names = c("Processed", 
    "Gmail", "Information"), class = "data.frame", row.names = 1L), 
    structure(list(Processed = "Valid", Gmail = "tt@gmail", Information = list(
        structure(list(Descrption = "D3, Y3", VID = "RT"), .Names = c("Descrption", 
        "VID"), class = "data.frame", row.names = 1L))), .Names = c("Processed", 
    "Gmail", "Information"), class = "data.frame", row.names = 1L), 
    structure(list(Processed = "Valid", Gmail = "pp@gmail", Information = list(
        structure(list(Descrption = "Y2, LE", VID = "UIU"), .Names = c("Descrption", 
        "VID"), class = "data.frame", row.names = 1L))), .Names = c("Processed", 
    "Gmail", "Information"), class = "data.frame", row.names = 1L))), row.names = c(NA, 
3L), class = "data.frame", .Names = c("Policy", "Item"))

Sample data frame

Policy             colval                                 Item     

Checked         list(PID="4",Bdetail ="ui,89")      list(Processed = "Valid", Gmail = "yy@gmail", Information = list(list(Descrption = "T1, R1", VID = "YUY")))

Sample          list(PID="7",Bdetail ="ju,78")      list(Processed = "Valid", Gmail = "tt@gmail", Information = list(list(Descrption = "D3, Y3", VID = "RT")))

Processed       list(PID ="8",Bdetail ="nj,45")     list(Processed = "Valid", Gmail = "pp@gmail", Information = list(list(Descrption = "Y2, LE", VID = "UIU")))

Answer 1

Here a solution in base R:

dd <- 
cbind(
  dx$Policy,
  do.call(rbind,
          lapply(seq_len(nrow(dx)), function(i)unlist(dx$Item[i]))
  )
)

colnames(dd) <- c("Policy","Processed","Gmail","Descrption","VID")

dd

#       Policy      Processed Gmail      Descrption VID  
# [1,] "Checked"   "Valid"   "yy@gmail" "T1, R1"   "YUY"
# [2,] "Sample"    "Valid"   "tt@gmail" "D3, Y3"   "RT" 
# [3,] "Processed" "Valid"   "pp@gmail" "Y2, LE"   "UIU"

Basically I am using unlist for each item. and Then joining them using the classic d.call(rbind,llist) .

edit

in case you want tu use the same names as the original sub lists you can do something like :

colnames(dd) <- c("Policy",gsub(".*[.]","",colnames(dd)[-1]))

data.table solution

library(data.table)
setDT(dx)
dx[, rbindlist(lapply(.SD,function(x)data.table(t(unlist(x))))),Policy]

Answer 2

Easily done with unnest from tidyr :

library(dplyr)
library(tidyr)

df %>%
  unnest() %>%
  unnest()

Result:

     Policy Processed    Gmail Descrption VID
1   Checked     Valid yy@gmail     T1, R1 YUY
2    Sample     Valid tt@gmail     D3, Y3  RT
3 Processed     Valid pp@gmail     Y2, LE UIU

Data:

df =     structure(list(Policy = c("Checked", "Sample", "Processed"), Item = list(
  structure(list(Processed = "Valid", Gmail = "yy@gmail", Information = list(
    structure(list(Descrption = "T1, R1", VID = "YUY"), .Names = c("Descrption", 
                                                                   "VID"), class = "data.frame", row.names = 1L))), .Names = c("Processed", 
                                                                                                                               "Gmail", "Information"), class = "data.frame", row.names = 1L), 
  structure(list(Processed = "Valid", Gmail = "tt@gmail", Information = list(
    structure(list(Descrption = "D3, Y3", VID = "RT"), .Names = c("Descrption", 
                                                                  "VID"), class = "data.frame", row.names = 1L))), .Names = c("Processed", 
                                                                                                                              "Gmail", "Information"), class = "data.frame", row.names = 1L), 
  structure(list(Processed = "Valid", Gmail = "pp@gmail", Information = list(
    structure(list(Descrption = "Y2, LE", VID = "UIU"), .Names = c("Descrption", 
                                                                   "VID"), class = "data.frame", row.names = 1L))), .Names = c("Processed", 
                                                                                                                               "Gmail", "Information"), class = "data.frame", row.names = 1L))), row.names = c(NA, 
                                                                                                                                                                                                               3L), class = "data.frame", .Names = c("Policy", "Item"))

Note:

Notice I used two passes of unnest because there are two levels of lists in your original dataframe. unnest automatically flattens all lists in the dataframe and reuses the names, but it does not do it recursively, so you will have to have as many unnest as there are list levels.