How to create new column from existing column value in R dynamically - r

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.