Packen Sie eine R-Datenrahmenspalte von Listen aus

Question

In R habe ich einen data.frame (oder data.table). Innerhalb dieses data.frames habe ich eine Spalte, in der jede Zelle aus einer Liste von Listen besteht (ein data.frame).

Ich kann diese Spalte in eine einzige data.frame von rbindlist(data$Subdocuments) konvertieren, aber hier sind die anderen Spalten der ursprünglichen data.frame fehlt.

Wie effizient entpacken Sie diese Spalte von Listen, aber behalten Sie die anderen Spalten an den neuen data.frame?

 library(data.table) 

    data <- structure(list(ID = c("1", "2", "3"), Country = c("Netherlands", 
"Germany", "Belgium"), Subdocuments = list(structure(list(Value = c("5", 
"5", "1", "3", "2", "1", "1", "1", "2", "5", "3", "2", "4", "5", 
"5", "2"), Label = c("Test1", "Test2", "Test3", "Test4", "Test5", 
"Test6", "Test7", "Test8", "Test9", "Test10", "Test11", "Test12", 
"Test13", "Test14", "Test15", "Test16"), Year = c(2001, 2002, 
2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013, 
2014, 2015, 2016)), .Names = c("Value", "Label", "Year"), class = "data.frame", row.names = c(NA, 
16L)), structure(list(Value = c("5", "4", "3", "2", "2", "2", 
"1", "1", "5", "4", "4", "4", "5", "1", "1", "3"), Label = c("Test1", 
"Test2", "Test3", "Test4", "Test5", "Test6", "Test7", "Test8", 
"Test9", "Test10", "Test11", "Test12", "Test13", "Test14", "Test15", 
"Test16"), Year = c(2001, 2002, 2003, 2004, 2005, 2006, 2007, 
2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016)), .Names = c("Value", 
"Label", "Year"), class = "data.frame", row.names = c(NA, 16L 
)), structure(list(Value = c("1", "2", "3", "1", "1", "4", "5", 
"1", "2", "3", "2", "2", "1", "1", "1", "5"), Label = c("Test1", 
"Test2", "Test3", "Test4", "Test5", "Test6", "Test7", "Test8", 
"Test9", "Test10", "Test11", "Test12", "Test13", "Test14", "Test15", 
"Test16"), Year = c(2001, 2002, 2003, 2004, 2005, 2006, 2007, 
2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016)), .Names = c("Value", 
"Label", "Year"), class = "data.table", row.names = c(NA, 16L 
)))), .Names = c("ID", "Country", "Subdocuments"), row.names = c(NA, 
-3L), class = "data.frame") 

Answer 1

mit würde ich

tun

setDT(data) 

dfcol = "Subdocuments" 
othcols = setdiff(names(data), dfcol) 

subs = rbindlist(data[[dfcol]], id=TRUE) 
subs[, (othcols) := data[.id, othcols, with=FALSE]] 

Wenn Sie setDT(data) nicht wollen, können Sie die letzte Zeile wie data[.id, othcols] ändern.

Answer 2

Kann dies

library(data.table) 
rbindlist(setNames(data[[3]], do.call(paste, data[1:2])), idcol=TRUE)[ 
     , c("ID", "Country") := tstrsplit(.id, " ")][, .id := NULL][] 
# Value Label Year ID  Country 
# 1:  5 Test1 2001 1 Netherlands 
# 2:  5 Test2 2002 1 Netherlands 
# 3:  1 Test3 2003 1 Netherlands 
# 4:  3 Test4 2004 1 Netherlands 
# 5:  2 Test5 2005 1 Netherlands 
# 6:  1 Test6 2006 1 Netherlands 
# 7:  1 Test7 2007 1 Netherlands 
# 8:  1 Test8 2008 1 Netherlands 
# 9:  2 Test9 2009 1 Netherlands 
#10:  5 Test10 2010 1 Netherlands 
#11:  3 Test11 2011 1 Netherlands 
#12:  2 Test12 2012 1 Netherlands 
#13:  4 Test13 2013 1 Netherlands 
#14:  5 Test14 2014 1 Netherlands 
#15:  5 Test15 2015 1 Netherlands 
#16:  2 Test16 2016 1 Netherlands 
#17:  5 Test1 2001 2  Germany 
#18:  4 Test2 2002 2  Germany 
#19:  3 Test3 2003 2  Germany 
#20:  2 Test4 2004 2  Germany 
#21:  2 Test5 2005 2  Germany 
#22:  2 Test6 2006 2  Germany 
#23:  1 Test7 2007 2  Germany 
#24:  1 Test8 2008 2  Germany 
#25:  5 Test9 2009 2  Germany 
#26:  4 Test10 2010 2  Germany 
#27:  4 Test11 2011 2  Germany 
#28:  4 Test12 2012 2  Germany 
#29:  5 Test13 2013 2  Germany 
#30:  1 Test14 2014 2  Germany 
#31:  1 Test15 2015 2  Germany 
#32:  3 Test16 2016 2  Germany 
#33:  1 Test1 2001 3  Belgium 
#34:  2 Test2 2002 3  Belgium 
#35:  3 Test3 2003 3  Belgium 
#36:  1 Test4 2004 3  Belgium 
#37:  1 Test5 2005 3  Belgium 
#38:  4 Test6 2006 3  Belgium 
#39:  5 Test7 2007 3  Belgium 
#40:  1 Test8 2008 3  Belgium 
#41:  2 Test9 2009 3  Belgium 
#42:  3 Test10 2010 3  Belgium 
#43:  2 Test11 2011 3  Belgium 
#44:  2 Test12 2012 3  Belgium 
#45:  1 Test13 2013 3  Belgium 
#46:  1 Test14 2014 3  Belgium 
#47:  1 Test15 2015 3  Belgium 
#48:  5 Test16 2016 3  Belgium 

HINWEIS hilft: Die 'Daten' ist aus der eigenen Post OP.

---

Oder dplyr

library(dplyr) 
bind_rows(data[[3]], .id="ID") %>% 
      left_join(data[-3], ., by = "ID")