[R] Assign a list to one column of data frame

[Marlin JL.M]

Dear Bert,

This is awesome, thanks a lot!

Best,
Marlin


On Sun, 2016-12-11 at 06:52 -0800, Bert Gunter wrote:
> Use list indexing, "[[" not "[" .
> 
> > df <- data.frame(a=1:3,b=letters[1:3])
> > x <- "new"
> > df[[x]]<-  I(list(1:5,g = "foo", abb = matrix(runif(6),nr=3)))
> > df
> 
>   a b          new
> 1 1 a 1, 2, 3,....
> 2 2 b          foo
> 3 3 c 0.248115....
> > df$new
> 
> [[1]]
> [1] 1 2 3 4 5
> 
> $g
> [1] "foo"
> 
> $abb
>           [,1]      [,2]
> [1,] 0.2481156 0.2138564
> [2,] 0.8598658 0.2898058
> [3,] 0.5854885 0.4084578
> 
> 
> Cheers,
> Bert
> 
> 
> Bert Gunter
> 
> "The trouble with having an open mind is that people keep coming
> along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
> 
> 
> On Sun, Dec 11, 2016 at 3:54 AM, Marlin JL.M <marlin- at gmx.cn> wrote:
> > 
> > If you see my previous example, I have tried something like
> > > df[,n3] <- I(mylist)
> > 
> > However, in my case, the name of the new column is in a variable
> > (by
> > user input) which can not be directly used by the dollar assign. On
> > the
> > other hand, "[<-" does not work correctly even if I wrap the list
> > into
> > "I()".
> > 
> > Perhaps the title of my email is a little unclear, sorry for the
> > case.
> > 
> > Best,
> > Marlin.
> > 
> > On Sun, 2016-12-11 at 03:03 -0800, Bert Gunter wrote:
> > > ?data.frame says:
> > > 
> > > "If a list or data frame or matrix is passed to data.frame it is
> > > as
> > > if
> > > each component or column had been passed as a separate argument
> > > (except for matrices of class "model.matrix" and those protected
> > > by
> > > I). "
> > > 
> > > So doing what Help says to do seems to do what you asked:
> > > 
> > > 
> > > > df <- data.frame(a=1:3,b=letters[1:3])
> > > > df
> > > 
> > >   a b
> > > 1 1 a
> > > 2 2 b
> > > 3 3 c
> > > 
> > > ## add a list column, protected by "I()"
> > > > df$new <- I(list(1:5,g = "foo", abb = matrix(runif(6),nr=3)))
> > > 
> > > ## works as advertised
> > > > df
> > > 
> > >   a b          new
> > > 1 1 a 1, 2, 3,....
> > > 2 2 b          foo
> > > 3 3 c 0.080349....
> > > 
> > > > df$new
> > > 
> > > [[1]]
> > > [1] 1 2 3 4 5
> > > 
> > > $g
> > > [1] "foo"
> > > 
> > > $abb
> > >            [,1]       [,2]
> > > [1,] 0.08034915 0.83671591
> > > [2,] 0.43938440 0.06067429
> > > [3,] 0.88196881 0.33461234
> > > 
> > > 
> > > Cheers,
> > > Bert
> > > 
> > > 
> > > Bert Gunter
> > > 
> > > "The trouble with having an open mind is that people keep coming
> > > along
> > > and sticking things into it."
> > > -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip
> > > )
> > > 
> > > 
> > > On Sat, Dec 10, 2016 at 7:18 PM, Marlin JL.M <marlin- at gmx.cn>
> > > wrote:
> > > > Dear all,
> > > > 
> > > > I want to assign a list to one column of data.frame where the
> > > > name
> > > > of the column
> > > > is a variable. I tried the following:
> > > > 
> > > > Using R version 3.3.2
> > > > 
> > > > > df <- iris[1:3, ]
> > > > > df
> > > > 
> > > > # Sepal.Length Sepal.Width Petal.Length Petal.Width Species
> > > > # 1          5.1         3.5          1.4         0.2  setosa
> > > > # 2          4.9         3.0          1.4         0.2  setosa
> > > > # 3          4.7         3.2          1.3         0.2  setosa
> > > > 
> > > > > mylist <- list(c(1,2,3),c(1),c())
> > > > > mylist
> > > > 
> > > > # [[1]]
> > > > # [1] 1 2 3
> > > > #
> > > > # [[2]]
> > > > # [1] 1
> > > > #
> > > > # [[3]]
> > > > # NULL
> > > > 
> > > > > n1 <- 'new1'
> > > > > df$n1 <- mylist
> > > > > n2 <- 'new2'
> > > > > df[,n2] <- mylist
> > > > 
> > > > # Warning message:
> > > > # In `[<-.data.frame`(`*tmp*`, , "new4", value = list(c(1, 2,
> > > > 3),  :
> > > > # provided 3 variables to replace 1 variables
> > > > 
> > > > > n3 <- 'new3'
> > > > > df[,n3] <- I(mylist)
> > > > 
> > > > # Warning message:
> > > > # In `[<-.data.frame`(`*tmp*`, , "new3", value = list(c(1, 2,
> > > > 3),  :
> > > > # provided 3 variables to replace 1 variables
> > > > 
> > > > > eval(parse(text = paste0("df$","new4","<- mylist")))
> > > > > df
> > > > 
> > > > # Sepal.Length Sepal.Width Petal.Length Petal.Width
> > > > Species      n1
> > > > new2 new3    new4
> > > > # 1          5.1         3.5          1.4         0.2  setosa
> > > > 1, 2,
> > > > 3    1    1 1, 2, 3
> > > > #
> > > > 2          4.9         3.0          1.4         0.2  setosa    
> > > >    1
> > > >     2    2       1
> > > > #
> > > > 3          4.7         3.2          1.3         0.2  setosa    
> > > > NULL
> > > >     3    3    NULL
> > > > 
> > > > 
> > > > The "eval" works correctly, however, if I want to avoid using
> > > > "eval", what
> > > > should I do?
> > > > 
> > > > Thanks!
> > > > 
> > > > ______________________________________________
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