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!
?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]) > dfa b 1 1 a 2 2 b 3 3 c ## add a list column, protected by "I()"> df$new <- I(list(1:5,g = "foo", abb = matrix(runif(6),nr=3)))## works as advertised> dfa b new 1 1 a 1, 2, 3,.... 2 2 b foo 3 3 c 0.080349....> df$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$","new4","<- mylist"))) > >> df > # Sepal.Length Sepal.Width Petal.Length Petal.Width Species n1 new2 new3 new4 > # 1 5.1 3.5 1.4 0.2 setosa 1, 2, 3 1 1 1, 2, 3 > # 2 4.9 3.0 1.4 0.2 setosa 1 2 2 1 > # 3 4.7 3.2 1.3 0.2 setosa NULL 3 3 NULL > > > The "eval" works correctly, however, if I want to avoid using "eval", what > should I do? > > Thanks! > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.
If you see my previous example, I have tried something like> df[,n3] <- I(mylist)However, in my case, the name of the new column is in a variable (by user input) which can not be directly used by the dollar assign. On the other hand, "[<-" does not work correctly even if I wrap the list into "I()". Perhaps the title of my email is a little unclear, sorry for the case. Best, Marlin. On Sun, 2016-12-11 at 03:03 -0800, Bert Gunter wrote:> ?data.frame says: > > "If a list or data frame or matrix is passed to data.frame it is as > if > each component or column had been passed as a separate argument > (except for matrices of class "model.matrix" and those protected by > I). " > > So doing what Help says to do seems to do what you asked: > > > > df <- data.frame(a=1:3,b=letters[1:3]) > > df > > ? a b > 1 1 a > 2 2 b > 3 3 c > > ## add a list column, protected by "I()" > > df$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! > > > > ______________________________________________ > > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide http://www.R-project.org/posting-g > > uide.html > > and provide commented, minimal, self-contained, reproducible code.