R program: list in one column -> data-frame with dimension

Question

I have a dataframe with 3840 rows (only one value in columnA). I would like to convert this into a dataframe (40 rows & 96 columns). I tried to use the following code.

a1<-a[1:40,]
a2<-cbind(a1,a[41:80,])
a3<-cbind(a2,a[81:120,])
a4<-cbind(a3,a[121:160,])
a5<-cbind(a4,a[161:200,])
a6<-cbind(a5,a[201:240,])
......
a96<-cbind(a95,a[3801:3840,])

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This is too tedious work. Can I use either while loop if loop? If anyone can help me, it will be a great. Thanks.

Answer 1

Just redim

`dim<-`(a[,1],c(40,96))

and you are done (wrap with as.data.frame as desired)

Answer 2

我们可以使用matrix

 matrix(a[,1], nrow=40, ncol=96)

Answer 3

You could do this with data.table like so (noting that this has the advantage that it is easily extended to handle the case when a has two or many columns that you'd like to reshape):

library(data.table) #1.9.7 + 
setDT(a)[, ct := 1:40]
dcast(a, rowid(ct) ~ ct, value.var = "columnA")

(instructions for installing the development version, which has the rowid function, here )

Test data:

set.seed(10932)
a <- data.table(columnA = rnorm(3840))

---

If you're feeling parsimonious, here's a one-line version:

dcast(a, rowid(ct <<- rep_len(1:40, nrow(a))) ~ ct, value.var = "columnA")

And another (this viable in reshape2 and the current CRAN version of data.table ):

dcast(a, rep(1:ceiling((nrow(a) / 40)), each = 40, length.out = nrow(a)) ~
        rep_len(1:40, nrow(a)), value.var = "columnA")

Answer 4

Here's a base R method:

# Fake data
dat = data.frame(v1 = rnorm(3840))

# Split into columns of 40 rows each
dat.new = as.data.frame(mapply(function(i,j) {dat[i:j,]},
  which(1:nrow(dat) %% 40 == 1), which(1:nrow(dat) %% 40 == 0)))