R逐行堆栈数据

Question

I am trying to do stack. My data is

set.seed(1)
x<-runif(5)
y<-runif(5)
dat<-cbind(x,y)
dat<-as.data.frame(dat)
dat
   x          y
1 0.2655087 0.89838968
2 0.3721239 0.94467527
3 0.5728534 0.66079779
4 0.9082078 0.62911404
5 0.2016819 0.06178627


stack(dat)

   values ind
1  0.26550866   x
2  0.37212390   x
3  0.57285336   x
4  0.90820779   x
5  0.20168193   x
6  0.89838968   y
7  0.94467527   y
8  0.66079779   y
9  0.62911404   y
10 0.06178627   y

However, this stacks by column ie it takes y column and puts it below x . What I want to do is to stack it by row like this:

0.2655087    x
0.89838968   y
0.3721239    x
0.94467527   y
0.5728534    x
0.66079779   y
0.9082078    x
0.62911404   y
0.2016819    x
0.06178627   y

How can this be done using stack ?

Thanks

Answer 1

A base R method that exploits the column dominant storage of matrices. The columns x and y are turned into a matrix, which is transposed and then unwrapped into a vector. Since we know the structure (ordering) of the resulting vector, we build the xy names into a new variable:

data.frame(values=c(t(data.matrix(dat))), ind=I(rep(colnames(dat), nrow(dat))))

Which returns

       values ind
1  0.26550866   x
2  0.89838968   y
3  0.37212390   x
4  0.94467527   y
5  0.57285336   x
6  0.66079779   y
7  0.90820779   x
8  0.62911404   y
9  0.20168193   x
10 0.06178627   y

I wrapped the xy vector in I to "insulate" it, so that it would return as a character vector within the data.frame function rather than as a factor, which is the default. Using the stringsAsFactors=TRUE argument in data.frame would also return the xy vector as a character type.

Answer 2

Why do you have to use stack()? This will do the trick:

# Creating your data frame
set.seed(1)
x<-runif(5)
y<-runif(5)
dat<-cbind(x,y)
dat<-as.data.frame(dat)
dat

# Stacking the data
dat2 <- rbind(data.frame("Value"=dat$x,"Ind"="x","Row"=seq(nrow(dat))),
      data.frame("Value"=dat$y,"Ind"="y","Row"=seq(nrow(dat))))

# Ordering the data
dat2 <- dat2[order(dat2$Row),setdiff(names(dat2),"Row")]

        Value Ind
1  0.26550866   x
6  0.89838968   y
2  0.37212390   x
7  0.94467527   y
3  0.57285336   x
8  0.66079779   y
4  0.90820779   x
9  0.62911404   y
5  0.20168193   x
10 0.06178627   y

Answer 3

We could simply do:

data.frame(values=matrix(t(dat)), ind=colnames(dat))

       # values ind
# 1  0.26550866   x
# 2  0.89838968   y
# 3  0.37212390   x
# 4  0.94467527   y
# 5  0.57285336   x
# 6  0.66079779   y
# 7  0.90820779   x
# 8  0.62911404   y
# 9  0.20168193   x
# 10 0.06178627   y