Given an R dataframe with column A, how do I create two new columns containing all ordered combinations of A_问答_开发者

I have a data.frame with one id column (x below), and a number of variables (y1,y2 below).

    x y1 y2
1   1 43 55
2   2 51 53
[...]

What I would like to generate from this is a dataframe where the first two columns cover every ordered combination of x (except where they are equal) along with c开发者_开发问答olumns for each variable related to the order. The data frame header and first two rows would look like this (did this by hand, excuse errors):

xi xj y1i y1j y2i y2j
 1  2  43  51  55  53
 2  1  51  43  53  55
[...]

So each row would container a source and destination (i and j) and then values for y1 at each source and destination.

I'm slowly learning R data manipulation, but this one is stumping me. Kudos for the one line does-it-all answer, as well as a more readable didactic answer.

This works (apart perhaps from order)

firstdf  <- data.frame(x  = c( 1, 2, 4, 5), 
                       y1 = c(43,51,57,49), y2 = c(55,53,47,44)) 
co       <- combn(firstdf$x,2)
seconddf <- data.frame(xi = c(co[1,], co[2,]), xj = c(co[2,], co[1,]))
thirddf  <- merge(merge(seconddf, firstdf, by.x = "xj", by.y = "x" ),
                  firstdf, by.x = "xi", by.y = "x", suffixes = c("j", "i") )

to produce

> thirddf
   xi xj y1j y2j y1i y2i
1   1  2  51  53  43  55
2   1  5  49  44  43  55
3   1  4  57  47  43  55
4   2  4  57  47  51  53
5   2  1  43  55  51  53
6   2  5  49  44  51  53
7   4  5  49  44  57  47
8   4  1  43  55  57  47
9   4  2  51  53  57  47
10  5  1  43  55  49  44
11  5  2  51  53  49  44
12  5  4  57  47  49  44

where the first and fifth rows match your example.

If you take firstdf as given and insist on one line then you can turn this into

merge(merge(data.frame(xi = c(combn(firstdf$x,2)[1,], combn(firstdf$x,2)[2,]), xj = c(combn(firstdf$x,2)[2,], combn(firstdf$x,2)[1,])), firstdf, by.x = "xj", by.y = "x" ), firstdf, by.x = "xi", by.y = "x", suffixes = c("j", "i") )

but I don't really see the point

Two lines is the best I can do and still keep it sensible: (Edit: see bottom of answer for one-liner.)

Create some data:

n <- 4
a <- cbind(x=LETTERS[1:n], y=letters[1:n])
a

     x   y  
[1,] "A" "a"
[2,] "B" "b"
[3,] "C" "c"
[4,] "D" "d"

The code:

f <- function(x, i){cbind(i, x[i[,1],], x[i[,2],])}
f(a, t(combn(seq_len(nrow(a)), 2)))

The results:

             x   y   x   y  
[1,] "1" "2" "A" "a" "B" "b"
[2,] "1" "3" "A" "a" "C" "c"
[3,] "1" "4" "A" "a" "D" "d"
[4,] "2" "3" "B" "b" "C" "c"
[5,] "2" "4" "B" "b" "D" "d"
[6,] "3" "4" "C" "c" "D" "d"

EDIT

This can be turned into a one-liner by making use of anonymous functions:

(function(x, i=t(combn(seq_len(nrow(a)), 2))){cbind(i, x[i[,1],], x[i[,2],])})(a)

             x   y   x   y  
[1,] "1" "2" "A" "a" "B" "b"
[2,] "1" "3" "A" "a" "C" "c"
[3,] "1" "4" "A" "a" "D" "d"
[4,] "2" "3" "B" "b" "C" "c"
[5,] "2" "4" "B" "b" "D" "d"
[6,] "3" "4" "C" "c" "D" "d"

I'm not sure what you exactly want in general, but as far as my understanding, this may be close to what you want:

> library(combinat) # for permn
> library(plyr) # for llply
> 
> # sample data
> d <- data.frame(x = 1:3, y1 = rnorm(3), y2 = rnorm(3))
> d
  x          y1         y2
1 1 -0.17525893 -1.1660321
2 2 -0.05585689 -0.2059244
3 3  0.90500983 -1.3067601
> 
> # permutation of rows
> idx <- permn(nrow(d))
> idx
[[1]]
[1] 1 2 3

... snip ...

[[6]]
[1] 2 1 3

> 
> # a list of perm-ed data.frame
> d2 <- llply(idx, function(i)data.frame(idx = 1:nrow(d), d[i,]))
> d2
[[1]]
  idx x          y1         y2
1   1 1 -0.17525893 -1.1660321
2   2 2 -0.05585689 -0.2059244
3   3 3  0.90500983 -1.3067601

... snip ...

[[6]]
  idx x          y1         y2
2   1 2 -0.05585689 -0.2059244
1   2 1 -0.17525893 -1.1660321
3   3 3  0.90500983 -1.3067601

> 
> # merge htam
> d3 <- subset(Reduce(function(df1, df2) merge(df1, df2, by="idx"), d2), select = -c(idx))
> d3
  x.x        y1.x       y2.x x.y        y1.y       y2.y x.x.1      y1.x.1     y2.x.1 x.y.1      y1.y.1     y2.y.1 x.x.2      y1.x.2     y2.x.2 x.y.2
1   1 -0.17525893 -1.1660321   1 -0.17525893 -1.1660321     3  0.90500983 -1.3067601     3  0.90500983 -1.3067601     2 -0.05585689 -0.2059244     2
2   2 -0.05585689 -0.2059244   3  0.90500983 -1.3067601     1 -0.17525893 -1.1660321     2 -0.05585689 -0.2059244     3  0.90500983 -1.3067601     1
3   3  0.90500983 -1.3067601   2 -0.05585689 -0.2059244     2 -0.05585689 -0.2059244     1 -0.17525893 -1.1660321     1 -0.17525893 -1.1660321     3
       y1.y.2     y2.y.2
1 -0.05585689 -0.2059244
2 -0.17525893 -1.1660321
3  0.90500983 -1.3067601
> 
> # and here is the one-liner version
> subset(Reduce(function(df1, df2) merge(df1, df2, by="idx"), llply(permn(nrow(d)), function(i)data.frame(idx=1:nrow(d), d[i,]))), select=-c(idx))
  x.x        y1.x       y2.x x.y        y1.y       y2.y x.x.1      y1.x.1     y2.x.1 x.y.1      y1.y.1     y2.y.1 x.x.2      y1.x.2     y2.x.2 x.y.2
1   1 -0.17525893 -1.1660321   1 -0.17525893 -1.1660321     3  0.90500983 -1.3067601     3  0.90500983 -1.3067601     2 -0.05585689 -0.2059244     2
2   2 -0.05585689 -0.2059244   3  0.90500983 -1.3067601     1 -0.17525893 -1.1660321     2 -0.05585689 -0.2059244     3  0.90500983 -1.3067601     1
3   3  0.90500983 -1.3067601   2 -0.05585689 -0.2059244     2 -0.05585689 -0.2059244     1 -0.17525893 -1.1660321     1 -0.17525893 -1.1660321     3
       y1.y.2     y2.y.2
1 -0.05585689 -0.2059244
2 -0.17525893 -1.1660321
3  0.90500983 -1.3067601

If you provide information in more detail, probably you can get better answers.

Well, it's nowhere close to a one-liner (which I kind of doubt is possible) but here's a 'naive' approach:

dat <- data.frame(x=1:5,y1=6:10,y2=11:15)

#Collect all ordered pairs of elements of x
tmp <- expand.grid(dat$x,dat$x)
tmp <- tmp[tmp[,1] != tmp[,2],]

#Init a matrix to hold the results
rs <- as.matrix(cbind(tmp,matrix(NA,nrow(tmp),4)))

#Loop through each ordered pair
for (i in 1:nrow(rs)){
    rs[i,3:6] <- c(dat$y1[rs[i,1:2]],dat$y2[rs[i,1:2]])
}

I didn't name the columns, but that's easily done after the fact.

Not very elegant, but maybe something to get you started...