I have a data.frame with one id column (x below), and a number of variables (y1,y2 below).
我有一个id列(x下)和一些变量(y1,y2)的数据。
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 columns 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):
我想从中生成一个dataframe，其中前两列包含x的每个有序组合(除了它们是相等的)，以及与顺序相关的每个变量的列。数据帧头和前两行看起来是这样的(手工操作，请原谅错误):
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.
因此，每一行将包含一个源和目的地(i和j)，然后在每个源和目的地为y1设置值。
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.
我正在慢慢地学习R数据处理，但是这个却让我困惑。这句话是“一言既出，驷马难追”的回答，以及可读性更强的说教性回答。
4 个解决方案
#1
4
This works (apart perhaps from order)
这是有效的(除了顺序)
firstdf <- data.frame(x = c( 1, 2, 4, 5),
y1 = c(43,51,57,49), y2 = c(55,53,47,44))
co <- combn(firstdf$x,2)
seconddf <- data.frame(xi = c(co[1,], co[2,]), xj = c(co[2,], co[1,]))
thirddf <- merge(merge(seconddf, firstdf, by.x = "xj", by.y = "x" ),
firstdf, by.x = "xi", by.y = "x", suffixes = c("j", "i") )
to produce
生产
> thirddf
xi xj y1j y2j y1i y2i
1 1 2 51 53 43 55
2 1 5 49 44 43 55
3 1 4 57 47 43 55
4 2 4 57 47 51 53
5 2 1 43 55 51 53
6 2 5 49 44 51 53
7 4 5 49 44 57 47
8 4 1 43 55 57 47
9 4 2 51 53 57 47
10 5 1 43 55 49 44
11 5 2 51 53 49 44
12 5 4 57 47 49 44
where the first and fifth rows match your example.
第一行和第5行与示例匹配的地方。
If you take firstdf as given and insist on one line then you can turn this into
如果你把firstdf作为给定的并且坚持一行，那么你可以把它变成。
merge(merge(data.frame(xi = c(combn(firstdf$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
但我不太明白这一点
#2
4
Two lines is the best I can do and still keep it sensible: (Edit: see bottom of answer for one-liner.)
两行是我能做的最好的，但仍然保持合理的:
Create some data:
创建一些数据:
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:
代码:
f <- function(x, i){cbind(i, x[i[,1],], x[i[,2],])}
f(a, t(combn(seq_len(nrow(a)), 2)))
The results:
结果:
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
编辑
This can be turned into a one-liner by making use of anonymous functions:
通过使用匿名函数，可以将其转变为一行代码:
(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"
#3
2
I'm not sure what you exactly want in general, but as far as my understanding, this may be close to what you want:
我不确定你到底想要什么，但就我的理解而言，这可能接近你想要的:
> 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.
如果你提供更详细的信息，你可能会得到更好的答案。
#4
1
Well, it's nowhere close to a one-liner (which I kind of doubt is possible) but here's a 'naive' approach:
好吧，这根本就不可能是一艘客轮(我怀疑这是可能的)，但这里有一个“天真”的方法:
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$y2[rs[i,1:2]])
}
I didn't name the columns, but that's easily done after the fact.
我没有命名列，但这很容易做到。
Not very elegant, but maybe something to get you started...
不是很优雅，但也许你可以开始……
#1
4
This works (apart perhaps from order)
这是有效的(除了顺序)
firstdf <- data.frame(x = c( 1, 2, 4, 5),
y1 = c(43,51,57,49), y2 = c(55,53,47,44))
co <- combn(firstdf$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
生产
> 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.
第一行和第5行与示例匹配的地方。
If you take firstdf as given and insist on one line then you can turn this into
如果你把firstdf作为给定的并且坚持一行，那么你可以把它变成。
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
但我不太明白这一点
#2
4
Two lines is the best I can do and still keep it sensible: (Edit: see bottom of answer for one-liner.)
两行是我能做的最好的，但仍然保持合理的:
Create some data:
创建一些数据:
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:
代码:
f <- function(x, i){cbind(i, x[i[,1],], x[i[,2],])}
f(a, t(combn(seq_len(nrow(a)), 2)))
The results:
结果:
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
编辑
This can be turned into a one-liner by making use of anonymous functions:
通过使用匿名函数，可以将其转变为一行代码:
(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"
#3
2
I'm not sure what you exactly want in general, but as far as my understanding, this may be close to what you want:
我不确定你到底想要什么，但就我的理解而言，这可能接近你想要的:
> 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.
如果你提供更详细的信息，你可能会得到更好的答案。
#4
1
Well, it's nowhere close to a one-liner (which I kind of doubt is possible) but here's a 'naive' approach:
好吧，这根本就不可能是一艘客轮(我怀疑这是可能的)，但这里有一个“天真”的方法:
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.
我没有命名列，但这很容易做到。
Not very elegant, but maybe something to get you started...
不是很优雅，但也许你可以开始……