R-生成二进制向量的所有可能的成对组合 [英] R - generate all possible pairwise combinations of binary vectors

问题描述

我正在寻找一种聪明的方法来生成长度为n的两个向量的所有成对组合，其中只有一个值不为零.

就目前而言，我正在通过以下组合对每个循环进行绝望的操作:n<-3; z<-rep(0，n); m<-apply(combn(1:n，1)，2，function(k){z [k] = 1; z}) 但是必须有一种更好的无循环方法吗?

这就是我想要的，例如n = 3:

     [,1] [,2] [,3]
[1,]    1    0    0
[2,]    0    1    0  

[1,]    1    0    0
[2,]    0    0    1

[1,]    0    1    0
[2,]    1    0    0

[1,]    0    1    0
[2,]    0    0    1

[1,]    0    0    1
[2,]    1    0    0

[1,]    0    0    1
[2,]    0    1    0

非常感谢您的帮助.

解决方案

像这样吗?

n <- 3
g <- 2 # g must be < n 
m <- combn(n, g)
mm <- as.numeric(m)
mat <- matrix(0, nrow = g * ncol(m), ncol = n)
mat[ cbind(1:nrow(mat), mm)] <- 1

mat
#       [,1] [,2] [,3]
#[1,]    1    0    0
#[2,]    0    1    0

#[3,]    1    0    0
#[4,]    0    0    1

#[5,]    0    1    0
#[6,]    0    0    1

# mat is half the answer :)
# the other half is
mat[nrow(mat):1, ]

#      [,1] [,2] [,3]
#[1,]    0    0    1
#[2,]    0    1    0

#[3,]    0    0    1
#[4,]    1    0    0

#[5,]    0    1    0
#[6,]    1    0    0

soln <- rbind(mat, mat[nrow(mat):1, ])

# as suggested by the OP to split the soln 
d <- split(data.frame(soln), rep(1:(nrow(soln)/g), each=g))

I am looking for a smart way to generate all pairwise combinations of two vectors of length n, where only one value is not zero.

For now I am doing something quite desperate with loops through each combination with: n <- 3; z <- rep(0,n); m <- apply(combn(1:n,1),2,function(k) {z[k]=1;z}) but there must be a better way without loops?

This is what I'm after for example for n=3:

     [,1] [,2] [,3]
[1,]    1    0    0
[2,]    0    1    0  

[1,]    1    0    0
[2,]    0    0    1

[1,]    0    1    0
[2,]    1    0    0

[1,]    0    1    0
[2,]    0    0    1

[1,]    0    0    1
[2,]    1    0    0

[1,]    0    0    1
[2,]    0    1    0

Thanks so much for the help.

解决方案

Something like this?

n <- 3
g <- 2 # g must be < n 
m <- combn(n, g)
mm <- as.numeric(m)
mat <- matrix(0, nrow = g * ncol(m), ncol = n)
mat[ cbind(1:nrow(mat), mm)] <- 1

mat
#       [,1] [,2] [,3]
#[1,]    1    0    0
#[2,]    0    1    0

#[3,]    1    0    0
#[4,]    0    0    1

#[5,]    0    1    0
#[6,]    0    0    1

# mat is half the answer :)
# the other half is
mat[nrow(mat):1, ]

#      [,1] [,2] [,3]
#[1,]    0    0    1
#[2,]    0    1    0

#[3,]    0    0    1
#[4,]    1    0    0

#[5,]    0    1    0
#[6,]    1    0    0

soln <- rbind(mat, mat[nrow(mat):1, ])

# as suggested by the OP to split the soln 
d <- split(data.frame(soln), rep(1:(nrow(soln)/g), each=g))

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