我有这个矩阵,我想获得一个2列的矩阵,其中一列有rowname和其他列,细胞与值1 colname的
x
X1 X2 X3
X1 1 0 1
X2 0 1 0
X3 0 1 1
X4 1 0 0
str(x)
num [1:886, 1:886] 1 0 1 1 1 0 1 1 1 1
我想这样的矩阵
# X1 X1
# X1 X3
# X2 X2
# X3 X2
# X3 X3
# X4 X1
其是具有值的双= 1
在预先感谢A.
这里的另一种选择:
mm <- expand.grid(rownames(mat),colnames(mat))[as.vector(mat==1),]
Var1 Var2
1 X1 X1
4 X4 X1
6 X2 X2
7 X3 X2
9 X1 X3
11 X3 X3
并获得了OP显示,我们为了在第一列:
mm[order(mm$Var1),]
Var1 Var2
1 X1 X1
9 X1 X3
6 X2 X2
7 X3 X2
11 X3 X3
4 X4 X1
在这里,我垫的开关输入,那我重现:
mat <- data.frame(X1=c(1,0,0,1),X2=c(0,1,1,0),X3=c(1,0,1,0))
rownames(mat)= paste0('X',1:4)
X1 X2 X3
X1 1 0 1
X2 0 1 0
X3 0 1 1
X4 1 0 0
你可以这样做:
mat <- which(x==1, arr.ind=TRUE)
mat[,"col"] <- names(x)[mat[,"col"]]
mat[,"row"] <- rownames(mat)
这将给:
row col
X1 "X1" "X1"
X4 "X4" "X1"
X2 "X2" "X2"
X3 "X3" "X2"
X1 "X1" "X3"
X3 "X3" "X3"
下面是一行答案
x
## X1 X2 X3
## X1 1 0 1
## X2 0 1 0
## X3 0 1 1
## X4 1 0 0
cbind(rownames(x)[row(x) * x], colnames(x)[col(x) * x])
## [,1] [,2]
## [1,] "X1" "X1"
## [2,] "X4" "X1"
## [3,] "X2" "X2"
## [4,] "X3" "X2"
## [5,] "X1" "X3"
## [6,] "X3" "X3"