即使你有丢失的对，它也能工作，而且它不需要排序(as.matrix(dat1)[，1:2]可以用cbind(dat1[，1]，dat1[，2])替换):
> set.seed(45);dat1=data.frame(name=rep(c("firstName","secondName"),each=4),numbers=rep(1:4,2),value=rnorm(8))
> u1=unique(dat1[,1]);u2=unique(dat1[,2])
> m=matrix(nrow=length(u1),ncol=length(u2),dimnames=list(u1,u2))
> m[as.matrix(dat1)[,1:2]]=dat1[,3]
> m
1 2 3 4
firstName 0.3407997 -0.7033403 -0.3795377 -0.7460474
secondName -0.8981073 -0.3347941 -0.5013782 -0.1745357
如果你有缺少的对并且需要排序，这是行不通的，但如果对已经排序了，它会更短一些:
> u1=unique(dat1[,1]);u2=unique(dat1[,2])
> dat1=dat1[order(dat1[,1],dat1[,2]),] # not actually needed in this case
> matrix(dat1[,3],length(u1),,T,list(u1,u2))
1 2 3 4
firstName 0.3407997 -0.7033403 -0.3795377 -0.7460474
secondName -0.8981073 -0.3347941 -0.5013782 -0.1745357
下面是第一个方法的函数版本(添加as.data.frame使其与tibbles一起工作):
l2w=function(x,row=1,col=2,val=3,sort=F){
u1=unique(x[,row])
u2=unique(x[,col])
if(sort){u1=sort(u1);u2=sort(u2)}
out=matrix(nrow=length(u1),ncol=length(u2),dimnames=list(u1,u2))
out[cbind(x[,row],x[,col])]=x[,val]
out
}
或者如果你只有下三角形的值，你可以这样做:
> euro=as.matrix(eurodist)[1:3,1:3]
> lower=data.frame(V1=rownames(euro)[row(euro)[lower.tri(euro)]],V2=colnames(euro)[col(euro)[lower.tri(euro)]],V3=euro[lower.tri(euro)])
> lower
V1 V2 V3
1 Barcelona Athens 3313
2 Brussels Athens 2963
3 Brussels Barcelona 1318
> n=unique(c(lower[,1],lower[,2]))
> full=rbind(lower,setNames(lower[,c(2,1,3)],names(lower)),data.frame(V1=n,V2=n,V3=0))
> full
V1 V2 V3
1 Barcelona Athens 3313
2 Brussels Athens 2963
3 Brussels Barcelona 1318
4 Athens Barcelona 3313
5 Athens Brussels 2963
6 Barcelona Brussels 1318
7 Athens Athens 0
8 Barcelona Barcelona 0
9 Brussels Brussels 0
> l2w(full,sort=T)
Athens Barcelona Brussels
Athens 0 3313 2963
Barcelona 3313 0 1318
Brussels 2963 1318 0
或者还有另一种方法:
> rc=as.matrix(lower[-3])
> n=sort(unique(c(rc)))
> m=matrix(0,length(n),length(n),,list(n,n))
> m[rc]=lower[,3]
> m[rc[,2:1]]=lower[,3]
> m
Athens Barcelona Brussels
Athens 0 3313 2963
Barcelona 3313 0 1318
Brussels 2963 1318 0
base R中的另一个简单方法是使用xtabs。xtabs的结果基本上只是一个带有花哨类名的矩阵，但你可以让它看起来像一个普通的矩阵，class(x)=NULL;attr(x，"call")=NULL;dimnames(x)=unname(dimnames(x)):
> x=xtabs(value~name+numbers,dat1);x
numbers
name 1 2 3 4
firstName 0.3407997 -0.7033403 -0.3795377 -0.7460474
secondName -0.8981073 -0.3347941 -0.5013782 -0.1745357
> str(x)
'xtabs' num [1:2, 1:4] 0.341 -0.898 -0.703 -0.335 -0.38 ...
- attr(*, "dimnames")=List of 2
..$ name : chr [1:2] "firstName" "secondName"
..$ numbers: chr [1:4] "1" "2" "3" "4"
- attr(*, "call")= language xtabs(formula = value ~ name + numbers, data = dat1)
> class(x)
[1] "xtabs" "table"
> class(as.matrix(x)) # `as.matrix` has no effect because `x` is already a matrix
[1] "xtabs" "table"
> class(x)=NULL;class(x)
[1] "matrix" "array"
> attr(x,"call")=NULL;dimnames(x)=unname(dimnames(x))
> x # now it looks like a regular matrix
1 2 3 4
firstName 0.3407997 -0.7033403 -0.3795377 -0.7460474
secondName -0.8981073 -0.3347941 -0.5013782 -0.1745357
> str(x)
num [1:2, 1:4] 0.341 -0.898 -0.703 -0.335 -0.38 ...
- attr(*, "dimnames")=List of 2
..$ : chr [1:2] "firstName" "secondName"
..$ : chr [1:4] "1" "2" "3" "4"
通常as.data.frame(x)将xtab的结果转换回长格式，但你可以使用class(x)=NULL来避免这种情况:
> x=xtabs(value~name+numbers,dat1);as.data.frame(x)
name numbers Freq
1 firstName 1 0.3407997
2 secondName 1 -0.8981073
3 firstName 2 -0.7033403
4 secondName 2 -0.3347941
5 firstName 3 -0.3795377
6 secondName 3 -0.5013782
7 firstName 4 -0.7460474
8 secondName 4 -0.1745357
> class(x)=NULL;as.data.frame(x)
1 2 3 4
firstName 0.3407997 -0.7033403 -0.3795377 -0.7460474
secondName -0.8981073 -0.3347941 -0.5013782 -0.1745357
这将宽格式的数据转换为长格式(unlist将数据帧转换为向量，c将矩阵转换为向量):
w2l=function(x)data.frame(V1=rownames(x)[row(x)],V2=colnames(x)[col(x)],V3=unname(c(unlist(x))))
