获取矩阵条目的排名?

问题描述：

假设一个矩阵:

> a <- matrix(c(100, 90, 80, 20), 2, 2)
> a
     [,1] [,2]
[1,]  100   80
[2,]   90   20

假设我想将矩阵的元素转换为秩:

Suppose I want to convert the elements of the matrix to ranks:

>rank.a <- rank(a)
> rank.a
[1] 4 3 2 1

这将返回一个向量，即矩阵结构丢失.是否可以对矩阵进行排序，以使输出形式为:

This returns a vector, i.e. the matrix structure is lost. Is it possible to rank a matrix such that the output will be of the form:

     [,1] [,2]
[1,]   4    2 
[2,]   3    1

答

@EDi答案的另一种方法是复制a，然后将rank(a)的输出直接分配到a副本的元素中:

An alternative to @EDi's Answer is to copy a and then assign the output of rank(a) directly into the elements of the copy of a:

> a <- matrix(c(100, 90, 80, 20), 2, 2)
> rank.a <- a
> rank.a[] <- rank(a)
> rank.a
     [,1] [,2]
[1,]    4    2
[2,]    3    1

通过查询输入矩阵的尺寸，可以避免重建矩阵.

That saves you from rebuilding a matrix by interrogating the dimensions of the input matrix.

请注意(如@Andrie在评论中所述)，仅当要保留原始a时才需要复制a.需要注意的主要点是，由于a已经具有适当的尺寸，因此我们可以将其视为向量，并用a等级的向量替换a的内容.

Note that (as @Andrie mentions in the comments) the copying of a is only required if one wants to keep the original a. The main point to note is that because a is already of the appropriate dimensions, we can treat it like a vector and replace the contents of a with the vector of ranks of a.

获取矩阵条目的排名?

相关推荐