23

我有一个矩阵

df<-matrix(data=c(3,7,5,0,1,0,0,0,0,8,0,9), ncol=2)
rownames(df)<-c("a","b","c","d","e","f")

[,1] [,2]
a    3    0
b    7    0
c    5    0
d    0    8
e    1    0
f    0    9

我想先按第 1 列按降序排列矩阵,然后按第 2 列按降序排列,得到矩阵

df.ordered<-matrix(data=c(7,5,3,1,0,0,0,0,0,0,9,8),ncol=2)
rownames(df.ordered)<-c("b","c","a","e","f","d")

   [,1] [,2]
 b    7    0
 c    5    0
 a    3    0
 e    1    0
 f    0    9
 d    0    8

关于如何实现这一目标的任何建议?谢谢。

4

3 回答 3

31

order功能应该做到这一点。

df[order(df[,1],df[,2],decreasing=TRUE),]
于 2012-08-22T16:28:26.367 回答
17

要完成主要答案,这是一种以编程方式完成的方法,而无需手动指定列:

set.seed(2013) # preparing my example
mat <- matrix(sample.int(10,size = 30, replace = T), ncol = 3)
mat
      [,1] [,2] [,3]
 [1,]    5    1    6
 [2,]   10    3    1
 [3,]    8    8    1
 [4,]    8    9    9
 [5,]    3    7    3
 [6,]    8    8    5
 [7,]   10   10    2
 [8,]    8   10    7
 [9,]   10    1    9
[10,]    9    4    5

作为一个简单的例子,假设我想使用所有列的出现顺序来对矩阵的行进行排序:(可以很容易地为矩阵提供一个索引向量)

mat[do.call(order, as.data.frame(mat)),]   #could be ..as.data.frame(mat[,index_vec])..
      [,1] [,2] [,3]
 [1,]    3    7    3
 [2,]    5    1    6
 [3,]    8    8    1
 [4,]    8    8    5
 [5,]    8    9    9
 [6,]    8   10    7
 [7,]    9    4    5
 [8,]   10    1    9
 [9,]   10    3    1
[10,]   10   10    2
于 2013-07-12T05:26:07.497 回答
5

order功能会帮助你,试试这个:

df[order(-df[,1],-df[,2]),] 
  [,1] [,2]
b    7    0
c    5    0
a    3    0
e    1    0
f    0    9
d    0    8

前面的减号df表示订单正在减少。您将获得相同的结果设置decreasing=TRUE

df[order(df[,1],df[,2],decreasing=TRUE),]
于 2012-08-22T16:33:18.763 回答