我有一个简单的矩阵,如:
> a = matrix(c(c(1:10),c(10:1)), ncol=2)
> a
[,1] [,2]
[1,] 1 10
[2,] 2 9
[3,] 3 8
[4,] 4 7
[5,] 5 6
[6,] 6 5
[7,] 7 4
[8,] 8 3
[9,] 9 2
[10,] 10 1
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我想得到这个结果:
[,1] [,2]
[1,] 10 1
[2,] 9 2
[3,] 8 3
[4,] 7 4
[5,] 6 5
[6,] 5 6
[7,] 4 7
[8,] 3 8
[9,] 2 9
[10,] 1 10
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矩阵的确切反转.我怎么才能得到它?谢谢
Dir*_*tel 27
尝试rev用apply:
> a <- matrix(c(1:10,10:1), ncol=2)
> a
[,1] [,2]
[1,] 1 10
[2,] 2 9
[3,] 3 8
[4,] 4 7
[5,] 5 6
[6,] 6 5
[7,] 7 4
[8,] 8 3
[9,] 9 2
[10,] 10 1
> b <- apply(a, 2, rev)
> b
[,1] [,2]
[1,] 10 1
[2,] 9 2
[3,] 8 3
[4,] 7 4
[5,] 6 5
[6,] 5 6
[7,] 4 7
[8,] 3 8
[9,] 2 9
[10,] 1 10
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小智 22
a[nrow(a):1,]
# [,1] [,2]
# [1,] 10 1
# [2,] 9 2
# [3,] 8 3
# [4,] 7 4
# [5,] 6 5
# [6,] 5 6
# [7,] 4 7
# [8,] 3 8
# [9,] 2 9
# [10,] 1 10
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Jos*_*ien 11
这是一种方式:
a[, rev(seq_len(ncol(a)))]
[,1] [,2]
[1,] 10 1
[2,] 9 2
[3,] 8 3
[4,] 7 4
[5,] 6 5
[6,] 5 6
[7,] 4 7
[8,] 3 8
[9,] 2 9
[10,] 1 10
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