我想基于行/列名称对方阵内的块/子矩阵进行归一化（即，0-1）。重要的是归一化矩阵对应于原始矩阵。以下代码提取块，例如所有col / row名称==＆＃34; A＆＃34;并将其标准化为最大值。如何将标准化块的矩阵重新组合在一起，使其对应于原始矩阵，使得标准化块的每个单值与原始矩阵位于同一位置。即你不能把块放在一起，然后例如通过原始矩阵行/列名称对标准化矩阵进行排序。
#dummy code
mat <- matrix(round(runif(90, 0, 50),),9,9)
rownames(mat) <- rep(LETTERS[1:3],3)
colnames(mat) <- rep(LETTERS[1:3],3)
mat.n <- matrix(0,nrow(mat),ncol(mat), dimnames = list(rownames(mat),colnames(mat)))
for(i in 1:length(LETTERS[1:3])){
? <- mat[rownames(mat)==LETTERS[1:3][i],colnames(mat)==LETTERS[1:3][i]] / max(mat[rownames(mat)==LETTERS[1:3][i],colnames(mat)==LETTERS[1:3][i]])
#For example,
mat.n[rownames(mat)==LETTERS[1:3][i],colnames(mat)==LETTERS[1:3][i]] <- # doesn't work
}
更新
使用ave()作为@G。格洛腾迪克建议为这些街区工作，但我不确定它是如何正常化的。
mat.n <- mat / ave(mat, rownames(mat)[row(mat)], colnames(mat)[col(mat)], FUN = max)
在块内，规范化有效，例如
mat[rownames(mat)=="A",colnames(mat)=="A"]
A A A
A 13 18 15
A 38 33 41
A 12 18 47
mat.n[rownames(mat.n)=="A",colnames(mat.n)=="A"]
A A A
A 0.2765957 0.3829787 0.3191489
A 0.8085106 0.7021277 0.8723404
A 0.2553191 0.3829787 1.0000000
但除此之外，它看起来很奇怪。
> round(mat.n,1)
A B C A B C A B C
A 0.3 0.2 0.1 0.4 0.2 1.0 0.3 0.9 1.0
B 0.9 0.8 0.9 0.4 0.5 0.4 0.4 0.9 0.0
C 0.0 0.4 0.4 0.0 0.8 0.5 0.4 0.9 0.0
A 0.8 0.9 0.5 0.7 0.9 0.6 0.9 0.4 0.4
B 0.1 0.8 0.7 1.0 0.3 0.5 0.1 1.0 0.8
C 0.4 0.0 0.2 0.2 0.2 0.6 1.0 0.4 1.0
A 0.3 0.4 0.3 0.4 0.6 0.8 1.0 1.0 0.3
B 0.6 0.2 0.5 0.9 0.3 0.2 0.9 0.3 1.0
C 0.5 0.9 0.7 1.0 0.4 0.5 1.0 1.0 0.9
在这种情况下，我希望每个块的整个矩阵为1 1 -1。但是那里有10个，例如mat.n[3,2]，mat.n[1,9]。我不确定这个函数在块之间是如何规范化的。
更新2
#Original matrix.
#Suggested solution produces `NaN`
mat <- as.matrix(read.csv(text=",1.21,1.1,2.2,1.1,1.1,1.21,2.2,2.2,1.21,1.22,1.22,1.1,1.1,2.2,2.1,2.2,2.1,2.2,2.2,2.2,1.21,2.1,2.1,1.21,1.21,1.21,1.21,1.21,2.2,1.21,2.2,1.1,1.22,1.22,1.22,1.22,1.21,1.22,2.1,2.1,2.1,1.22
1.21,0,0,0,0,0,0,0,0,292,13,0,0,0,0,0,0,0,0,0,0,22,0,0,94,19,79,0,9,0,126,0,0,0,0,0,0,0,0,0,0,0,0
1.1,0,0,0,155,166,0,0,0,0,0,0,4,76,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,34,0,0,0,0,0,0,0,0,0,0
2.2,0,0,0,0,0,0,0,0,0,0,0,0,0,6,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
1.1,0,201,0,0,79,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
1.1,0,33,0,91,0,0,0,0,0,0,0,0,9,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
1.21,8,0,0,0,0,0,0,0,404,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,37,26,18,8,0,0,0,0,0,0,0,0,0,0,0,0,0,0
2.2,0,0,0,0,0,0,0,9,0,0,0,0,0,0,0,0,0,162,79,1,0,0,0,0,0,0,0,0,10,0,27,0,0,0,0,0,0,0,0,0,0,0
2.2,0,0,0,0,0,0,9,0,0,0,0,0,0,0,0,0,0,33,17,0,0,0,0,0,0,0,0,0,4,0,0,0,0,0,0,0,0,0,0,0,0,0
1.21,207,0,0,0,0,1644,0,0,0,0,0,0,0,0,0,0,0,0,0,0,8,0,0,16,17,402,0,0,0,606,0,0,0,0,0,0,0,0,0,0,0,0
1.22,13,0,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,26,0,0,15,0,0,0,0,0
1.22,0,0,0,0,0,0,0,0,0,71,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,3,374,6,121,6,21,0,0,0,0
1.1,0,0,0,44,0,0,0,0,0,0,0,0,103,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,33,0,0,0,0,0,0,0,0,0,0
1.1,0,0,0,24,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,0,0,0,10,0,0,0,0,0,0,0,0,0,0
2.2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,7,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
2.1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,18,0,0,0,0,353,116,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,29,0,5,0
2.2,0,0,0,0,0,0,0,37,0,0,0,0,0,4,0,0,0,36,46,62,0,0,0,0,0,0,0,0,0,0,73,0,0,0,0,0,0,1,0,0,0,0
2.1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,61,0,0,0,0,0,0,0,38,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0
2.2,17,0,23,0,0,0,444,65,0,0,0,0,0,0,0,78,0,0,42,30,15,0,0,0,0,0,0,0,4,0,18,0,0,0,0,0,0,0,0,0,0,0
2.2,0,0,0,0,0,0,75,8,0,0,0,0,0,0,0,87,0,74,0,85,0,0,0,0,0,0,0,0,1,0,19,0,25,0,0,0,0,0,0,0,0,0
2.2,0,0,13,0,0,0,12,20,0,0,0,0,0,0,0,118,0,29,92,0,25,0,0,0,0,0,0,0,0,0,16,0,48,0,0,0,0,0,0,0,0,0
1.21,14,0,1,0,0,0,0,0,17,0,0,0,0,0,0,0,0,0,0,14,0,0,0,0,0,0,0,0,3,0,20,0,0,0,0,0,0,0,0,0,0,0
2.1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,204,0,0,0,0,0,0,0,133,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,44,0,0
2.1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,67,0,0,0,0,0,0,143,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,3,12,15,0
1.21,79,0,0,0,0,0,0,0,34,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,38,26,6,9,0,112,0,0,0,0,0,0,0,0,0,0,0,0
1.21,11,0,0,0,0,17,0,0,49,0,0,0,0,0,0,0,0,0,0,0,0,0,0,28,0,0,0,32,0,0,0,0,0,0,0,0,0,0,0,0,0,0
1.21,40,0,0,0,0,0,0,0,122,0,0,0,0,0,0,0,0,0,0,0,3,0,0,24,11,0,887,20,0,389,0,0,0,0,0,0,0,0,0,0,0,0
1.21,14,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,8,0,50,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0
1.21,34,0,0,0,0,26,0,0,56,0,0,0,0,0,0,0,0,0,0,0,0,0,0,54,9,297,13,0,0,16,0,0,0,0,0,0,0,0,0,0,0,0
2.2,0,0,0,0,0,0,39,0,0,0,0,0,0,0,0,25,0,17,12,20,25,0,0,0,0,0,0,0,0,0,393,0,7,0,0,0,0,0,0,0,0,0
1.21,177,0,0,0,0,8,0,0,775,0,0,0,0,0,0,0,0,0,0,0,0,0,0,113,0,227,0,6,0,0,0,0,0,0,0,0,0,0,0,0,0,0
2.2,0,0,0,0,0,0,21,17,0,0,0,0,0,0,0,0,0,42,30,16,0,0,0,0,0,0,0,0,165,0,0,0,0,0,0,0,0,0,0,0,0,0
1.1,0,6,0,28,0,0,0,0,0,0,0,9,30,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
1.22,0,0,0,0,0,0,0,0,0,0,4,0,0,0,0,0,0,0,4,37,0,0,0,0,0,0,0,0,3,0,0,0,0,14,7,0,0,18,0,0,0,0
1.22,0,0,0,0,0,0,0,0,0,44,785,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,21,0,44,177,13,24,0,0,0,0
1.22,0,0,0,0,0,0,30,0,0,182,9,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,7,12,0,1231,135,17,0,0,0,0
1.22,0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,73,1308,0,669,16,0,0,0,8
1.21,0,0,0,0,0,0,0,0,0,15,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,13,33,197,626,0,44,0,0,0,0
1.22,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,24,37,12,80,0,0,0,0,16
2.1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,24,0,6,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,24,54,0
2.1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,10,0,0,0,0,0,0,27,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,75,0,0,0
2.1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,58,0,1,0,0,0,0,28,24,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,61,2,0,0
1.22,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,31,9,0,0,0,0"))
ids <- read.csv(text=",x
1,1.21
2,1.1
3,2.2
4,1.1
5,1.1
6,1.21
7,2.2
8,2.2
9,1.21
10,1.22
11,1.22
12,1.1
13,1.1
14,2.2
15,2.1
16,2.2
17,2.1
18,2.2
19,2.2
20,2.2
21,1.21
22,2.1
23,2.1
24,1.21
25,1.21
26,1.21
27,1.21
28,1.21
29,2.2
30,1.21
31,2.2
32,1.1
33,1.22
34,1.22
35,1.22
36,1.22
37,1.21
38,1.22
39,2.1
40,2.1
41,2.1
42,1.22")
mat <- mat[,-1]
rownames(mat) <- ids$x
colnames(mat) <- ids$x
ans <- mat / ave(mat, rownames(mat)[row(mat)], colnames(mat)[col(mat)], FUN = max)
非常感谢任何帮助，谢谢。
答案 0 :(得分：1)
使用ave获取最大值：
mat / ave(mat, rownames(mat)[row(mat)], colnames(mat)[col(mat)], FUN = max)
例如，正如预期的那样，有9个，并且每个块中也有1个如预期的那样。 （如果矩阵在一个或多个块中碰巧有多个最大值，那么可能会超过9个，但不得少于9个。）
set.seed(123)
mat <- matrix(round(runif(90, 0, 50),),9,9)
rownames(mat) <- rep(LETTERS[1:3],3)
colnames(mat) <- rep(LETTERS[1:3],3)
ans <- mat / ave(mat, rownames(mat)[row(mat)], colnames(mat)[col(mat)], FUN = max)
sum(ans == 1)
## [1] 9
# there are no duplicates (i.e. a block showing up more than once) hence
# there is exactly one 1 in each block
w <- which(ans == 1, arr = TRUE)
anyDuplicated(cbind(rownames(mat)[w[, 1]], colnames(mat)[w[, 2]]))
## [1] 0
ADDED
如果某些块完全为零（在UPDATE 2中就是这种情况），那么您将获得这些块的NaN。如果你想要0而不是全零块，试试这个：
xmax <- function(x) if (all(x == 0)) 0 else x/max(x)
ave(mat, rownames(mat)[row(mat)], colnames(mat)[col(mat)], FUN = xmax)