R 子集非NA
我有一个矩阵,其中每行至少有一个NA单元格,每列也至少有一个NA单元格。我需要的是找到此矩阵中不包含NAs的最大子集 例如,对于该矩阵R 子集非NA,r,matrix,subset,na,R,Matrix,Subset,Na,我有一个矩阵,其中每行至少有一个NA单元格,每列也至少有一个NA单元格。我需要的是找到此矩阵中不包含NAs的最大子集 例如,对于该矩阵A A <- structure(c(NA, NA, NA, NA, 2L, NA, 1L, 1L, 1L, 0L, NA, NA, 1L, 8L, NA, 1L, 1L, NA, NA, 1L, 1L, 6L, 1L, 3L, NA,
A
A <-
structure(c(NA, NA, NA, NA, 2L, NA,
1L, 1L, 1L, 0L, NA, NA,
1L, 8L, NA, 1L, 1L, NA,
NA, 1L, 1L, 6L, 1L, 3L,
NA, 1L, 5L, 1L, 1L, NA),
.Dim = c(6L, 5L),
.Dimnames =
list(paste0("R", 1:6),
paste0("C", 1:5)))
A
C1 C2 C3 C4 C5
R1 NA 1 1 NA NA
R2 NA 1 8 1 1
R3 NA 1 NA 1 5
R4 NA 0 1 6 1
R5 2 NA 1 1 1
R6 NA NA NA 3 NA
A我有一个解决方案,但它不能很好地扩展:
findBiggestSubmatrixNonContiguous <- function(A) {
A <- !is.na(A); ## don't care about non-NAs
howmany <- expand.grid(nr=seq_len(nrow(A)),nc=seq_len(ncol(A)));
howmany <- howmany[order(apply(howmany,1L,prod),decreasing=T),];
for (ri in seq_len(nrow(howmany))) {
nr <- howmany$nr[ri];
nc <- howmany$nc[ri];
rcom <- combn(nrow(A),nr);
ccom <- combn(ncol(A),nc);
comcom <- expand.grid(ri=seq_len(ncol(rcom)),ci=seq_len(ncol(ccom)));
for (comi in seq_len(nrow(comcom)))
if (all(A[rcom[,comcom$ri[comi]],ccom[,comcom$ci[comi]]]))
return(list(ri=rcom[,comcom$ri[comi]],ci=ccom[,comcom$ci[comi]]));
}; ## end for
NULL;
}; ## end findBiggestSubmatrixNonContiguous()
我不知道一个简单的方法来验证上面的结果是否正确,但我觉得这很好。但产生这个结果几乎花了9秒。在中等规模的矩阵上运行该函数,特别是在77x132矩阵上,可能是一个失败的原因
等待着看是否有人能想出一个出色有效的解决方案……1)optim在这种方法中,我们将问题放松为一个连续优化问题,我们使用optim
解决这个问题
目标函数是f
,其输入是一个0-1向量,其第一个nrow(a)
条目对应于行,其余条目对应于列f
使用矩阵Ainf
,该矩阵由a
派生而来,方法是将NAs替换为大负数,将非NAs替换为1。就Ainf
而言,对应于x
的行和列矩形中元素数量的负数为-x[seq(6)]%*%Ainf%*$x[-seq(6)]
,我们将其最小化为x
的函数,但x
的每个分量都在0和1之间
尽管这是将原始问题松弛为连续优化,但不管怎样,我们似乎得到了所需的整数解
实际上,下面的大部分代码只是为了获取起始值。为此,我们首先应用系列化。这会排列行和列,给出更块状的结构,然后在排列矩阵中我们找到最大的平方子矩阵
对于问题中的特定A
,最大的矩形子矩阵恰好是正方形,并且起始值已经足够好,可以产生最佳值,但我们将以任何方式执行优化,因此它通常有效。如果您愿意,您可以使用不同的起始值进行游戏。例如,在largestSquare
中,将k
从1更改为更高的数字,在这种情况下largestSquare
将返回k
列,给出k
起始值,可用于k
运行optim
以获得最佳效果
如果起始值足够好,则应产生最佳值
library(seriation) # only used for starting values
A.na <- is.na(A) + 0
Ainf <- ifelse(A.na, -prod(dim(A)), 1) # used by f
nr <- nrow(A) # used by f
f <- function(x) - c(x[seq(nr)] %*% Ainf %*% x[-seq(nr)])
# starting values
# Input is a square matrix of zeros and ones.
# Output is a matrix with k columns such that first column defines the
# largest square submatrix of ones, second defines next largest and so on.
# Based on algorithm given here:
# http://www.geeksforgeeks.org/maximum-size-sub-matrix-with-all-1s-in-a-binary-matrix/
largestSquare <- function(M, k = 1) {
nr <- nrow(M); nc <- ncol(M)
S <- 0*M; S[1, ] <- M[1, ]; S[, 1] <- M[, 1]
for(i in 2:nr)
for(j in 2:nc)
if (M[i, j] == 1) S[i, j] = min(S[i, j-1], S[i-1, j], S[i-1, j-1]) + 1
o <- head(order(-S), k)
d <- data.frame(row = row(M)[o], col = col(M)[o], mx = S[o])
apply(d, 1, function(x) {
dn <- dimnames(M[x[1] - 1:x[3] + 1, x[2] - 1:x[3] + 1])
out <- c(rownames(M) %in% dn[[1]], colnames(M) %in% dn[[2]]) + 0
setNames(out, unlist(dimnames(M)))
})
}
s <- seriate(A.na)
p <- permute(A.na, s)
# calcualte largest square submatrix in p of zeros rearranging to be in A's order
st <- largestSquare(1-p)[unlist(dimnames(A)), 1]
res <- optim(st, f, lower = 0*st, upper = st^0, method = "L-BFGS-B")
给予:
> res
$par
R1 R2 R3 R4 R5 R6 C1 C2 C3 C4 C5
0 1 1 1 0 0 0 1 0 1 1
$value
[1] -9
$counts
function gradient
1 1
$convergence
[1] 0
$message
[1] "CONVERGENCE: NORM OF PROJECTED GRADIENT <= PGTOL"
> setNames(resSA$par, unlist(dimnames(A)))
R1 R2 R3 R4 R5 R6 C1 C2 C3 C4 C5
0 1 1 1 0 0 0 1 0 1 1
> resSA$value
[1] -9
A[c(2,4,5),3:5]
不是最好的解决方案吗?对于矩阵77x132,您正在考虑大约2^(77+132)~8.2E62可能的子矩阵。我很想知道如何解决这个问题…@bgoldst或就这件事而言A[2:4,c(2,4,5)]
@Frank我怀疑我们可以通过首先识别所有NA
s来显著降低维度。。。但除此之外,如果只允许连续矩阵,则更易于管理。允许行或列跳过的问题要困难得多,您可以扩展seriate
的功能吗?帮助文件过于行话化,无法根据方法
参数对输出排列的行和列进行排列。。我们使用默认值。您可以使用该参数尝试不同的方法。它跑得很快,但不一定能给你想要的东西,所以你必须玩一玩。这与其说是一个完成的解决方案,不如说是一个起点,尽管它似乎确实能解决问题中的小问题。谢谢!对于示例矩阵和一些测试矩阵,这种方法运行良好且速度非常快,但并非对所有矩阵都有效。对于我的实际矩阵和许多随机测试矩阵,它返回一个单元格。我将在下一篇评论中给出一个例子。A我已经完全修改了答案。
randTest(11L,3L,4/5);
## [,1] [,2] [,3]
## [1,] NA NA NA
## [2,] NA NA NA
## [3,] NA NA NA
## [4,] 2 NA NA
## [5,] NA NA NA
## [6,] 5 NA NA
## [7,] 8 0 4
## [8,] NA NA NA
## [9,] NA NA NA
## [10,] NA 7 NA
## [11,] NA NA NA
## user system elapsed
## 0.297 0.000 0.300
## $ri
## [1] 4 6 7
##
## $ci
## [1] 1
##
## [,1]
## [1,] 2
## [2,] 5
## [3,] 8
randTest(10L,10L,1/3);
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
## [1,] NA NA 0 3 8 3 9 1 6 NA
## [2,] 1 NA NA 4 5 8 NA 8 2 NA
## [3,] 4 2 5 3 7 6 6 1 1 5
## [4,] 9 1 NA NA 4 NA NA 1 NA 9
## [5,] NA 7 NA 8 3 NA 5 3 7 7
## [6,] 9 3 1 2 7 NA NA 9 NA 7
## [7,] 0 2 NA 7 NA NA 3 8 2 6
## [8,] 5 0 1 NA 3 3 7 1 NA 6
## [9,] 5 1 9 2 2 5 NA 7 NA 8
## [10,] NA 7 1 6 2 6 9 0 NA 5
## user system elapsed
## 8.985 0.000 8.979
## $ri
## [1] 3 4 5 6 8 9 10
##
## $ci
## [1] 2 5 8 10
##
## [,1] [,2] [,3] [,4]
## [1,] 2 7 1 5
## [2,] 1 4 1 9
## [3,] 7 3 3 7
## [4,] 3 7 9 7
## [5,] 0 3 1 6
## [6,] 1 2 7 8
## [7,] 7 2 0 5
library(seriation) # only used for starting values
A.na <- is.na(A) + 0
Ainf <- ifelse(A.na, -prod(dim(A)), 1) # used by f
nr <- nrow(A) # used by f
f <- function(x) - c(x[seq(nr)] %*% Ainf %*% x[-seq(nr)])
# starting values
# Input is a square matrix of zeros and ones.
# Output is a matrix with k columns such that first column defines the
# largest square submatrix of ones, second defines next largest and so on.
# Based on algorithm given here:
# http://www.geeksforgeeks.org/maximum-size-sub-matrix-with-all-1s-in-a-binary-matrix/
largestSquare <- function(M, k = 1) {
nr <- nrow(M); nc <- ncol(M)
S <- 0*M; S[1, ] <- M[1, ]; S[, 1] <- M[, 1]
for(i in 2:nr)
for(j in 2:nc)
if (M[i, j] == 1) S[i, j] = min(S[i, j-1], S[i-1, j], S[i-1, j-1]) + 1
o <- head(order(-S), k)
d <- data.frame(row = row(M)[o], col = col(M)[o], mx = S[o])
apply(d, 1, function(x) {
dn <- dimnames(M[x[1] - 1:x[3] + 1, x[2] - 1:x[3] + 1])
out <- c(rownames(M) %in% dn[[1]], colnames(M) %in% dn[[2]]) + 0
setNames(out, unlist(dimnames(M)))
})
}
s <- seriate(A.na)
p <- permute(A.na, s)
# calcualte largest square submatrix in p of zeros rearranging to be in A's order
st <- largestSquare(1-p)[unlist(dimnames(A)), 1]
res <- optim(st, f, lower = 0*st, upper = st^0, method = "L-BFGS-B")
> res
$par
R1 R2 R3 R4 R5 R6 C1 C2 C3 C4 C5
0 1 1 1 0 0 0 1 0 1 1
$value
[1] -9
$counts
function gradient
1 1
$convergence
[1] 0
$message
[1] "CONVERGENCE: NORM OF PROJECTED GRADIENT <= PGTOL"
library(GenSA)
resSA <- GenSA(lower = rep(0, sum(dim(A))), upper = rep(1, sum(dim(A))), fn = f)
> setNames(resSA$par, unlist(dimnames(A)))
R1 R2 R3 R4 R5 R6 C1 C2 C3 C4 C5
0 1 1 1 0 0 0 1 0 1 1
> resSA$value
[1] -9