拥有这样的数据框:
df <- data.frame(a=c(31, 18, 0, 1, 20, 2),
b=c(1, 0, 0, 3, 1, 1),
c=c(12, 0, 9, 8, 10, 3))
> df
a b c
1 31 1 12
2 18 0 0
3 0 0 9
4 1 3 8
5 20 1 10
6 2 1 3
如何进行随机子集,使行和列的总和等于一个值,即100?
答案 0 :(得分:2)
据我了解您的问题,您正在尝试对矩阵的行和列的子集进行采样,以便它们总和为目标值。
您可以使用整数优化来完成此任务。您将为每个行,列和单元格提供二元决策变量,并使用约束来强制单元格值等于行值和列值的乘积。我将使用lpSolve
包来执行此操作,因为它有一个方便的机制来获得多个最佳解决方案。然后我们可以使用sample
函数在它们之间进行选择:
library(lpSolve)
get.subset <- function(dat, target) {
nr <- nrow(dat)
nc <- ncol(dat)
nvar <- nr + nc + nr*nc
# Cells upper bounded by row and column variable values (r and c) and lower bounded by r+c-1
mat <- as.matrix(do.call(rbind, apply(expand.grid(seq(nr), seq(nc)), 1, function(x) {
r <- x[1]
c <- x[2]
pos <- nr + nc + (r-1)*nc + c
ltc <- rep(0, nvar)
ltc[nr + c] <- 1
ltc[pos] <- -1
ltr <- rep(0, nvar)
ltr[r] <- 1
ltr[pos] <- -1
gtrc <- rep(0, nvar)
gtrc[nr + c] <- 1
gtrc[r] <- 1
gtrc[pos] <- -1
return(as.data.frame(rbind(ltc, ltr, gtrc)))
})))
dir <- rep(c(">=", ">=", "<="), nr*nc)
rhs <- rep(c(0, 0, 1), nr*nc)
# Sum of selected cells equals target
mat <- rbind(mat, c(rep(0, nr+nc), as.vector(t(dat))))
dir <- c(dir, "=")
rhs <- c(rhs, target)
res <- lp(objective.in=rep(0, nvar), # Feasibility problem
const.mat=mat,
const.dir=dir,
const.rhs=rhs,
all.bin=TRUE,
num.bin.solns=100 # Number of feasible solutions to get
)
if (res$status != 0) {
return(list(rows=NA, cols=NA, subset=NA, num.sol=0))
}
sol.num <- sample(res$num.bin.solns, 1)
vals <- res$solution[seq((sol.num-1)*nvar+1, sol.num*nvar)]
rows <- which(vals[seq(nr)] >= 0.999)
cols <- which(vals[seq(nr+1, nr+nc)] >= 0.999)
return(list(rows=rows, cols=cols, subset=dat[rows,cols], num.sol=res$num.bin.solns))
}
该函数返回具有该总和的子集数,并返回随机选择的子集:
set.seed(144)
get.subset(df, 1)
# $rows
# [1] 1
# $cols
# [1] 2
# $subset
# [1] 1
# $num.sol
# [1] 14
get.subset(df, 100)
# $rows
# [1] 1 2 4 5
# $cols
# [1] 1 3
# $subset
# a c
# 1 31 12
# 2 18 0
# 4 1 8
# 5 20 10
# $num.sol
# [1] 2
get.subset(df, 10000)
# $rows
# [1] NA
# $cols
# [1] NA
# $subset
# [1] NA
# $num.sol
# [1] 0