Question

我需要从共同的30个项目中选择10个不同的15个项目组合，并有几个约束条件。

第一个约束：对于10个组合中的每一个，我需要删除在该特定组合中不应存在的三个项目。让我们说第一个组合的wf01，wf02和wf03;第二个组合的wf04，wf05和wf06等等。

第二个约束：在十个组合中，每个项目应该存在相同的次数（5）。

第三个限制：十个组合应尽可能彼此不同（即他们应尽量避免共享共同项目）。

我目前正在Excel中这样做，但我想在R中应该有一种更有效的方法！

到目前为止，这是我的编码：

# Data: 30 items (pictures to choose from)

PictureNames <- c("wf01", "wf02", "wf03", "wf04", "wf05", "wf06", "wf07", "wf08", "wf09", "wf10", "wf11", "wf12", "wf13", "wf14", "wf15", "wf16", "wf17", "wf18", "wf19", "wf20", "wf21", "wf22", "wf23", "wf24", "wf25", "wf26", "wf27", "wf28", "wf29", "wf30")

set.seed(10)

sample01 <- sample(PictureNames[4:30], 15)
sample02 <- sample(PictureNames[c(1:3, 7:30)], 15)
sample03 <- sample(PictureNames[c(1:6, 10:30)], 15)
sample04 <- sample(PictureNames[c(1:9, 13:30)], 15)
sample05 <- sample(PictureNames[c(1:12, 16:30)], 15)
sample06 <- sample(PictureNames[c(1:15, 19:30)], 15)
sample07 <- sample(PictureNames[c(1:18, 22:30)], 15)
sample08 <- sample(PictureNames[c(1:21, 25:30)], 15)
sample09 <- sample(PictureNames[c(1:24, 28:30)], 15)
sample10 <- sample(PictureNames[c(1:27)], 15)

到目前为止，我在Excel中手动实现了约束（我知道......）。首先，我确保每个项目在10种不同组合中仅出现5次。然后，我创建了一个矩阵，其中图片的名称为行和列。每个细胞计算在10种组合中相同组合中存在一对图片的次数。值范围从0（图片从不在一起）到5（图片总是在一起）。然后，通过在组合之间手动移动图片，我设法避免在此矩阵中存在0和5。之后，我实现了一个优化任务，其中每个图片应该与尽可能多的不同图片相关联，同时尊重先前的约束。为此，我为每张图片计算其列（或行）值的标准偏差。然后，我计算了这些标准偏差的标准偏差。在保持尊重其他约束的同时减少该值的任何移位。我设法达到0.036的标准偏差，但我不知道这是否是我能得到的最好结果。任何帮助将不胜感激!!!

提前致谢！

修改

在@ roman-luštrik的评论之后，这是我的新版本的代码。仍然可怕，并没有从优化矩阵那么近，但慢慢到达那里！

# Data: 30 items (pictures to choose from)

PictureNames <- data.frame(Names = c("wf01", "wf02", "wf03", "wf04", "wf05",
              "wf06", "wf07", "wf08", "wf09", "wf10", "wf11", "wf12", "wf13",
              "wf14", "wf15", "wf16", "wf17", "wf18", "wf19", "wf20", "wf21",
              "wf22", "wf23", "wf24", "wf25", "wf26", "wf27", "wf28", "wf29",
              "wf30"), stringsAsFactors = FALSE)

library(dplyr)

set.seed(10)

sample01 <- data.frame(Names = sort(sample(PictureNames$Names[c(4:30)], 15)))
sample02 <- anti_join(PictureNames, sample01)
sample03 <- data.frame(Names = sort(sample(PictureNames$Names[c(1:6, 10:30)], 15)))
sample04 <- anti_join(PictureNames, sample03)
sample05 <- data.frame(Names = sort(sample(PictureNames$Names[c(1:12, 16:30)], 15)))
sample06 <- anti_join(PictureNames, sample05)
sample07 <- data.frame(Names = sort(sample(PictureNames$Names[c(1:18, 22:30)], 15)))
sample08 <- anti_join(PictureNames, sample07)
sample09 <- data.frame(Names = sort(sample(PictureNames$Names[c(1:24, 28:30)], 15)))
sample10 <- anti_join(PictureNames, sample09)

AllSamples <- data.frame(sample01, sample02, sample03, sample04, sample05,
                     sample06, sample07, sample08, sample09, sample10)

colnames(AllSamples) <- c("sA", "sB", "sC", "sD", "sE", "sF", "sG", "sH", "sI", "sJ")

CoOccurMat <- matrix(nrow = length(PictureNames$Names), ncol = length(PictureNames$Names),
                 dimnames = c(PictureNames, PictureNames))

for(j in 1:length(PictureNames$Names)){
  for(i in 1:length(PictureNames$Names)){
CoOccurMat[i,j] <- sum(ifelse(sapply(PictureNames,
                       function(x)colSums(xor(AllSamples == PictureNames$Names[j],
                       AllSamples == PictureNames$Names[i]))) == 2, 1, 0))
ColSd <- apply(CoOccurMat, 2, sd)
OverallSd <- sd(ColSd)
  }
}

CoOccurMat
ColSd
OverallSd

如何选择公共向量的最不相似组合的子集？

修改

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