我的数据：

data=cbind(c(1,1,2,1,1,3),c(1,1,2,1,1,1),c(2,2,1,2,1,2))
colnames(data)=paste("item",1:3)
rownames(data)=paste("method",1:6)

我想这一点，与多数表决，有两个社区（他们的元素）的输出。是这样的：组别1 = {ITEM1，ITEM2} ，组2 = {项目3} 。

推荐答案

此函数传递一个矩阵，其中各列是一个项目，每行是对应于根据一个聚类方法的项目的一个分区成员资格向量。元素（数字）组成的每一行没有任何意义不是说明其他成员，并从排被回收到排。该函数返回多数票分区。当没有共识存在一个项目，由第一行中给出的分区获胜。这允许分区的顺序通过降低模块化的值，例如

This function is passed a matrix where each column is an item and each row is a membership vector corresponding to a partition of the items according to a clustering method. The elements (numbers) composing each row have no meaning other than indicating membership and are recycled from row to row. The function returns the majority vote partition. When no consensus exists for an item, the partition given by the first row wins. This allows ordering of the partitions by decreasing values of modularity, for instance.

    consensus.final <-
  function(data){
    output=list()
    for (i in 1:nrow(data)){
      row=as.numeric(data[i,])
      output.inner=list()
      for (j in 1:length(row)){
        group=character()
        group=c(group,colnames(data)[which(row==row[j])])
        output.inner[[j]]=group
      }
      output.inner=unique(output.inner)
      output[[i]]=output.inner
    }

    # gives the mode of the vector representing the number of groups found by each method
    consensus.n.comm=as.numeric(names(sort(table(unlist(lapply(output,length))),decreasing=TRUE))[1])

    # removes the elements of the list that do not correspond to this consensus solution
    output=output[lapply(output,length)==consensus.n.comm]

    # 1) find intersection 
    # 2) use majority vote for elements of each vector that are not part of the intersection

    group=list()

    for (i in 1:consensus.n.comm){ 
      list.intersection=list()
      for (p in 1:length(output)){
        list.intersection[[p]]=unlist(output[[p]][i])
      }

      # candidate group i
      intersection=Reduce(intersect,list.intersection)
      group[[i]]=intersection

      # we need to reinforce that group
      for (p in 1:length(list.intersection)){
        vector=setdiff(list.intersection[[p]],intersection)
        if (length(vector)>0){
          for (j in 1:length(vector)){
            counter=vector(length=length(list.intersection))
            for (k in 1:length(list.intersection)){
              counter[k]=vector[j]%in%list.intersection[[k]]
            }
            if(length(which(counter==TRUE))>=ceiling((length(counter)/2)+0.001)){
              group[[i]]=c(group[[i]],vector[j])
            }
          }
        }
      }
    }

    group=lapply(group,unique)

    # variables for which consensus has not been reached
    unclassified=setdiff(colnames(data),unlist(group))

    if (length(unclassified)>0){
      for (pp  in 1:length(unclassified)){
        temp=matrix(nrow=length(output),ncol=consensus.n.comm)
        for (i in 1:nrow(temp)){
          for (j in 1:ncol(temp)){
            temp[i,j]=unclassified[pp]%in%unlist(output[[i]][j])
          }
        }
        # use the partition of the first method when no majority exists (this allows ordering of partitions by decreasing modularity values for instance)
        index.best=which(temp[1,]==TRUE)
        group[[index.best]]=c(group[[index.best]],unclassified[pp])
      }
    }
    output=list(group=group,unclassified=unclassified)
  }

一些例子：

data=cbind(c(1,1,2,1,1,3),c(1,1,2,1,1,1),c(2,2,1,2,1,2))
colnames(data)=paste("item",1:3)
rownames(data)=paste("method",1:6)
data
consensus.final(data)$group

[[1]]
[1] "item 1" "item 2"

[[2]]
[1] "item 3"

data=cbind(c(1,1,1,1,1,3),c(1,1,1,1,1,1),c(1,1,1,2,1,2)) 
colnames(data)=paste("item",1:3) 
rownames(data)=paste("method",1:6)
data
consensus.final(data)$group

[[1]]
[1] "item 1" "item 2" "item 3"

data=cbind(c(1,3,2,1),c(2,2,3,3),c(3,1,1,2))
colnames(data)=paste("item",1:3)
rownames(data)=paste("method",1:4)
data
consensus.final(data)$group

[[1]]
[1] "item 1"

[[2]]
[1] "item 2"

[[3]]
[1] "item 3"