对于列表的n个元素中的匹配元素的数量的更一般的函数

Question

我试图计算列表元素中的公共条目数:

temp<-list(element1=c("a","b","c"), element2=c("b","c","d"), 
           element3=c("d","e","f"), element4=c("a","c","z"))

为了获得元素的所有成对组合之间的重叠,此函数有效:

calculate.overlap.2<-function(y){
  pw<-combn(y,2,FUN=function(x)sum(x[[1]]%in%x[[2]]))
  names(pw)<-combn(y,2,FUN=function(x)paste(names(x)[[1]],names(x)[[2]],sep="-"))
  return(pw)
}

为了获得元素的所有三向组合之间的重叠,此函数有效:

calculate.overlap.3<-function(y){
   pw<-combn(y,3,FUN=function(x)sum(x[[1]]%in%x[[2]]&x[[1]]%in%x[[3]]))
   names(pw)<-combn(y,3,FUN=function(x) paste(names(x)[[1]],names(x)[[2]],names(x)[[3]],sep="-"))
   return(pw)
}

但是从功能内部的数字可以看出,这不是一个优雅的解决方案.

将这两个函数概括为一个非常好,并且函数将每个重叠检查中的元素作为输入.也就是说,输入 number.of.elements.per.comparison=2将等于calculate.overlap.2上面,并且函数的输入number.of.elements.per.comparison=3将与之相同calculate.overlap.3.

我觉得有一个非常优雅的解决方案,但我只是看不到它.

Answer 1

这是一种方法:

ix = lapply(seq_along(temp), function(u) combn(seq_along(temp), u))

lapply(ix, function(m){
    res = apply(m,2, function(v) length(Reduce(intersect, temp[v])))
    setNames(res, apply(m, 2, paste, collapse='-'))
})

#[[1]]
#1 2 3 4 
#3 3 3 3 

#[[2]]
#1-2 1-3 1-4 2-3 2-4 3-4 
#  2   0   2   1   1   0 

#[[3]]
#1-2-3 1-2-4 1-3-4 2-3-4 
#    0     1     0     0 

#[[4]]
#1-2-3-4 
#      0 

Answer 2

calculate.overlap <- function(y, i){
  pw <- combn(seq_along(y), i, FUN= function(x) {
    res <- length(Reduce(intersect, y[x]))
    names(res) <- paste(names(y[x]), collapse = "-")
    res
  }, simplify = FALSE)
  do.call(c, pw)
}

calculate.overlap(temp, 3)
#element1-element2-element3 element1-element2-element4 element1-element3-element4 element2-element3-element4 
#                         0                          1                          0                          0