Question

我有以下数据集，我希望使用允许所有可能组合的循环策略来比较它们包含的元素中的相似性（即＆＃34; setA，setB，setC，setD＆＃34 ;;＆＃ 34; setA，setB，setC＆＃34 ;;＆＃34; setA，setB＆＃34;，＆＃34; setB，setC，setD＆＃34 ;;＆＃34; setC，setD＆＃34;＆＃34; setB，setD＆＃34; etc

数据集：

setA <- c("dog", "cat", "cow", "sheep", "dunkey")

setB <- c("fox", "cat", "cow", "snake")

setC <- c("dog", "cat", "cow", "sheep", "dunkey", "fox", "python")

setD <- c("dog", "cat", "lion", "sheep", "elephant", "fox")

不确定如何在R中对此进行编码，但这是我未能产生预期结果的尝试：

similar <- function(...){
     Reduce(intersect, list(...))

}


allSets <- list(setA, setB, setC, setD)

for(i in 1:length(allSets)){
   similar(allSets[[i]])
   similar(allSets[i-1])
   similar(allSets[i-2])
   similar(allSets[i-3])
}

任何人都可以帮忙吗？

Answer 1

以下是previous post获取所有交叉点的函数

## Build intersections, 'out' accumulates the result
intersects <- function(sets, out=NULL) {
    if (length(sets) < 2) return ( out )                               # return result
    len <- seq(length(sets))
    if (missing(out)) out <- list()                                    # initialize accumulator
    for (idx in split((inds <- combn(length(sets), 2)), col(inds))) {  # 2-way combinations
        ii <- len > idx[2] & !(len %in% idx)                           # indices to keep for next intersect
        out[[(n <- paste(names(sets[idx]), collapse="."))]] <- intersect(sets[[idx[1]]], sets[[idx[2]]])
        out <- intersects(append(out[n], sets[ii]), out=out)
    }
    out
}

## Put the sets in a list
sets <- mget(paste0("set", toupper(letters[1:4])))

intersects(sets)
# $setA.setB
# [1] "cat" "cow"
# 
# $setA.setB.setC
# [1] "cat" "cow"
# 
# $setA.setB.setC.setD
# [1] "cat"
# 
# $setA.setB.setD
# [1] "cat"
# 
# $setC.setD
# [1] "dog"   "cat"   "sheep" "fox"  
# 
# $setA.setC
# [1] "dog"    "cat"    "cow"    "sheep"  "dunkey"
# 
# $setA.setC.setD
# [1] "dog"   "cat"   "sheep"
# 
# $setA.setD
# [1] "dog"   "cat"   "sheep"
# 
# $setB.setC
# [1] "fox" "cat" "cow"
# 
# $setB.setC.setD
# [1] "fox" "cat"
# 
# $setB.setD
# [1] "fox" "cat"

循环遍历类似元素的数据集

1 个答案: