Question

我有一个数据框，其中包含在许多数据集中找到的元素对。对的顺序无关紧要，它们按字母顺序给出一次，但第一个实例可能在数据库之间有所不同，如示例所示。

data <- data.frame(i = c("b","b","b","c"), j = c("c","d","d","a"), +
        database = c(1,1,2,3))

我想为它们生成一个分数，以显示每个数据库中包含相同对的实例的比率。

我可以想象一个像这样粗略的功能：

# For each database that includes particular i or j,  test whether
# they have a connection to another particular element at j or i, 
# respectively. Count the number of successes.

# Divide it by:
# Count(number of databases that contain either of the members of the pair in i or j)

我期望从示例数据集（顺序不重要）得到的结果是：

a c 0.5
b c 0.33
b d 1

我可以看到这个原始循环系统如何工作，但我确信有一个更优雅的解决方案，任何人都可以提供帮助？也许在图库中有一个特定的功能。谢谢！

Answer 1

只是有点玩连接（即合并）

library(dplyr)

data <- data.frame(i = c("b","b","b","c"), j = c("c","d","d","a"),
                   database = c(1,1,2,3), stringsAsFactors = FALSE)

# Sort pairs lexicographic and count occurences of pairs
data2 <- mutate(data, x=pmin(i,j), y=pmax(i,j))
pairs_all <- summarize(group_by(data2, x, y), n_all = length(unique(database)))

# Introduce helper index to identify the pairs (for following joins)
pairs_all$pair_id <- 1:nrow(pairs_all)

# Count occurences of elements of pairs
r <- 
 merge(pairs_all, 
         summarize(group_by(merge(merge(pairs_all,
                                        transmute(data2, x, db1 = database)), 
                                  transmute(data2, y, db2 = database)), pair_id),
                   n_any = length(unique(union(db1,db2)))))

# Finally calculate the result
transmute(r, x, y, n_all/n_any)

Answer 2

哇，这真糟糕！但是我已经编写了我上面提到的hack。对于任何在未来同样模糊的即兴网络比较中绊脚石的人。如果有人仍然知道可以简化的引用，那么在网络节点对中找到这种类型的自然组会更加可靠，请告诉我。：）

#Calculate the score one row at a time
for (linenr in 1:length(data$i)){ 
  count_pair = 0
  count_one = 0
  # Loop through datasets
  for(setname in levels(data$database)){
    subset <- subset(data, database == setname)
    #Test whether either variable appears in dataset
    if(sum(c(as.character(data$i[linenr]),as.character(data$j[linenr])) %in%
             c(as.character(subset$i),as.character(subset$j))) > 0) 
      {count_one = count_one + 1}
    for (line2nr in 1:length(subset$i)){ 
    #Test whether dataset contains lines which have both elements of the original pair
      if(sum(c(as.character(data$i[linenr]),as.character(data$j[linenr])) %in%
               c(as.character(subset$i[line2nr]),as.character(subset$j[line2nr])))
         == 2) 
          {count_pair = count_pair + 1}
    }
  }
  #Simple ratio calculation
  data$score[linenr] <- count_pair/count_one
}

frame <- data.frame(data$i,data$j,data$score)
#Remove database duplicates
result <- frame[!duplicated(frame),]
#This still doesn't deal with changed order duplicates, but does the job now.

数据集之间元素对的R比率

2 个答案: