Question

我有以下输入数据：

#     [,1] [,2]
#[1,] "A"  "B" 
#[2,] "A"  "C" 
#[3,] "A"  "D" 
#[4,] "B"  "C" 
#[5,] "B"  "D" 
#[6,] "C"  "D"

接下来我想要排除第一个或第二个元素先前已经N次的行。例如，如果N = 2，则需要排除以下行：

#[3,] "A"  "D" - element "A" has been 2 times
#[5,] "B"  "D" - element "B" has been 2 times
#[6,] "C"  "D" - element "C" has been 2 times

注意：需要考虑立即排除结果。例如，如果元素已满足5次，并且在删除它之后只遇到1次，则需要使用此元素留下下一行。因为它现在遇到了2次。

示例（N = 2）：

输入数据：

      [,1] [,2]
 [1,] "A"  "B" 
 [2,] "A"  "C" 
 [3,] "A"  "D" 
 [4,] "A"  "E" 
 [5,] "B"  "C" 
 [6,] "B"  "D" 
 [7,] "B"  "E" 
 [8,] "C"  "D" 
 [9,] "C"  "E" 
[10,] "D"  "E"

输出数据：

     [,1] [,2]
 [1,] "A"  "B" 
 [2,] "A"  "C" 
 [5,] "B"  "C" 
[10,] "D"  "E"

Answer 1

可能有更优雅的解决方案......但这似乎有效：

v <- c("A", "B", "C", "D", "E")
cmb <- t(combn(v, 2))

n <- 2

# Go through each letter
for (l in v)
  {
  # Find the combinations using that letter 
  rows <- apply(cmb, 1, function(x){l %in% x})

  rows.2 <- which(rows==T)
  if (length(rows.2)>n)
    rows.2 <- rows.2[1:n]

  # Take the first n rows containing the letter,
  # then append all the ones not containing it
  cmb <- rbind(cmb[rows.2,], cmb[rows==F,])
  }

cmb

输出：

    [,1] [,2]
[1,] "D"  "E" 
[2,] "B"  "C" 
[3,] "A"  "C" 
[4,] "A"  "B"

排除先前已满足元素N次的行

1 个答案: