您可以：

val groupRDD = rdd
  .groupByKey()
  .filter(value => value._2.map(tuple => tuple._1).sum != value._2.size)
  .flatMapValues(list => list) // to get the result as you like, because right now, they are, e.g. (b, Seq((1, m), (1, n)))

我们首先通过groupByKey对键进行分组，然后通过对分组条目中的键求和来过滤filter，并检查sum是否与分组条目的大小相同。例如：

(a, Seq((1, m), (1, n))   -> grouped by key
(a, Seq((1, m), (1, n), 2 (the sum of 1 + 1), 2 (size of sequence))
2 = 2, filter this row out

最终结果：

(c,(1,m))
(b,(1,m))
(c,(5,m))
(b,(2,n))

祝你好运!

• • 编辑**

假设来自元组的key可以是任何字符串;假设rdd是包含以下内容的数据：

(a,(x,m))
(c,(x,m))
(c,(z,m))
(d,(x,m))
(b,(x,m))
(a,(x,n))
(d,(x,n))
(b,(y,n))

然后我们可以构造uniqueCount为：

val uniqueCount = rdd
  // we swap places, we want to check for combination of (a, 1), (b, a), (b, b), (c, a), etc.
  .map(entry => ((entry._1, entry._2._1), entry._2._2))
  // we count keys, meaning that (a, 1) gives us 2, (b, a) gives us 1, (b, b) gives us 1, etc.
  .countByKey()
  // we filter out > 2, because they are duplicates
  .filter(a => a._2 == 1)
  // we get the very keys, so we can filter below
  .map(a => a._1._1)
  .toList

然后这个：

val filteredRDD = rdd.filter(a => uniqueCount.contains(a._1))

给出以下输出：

(b,(y,n))
(c,(x,m))
(c,(z,m))
(b,(x,m))