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copied!<p><strong>Preamble</strong></p> <p>It occurred to me that the status could be seen as having a priority, and if given a sequence of <code>(agent,status)</code> pairs then the task is to select only the highest priority status for each agent. Unfortunately, status isn't strongly typed with an explicit ordering so defined, but... as it's a string with only two values we can safely use string ordering as having a 1:1 correspondence to the priority.</p> <p>Both my answers take advantage of two useful facts:</p> <ol> <li><p>In natural string ordering, <code>"FAIL" < "PASS"</code>, so:</p> <pre><code>List("PASS", "FAIL", "PASS").sorted.head = "FAIL" </code></pre></li> <li><p>For two tuples <code>(x,a)</code> and <code>(x,b)</code>, <code>(x,a) > (x, b)</code> if <code>(a > b)</code></p></li> </ol> <p><strong>UPDATED REPLY</strong></p> <pre><code>val solution = l.sorted.reverse.toMap </code></pre> <p>When converting a <code>Seq[(A,B)]</code> to a <code>Map[A,B]</code> via the <code>.toMap</code> method, each "key" in the original sequence of tuples can only appear in the resulting Map once. As it happens, the conversion uses the last such occurrence.</p> <pre><code>l.sorted.reverse = List( (Agent 2,PASS), // <-- Last "Agent 2" (Agent 1,FAIL), // <-- Last "Agent 1" (Agent,PASS), (Agent,PASS), (Agent,FAIL)) // <-- Last "Agent" l.sorted.reverse.toMap = Map( Agent 2 -> PASS, Agent 1 -> FAIL, Agent -> FAIL) </code></pre> <p><strong>ORIGINAL REPLY</strong></p> <p>Starting with the answer...</p> <pre><code>val oldSolution = (l groupBy (_._1)) mapValues {_.sorted.head._2} </code></pre> <p>...and then showing my working :)</p> <pre><code>//group l groupBy (_._1) = Map( Agent 2 -> List((Agent 2,PASS)), Agent 1 -> List((Agent 1,FAIL)), Agent -> List((Agent,PASS), (Agent,FAIL), (Agent,PASS)) ) //extract values (l groupBy (_._1)) mapValues {_.map(_._2)} = Map( Agent 2 -> List(PASS), Agent 1 -> List(FAIL), Agent -> List(PASS, FAIL, PASS)) //sort (l groupBy (_._1)) mapValues {_.map(_._2).sorted} = Map( Agent 2 -> List(PASS), Agent 1 -> List(FAIL), Agent -> List(FAIL, PASS, PASS)) //head (l groupBy (_._1)) mapValues {_.map(_._2).sorted.head} = Map( Agent 2 -> PASS, Agent 1 -> FAIL, Agent -> FAIL) </code></pre> <p>However, you can directly sort the <code>agent -> status</code> pairs without needing to first extract <code>_2</code>:</p> <pre><code>//group & sort (l groupBy (_._1)) mapValues {_.sorted} = Map( Agent 2 -> List((Agent 2,PASS)), Agent 1 -> List((Agent 1,FAIL)), Agent -> List((Agent,FAIL), (Agent,PASS), (Agent,PASS))) //extract values (l groupBy (_._1)) mapValues {_.sorted.head._2} = Map( Agent 2 -> PASS, Agent 1 -> FAIL, Agent -> FAIL) </code></pre> <hr> <p>In either case, feel free to convert back to a List of Pairs if you wish:</p> <pre><code>l.sorted.reverse.toMap.toList = List( (Agent 2, PASS), (Agent 1, FAIL), (Agent, FAIL)) </code></pre>

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