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<p>One topic that came up a lot when optimizing Scala data applications is the performance of standard collections, or the hidden cost of temporary copies. The collections <span class="caps">API</span> is easy to learn and maps well to many Python concepts where a lot of data engineers are familiar with. But the performance penalty can be pretty big when it’s repeated over millions of records in a <span class="caps">JVM</span> with limited heap.</p>
<h2>Mapping values</h2>
<p>Let’s take a look at one most naive example first, mapping the values of a <code>Map</code>.</p>
<div class="highlight"><pre><span></span><span class="k">val</span> <span class="n">m</span> <span class="k">=</span> <span class="nc">Map</span><span class="o">(</span><span class="s">"A"</span> <span class="o">-></span> <span class="mi">1</span><span class="o">,</span> <span class="s">"B"</span> <span class="o">-></span> <span class="mi">2</span><span class="o">,</span> <span class="s">"C"</span> <span class="o">-></span> <span class="mi">3</span><span class="o">)</span>
<span class="n">m</span><span class="o">.</span><span class="n">toList</span><span class="o">.</span><span class="n">map</span><span class="o">(</span><span class="n">t</span> <span class="k">=></span> <span class="o">(</span><span class="n">t</span><span class="o">.</span><span class="n">_1</span><span class="o">,</span> <span class="n">t</span><span class="o">.</span><span class="n">_2</span> <span class="o">+</span> <span class="mi">1</span><span class="o">)).</span><span class="n">toMap</span>
</pre></div>
<p>Looks simple enough but obviously not optimal. Two temporary <code>List[(String, Int)]</code> were created, one from <code>toList</code> and one from <code>map</code>. <code>map</code> also creates 3 copies of <code>(String, Int)</code>.</p>
<p>There are a few commonly seen variations. These don’t create temporary collections but still key-value tuples.</p>
<div class="highlight"><pre><span></span><span class="k">for</span> <span class="o">((</span><span class="n">k</span><span class="o">,</span> <span class="n">v</span><span class="o">)</span> <span class="k"><-</span> <span class="n">m</span><span class="o">)</span> <span class="k">yield</span> <span class="n">k</span> <span class="o">-></span> <span class="o">(</span><span class="n">v</span> <span class="o">+</span> <span class="mi">1</span><span class="o">)</span>
<span class="n">m</span><span class="o">.</span><span class="n">map</span> <span class="o">{</span> <span class="k">case</span> <span class="o">(</span><span class="n">k</span><span class="o">,</span> <span class="n">v</span><span class="o">)</span> <span class="k">=></span> <span class="n">k</span> <span class="o">-></span> <span class="o">(</span><span class="n">v</span> <span class="o">+</span> <span class="mi">1</span><span class="o">)</span> <span class="o">}</span>
</pre></div>
<p>If one reads the <a href="http://www.scala-lang.org/api/2.10.4/index.html#scala.collection.immutable.Map">ScalaDoc</a> closely, there’s a <code>mapValues</code> method already and it probably is the shortest and most performant.</p>
<div class="highlight"><pre><span></span><span class="n">m</span><span class="o">.</span><span class="n">mapValues</span><span class="o">(</span><span class="k">_</span> <span class="o">+</span> <span class="mi">1</span><span class="o">)</span>
</pre></div>
<h2>Java conversion</h2>
<p>Similar problem exists when converting between Scala and Java collections. We store data on <span class="caps">HDFS</span> in <a href="http://avro.apache.org/">Avro</a> and one most commonly used data structure is high dimensional vector, represented as <code>java.util.List[java.lang.Float]</code> in Avro and <code>Array[Float]</code> in Scala and <a href="https://github.com/scalanlp/breeze">breeze</a>.</p>
<p>Since there’s no direct conversion between Java <code>List</code> and Scala <code>Array</code>, the naive solution requires a temporary copy and looks like this:</p>
<div class="highlight"><pre><span></span><span class="k">import</span> <span class="nn">java.util.</span><span class="o">{</span> <span class="nc">List</span> <span class="k">=></span> <span class="nc">JList</span> <span class="o">}</span>
<span class="k">import</span> <span class="nn">java.lang.</span><span class="o">{</span> <span class="nc">Float</span> <span class="k">=></span> <span class="nc">JFloat</span> <span class="o">}</span>
<span class="k">import</span> <span class="nn">scala.collection.JavaConverters._</span>
<span class="k">def</span> <span class="n">javaToArray</span><span class="o">(</span><span class="n">v</span><span class="k">:</span> <span class="kt">JList</span><span class="o">[</span><span class="kt">JFloat</span><span class="o">])</span><span class="k">:</span> <span class="kt">Array</span><span class="o">[</span><span class="kt">Float</span><span class="o">]</span> <span class="k">=</span> <span class="n">v</span><span class="o">.</span><span class="n">asScala</span><span class="o">.</span><span class="n">asInstanceOf</span><span class="o">[</span><span class="kt">List</span><span class="o">[</span><span class="kt">Float</span><span class="o">]].</span><span class="n">toArray</span>
<span class="k">def</span> <span class="n">arrayToJava</span><span class="o">(</span><span class="n">v</span><span class="k">:</span> <span class="kt">Array</span><span class="o">[</span><span class="kt">Float</span><span class="o">])</span><span class="k">:</span> <span class="kt">JList</span><span class="o">[</span><span class="kt">JFloat</span><span class="o">]</span> <span class="k">=</span> <span class="n">v</span><span class="o">.</span><span class="n">toList</span><span class="o">.</span><span class="n">asInstanceOf</span><span class="o">[</span><span class="kt">List</span><span class="o">[</span><span class="kt">JFloat</span><span class="o">]].</span><span class="n">asJava</span>
</pre></div>
<p>The conversion is almost 10 times faster by doing the conversion manually in a loop.</p>
<div class="highlight"><pre><span></span><span class="k">def</span> <span class="n">javaToArray</span><span class="o">(</span><span class="n">v</span><span class="k">:</span> <span class="kt">JList</span><span class="o">[</span><span class="kt">JFloat</span><span class="o">])</span><span class="k">:</span> <span class="kt">Array</span><span class="o">[</span><span class="kt">Float</span><span class="o">]</span> <span class="k">=</span> <span class="o">{</span>
<span class="k">val</span> <span class="n">r</span> <span class="k">=</span> <span class="k">new</span> <span class="nc">Array</span><span class="o">[</span><span class="kt">Float</span><span class="o">](</span><span class="n">v</span><span class="o">.</span><span class="n">size</span><span class="o">)</span>
<span class="k">var</span> <span class="n">i</span> <span class="k">=</span> <span class="mi">0</span>
<span class="k">val</span> <span class="n">j</span> <span class="k">=</span> <span class="n">v</span><span class="o">.</span><span class="n">iterator</span>
<span class="k">while</span> <span class="o">(</span><span class="n">i</span> <span class="o"><</span> <span class="n">v</span><span class="o">.</span><span class="n">size</span><span class="o">)</span> <span class="o">{</span>
<span class="n">r</span><span class="o">(</span><span class="n">i</span><span class="o">)</span> <span class="k">=</span> <span class="n">j</span><span class="o">.</span><span class="n">next</span>
<span class="n">i</span> <span class="o">+=</span> <span class="mi">1</span>
<span class="o">}</span>
<span class="n">r</span>
<span class="o">}</span>
<span class="k">def</span> <span class="n">arrayToJava</span><span class="o">(</span><span class="n">v</span><span class="k">:</span> <span class="kt">Array</span><span class="o">[</span><span class="kt">Float</span><span class="o">])</span><span class="k">:</span> <span class="kt">JList</span><span class="o">[</span><span class="kt">JFloat</span><span class="o">]</span> <span class="k">=</span> <span class="o">{</span>
<span class="k">val</span> <span class="n">r</span> <span class="k">=</span> <span class="k">new</span> <span class="nc">JArrayList</span><span class="o">[</span><span class="kt">JFloat</span><span class="o">](</span><span class="n">v</span><span class="o">.</span><span class="n">size</span><span class="o">)</span>
<span class="k">var</span> <span class="n">i</span> <span class="k">=</span> <span class="mi">0</span>
<span class="k">while</span> <span class="o">(</span><span class="n">i</span> <span class="o"><</span> <span class="n">v</span><span class="o">.</span><span class="n">size</span><span class="o">)</span> <span class="o">{</span>
<span class="n">r</span><span class="o">.</span><span class="n">add</span><span class="o">(</span><span class="n">v</span><span class="o">(</span><span class="n">i</span><span class="o">))</span>
<span class="n">i</span> <span class="o">+=</span> <span class="mi">1</span>
<span class="o">}</span>
<span class="n">r</span>
<span class="o">}</span>
</pre></div>
<h2>Merging maps</h2>
<p>Same goes for merging 2 <code>Map[String, Double]</code> by summing up values of duplicate keys. Both solutions below require 2 copies of key sets, one union/intersect set,</p>
<div class="highlight"><pre><span></span><span class="k">val</span> <span class="n">m1</span> <span class="k">=</span> <span class="nc">Map</span><span class="o">(</span><span class="s">"A"</span> <span class="o">-></span> <span class="mf">1.0</span><span class="o">,</span> <span class="s">"B"</span> <span class="o">-></span> <span class="mf">2.0</span><span class="o">,</span> <span class="s">"C"</span> <span class="o">-></span> <span class="mf">3.0</span><span class="o">)</span>
<span class="k">val</span> <span class="n">m2</span> <span class="k">=</span> <span class="nc">Map</span><span class="o">(</span><span class="s">"A"</span> <span class="o">-></span> <span class="mf">1.5</span><span class="o">,</span> <span class="s">"B"</span> <span class="o">-></span> <span class="mf">2.5</span><span class="o">,</span> <span class="s">"D"</span> <span class="o">-></span> <span class="mf">3.5</span><span class="o">)</span>
<span class="c1">// solution 1</span>
<span class="k">val</span> <span class="n">i</span> <span class="k">=</span> <span class="n">m1</span><span class="o">.</span><span class="n">keySet</span> <span class="n">intersect</span> <span class="n">m2</span><span class="o">.</span><span class="n">keySet</span>
<span class="k">val</span> <span class="n">m</span> <span class="k">=</span> <span class="n">i</span><span class="o">.</span><span class="n">map</span> <span class="o">{</span> <span class="n">k</span> <span class="k">=></span> <span class="n">k</span> <span class="o">-></span> <span class="o">(</span><span class="n">m1</span><span class="o">(</span><span class="n">k</span><span class="o">)</span> <span class="o">+</span> <span class="n">m2</span><span class="o">(</span><span class="n">k</span><span class="o">))</span> <span class="o">}</span>
<span class="o">(</span><span class="n">m1</span> <span class="o">--</span> <span class="n">i</span><span class="o">)</span> <span class="o">++</span> <span class="o">(</span><span class="n">m2</span> <span class="o">--</span> <span class="n">i</span><span class="o">)</span> <span class="o">++</span> <span class="n">m</span> <span class="c1">// -- i unnecessary</span>
<span class="n">m1</span> <span class="o">++</span> <span class="n">m2</span> <span class="o">++</span> <span class="n">m</span> <span class="c1">// slightly better</span>
<span class="c1">// solution 2</span>
<span class="o">(</span><span class="n">m1</span><span class="o">.</span><span class="n">keySet</span> <span class="o">++</span> <span class="n">m2</span><span class="o">.</span><span class="n">keySet</span><span class="o">)</span> <span class="n">map</span> <span class="o">{</span> <span class="n">k</span> <span class="k">=></span>
<span class="n">k</span> <span class="o">-></span> <span class="o">(</span><span class="n">m1</span><span class="o">.</span><span class="n">getOrElse</span><span class="o">(</span><span class="n">k</span><span class="o">,</span> <span class="mf">0.0</span><span class="o">)</span> <span class="o">+</span> <span class="n">m2</span><span class="o">.</span><span class="n">getOrElse</span><span class="o">(</span><span class="n">k</span><span class="o">,</span> <span class="mf">0.0</span><span class="o">))</span>
<span class="o">}</span>
</pre></div>
<p>The best solution for this case also happens to be the shortest. It leverages the fact that pairs from the right-hand size of <code>++</code> overwrite those on the left-hand size. One could even compare size of the maps and put the smaller on the right-hand size to further reduce copies.</p>
<div class="highlight"><pre><span></span><span class="n">m1</span> <span class="o">++</span> <span class="n">m2</span><span class="o">.</span><span class="n">map</span> <span class="o">{</span> <span class="k">case</span> <span class="o">(</span><span class="n">k</span><span class="o">,</span> <span class="n">v</span><span class="o">)</span> <span class="k">=></span> <span class="n">k</span> <span class="o">-></span> <span class="o">(</span><span class="n">v</span> <span class="o">+</span> <span class="n">m1</span><span class="o">.</span><span class="n">getOrElse</span><span class="o">(</span><span class="n">k</span><span class="o">,</span> <span class="mf">0.0</span><span class="o">))</span> <span class="o">}</span>
</pre></div>
<h2>Cheating by mutation</h2>
<p>All of these optimizations are nice but sometimes they are just not enough. Imagine if the map merge operation is used in a <code>foldLeft</code> on the reducer size, over millions of users with one <code>Map</code> each, we’ll be basically creating that many temporary copies and sometimes <span class="caps">GC</span> just can’t keep up. Good thing is that in some cases we know for sure a temporary object won’t be reused ever again, and can mutate it in place and pass it along to further reduce copying. In the example below <span class="caps">LHS</span> is mutated in place and returned in the binary operation.</p>
<div class="highlight"><pre><span></span><span class="k">import</span> <span class="nn">scala.collection.mutable.</span> <span class="o">{</span> <span class="nc">Map</span> <span class="k">=></span> <span class="nc">MMap</span> <span class="o">}</span>
<span class="k">def</span> <span class="n">mergeMaps</span><span class="o">(</span><span class="n">m1</span><span class="k">:</span> <span class="kt">MMap</span><span class="o">[</span><span class="kt">String</span>, <span class="kt">Double</span><span class="o">],</span> <span class="n">m2</span><span class="k">:</span> <span class="kt">MMap</span><span class="o">[</span><span class="kt">String</span>, <span class="kt">Double</span><span class="o">])</span> <span class="k">=</span> <span class="o">{</span>
<span class="n">m2</span><span class="o">.</span><span class="n">foreach</span> <span class="o">{</span> <span class="k">case</span> <span class="o">(</span><span class="n">k</span><span class="o">,</span> <span class="n">v</span><span class="o">)</span> <span class="k">=></span> <span class="n">m1</span><span class="o">(</span><span class="n">k</span><span class="o">)</span> <span class="k">=</span> <span class="n">v</span> <span class="o">+</span> <span class="n">m1</span><span class="o">.</span><span class="n">getOrElse</span><span class="o">(</span><span class="n">k</span><span class="o">,</span> <span class="mf">0.0</span><span class="o">)</span> <span class="o">}</span>
<span class="n">m1</span>
<span class="o">}</span>
</pre></div>
<p>And in this <code>foldLeft</code> operation, <span class="caps">LHS</span> is always the same object as initial value <code>z</code> and mutated throughout the group.</p>
<div class="highlight"><pre><span></span><span class="k">val</span> <span class="n">group</span><span class="k">:</span> <span class="kt">Grouped</span><span class="o">[</span><span class="kt">String</span>, <span class="kt">MMap</span><span class="o">[</span><span class="kt">String</span>, <span class="kt">Double</span><span class="o">]]</span> <span class="c1">// = ...</span>
<span class="n">group</span><span class="o">.</span><span class="n">foldLeft</span><span class="o">(</span><span class="nc">MMap</span><span class="o">[</span><span class="kt">String</span>, <span class="kt">Double</span><span class="o">]())(</span><span class="n">mergeMaps</span><span class="o">)</span>
</pre></div>
<p>Some numeric libraries, like <a href="http://mikiobraun.github.io/jblas/"><span class="caps">JBLAS</span></a> offer operators in both immutable (by copying) and mutable (by mutating in place) fashions. It’s probably best to start with the immutable ones to ensure correctness, and drop down to mutating in place when running into performance problems.</p>
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