lambda表现的差异？

时间：2015-10-17 02:27:00

标签： java performance for-loop lambda java-8

This与我的问题不重复。我检查了它，它更多的是关于内部匿名类。

我对Lambda表达式感到好奇并测试了以下内容：

  • 给定一万个条目的数组，删除某些索引的速度会更快：Lamba表达式还是带有if测试的For-Loop？

在我不知道自己会想出什么的事实中，第一批结果并不令人惊讶：

final int NUMBER_OF_LIST_INDEXES = 10_000;
List<String> myList = new ArrayList<>();
String[] myWords = "Testing Lamba expressions with this String array".split(" ");

for (int i = 0 ; i < NUMBER_OF_LIST_INDEXES ; i++){
    myList.add(myWords[i%6]);
}

long time = System.currentTimeMillis();

// BOTH TESTS WERE RUN SEPARATELY OF COURSE
// PUT THE UNUSED ONE IN COMMENTS WHEN THE OTHER WAS WORKING

// 250 milliseconds for the Lambda Expression
myList.removeIf(x -> x.contains("s"));

// 16 milliseconds for the traditional Loop
for (int i = NUMBER_OF_LIST_INDEXES - 1 ; i >= 0 ; i--){
    if (myList.get(i).contains("s")) myList.remove(i);
}
System.out.println(System.currentTimeMillis() - time + " milliseconds");

但是，我决定将常数NUMBER_OF_LIST_INDEXES更改为一百万，结果如下：

final int NUMBER_OF_LIST_INDEXES = 1_000_000;
List<String> myList = new ArrayList<>();
String[] myWords = "Testing Lamba expressions with this String array".split(" ");

for (int i = 0 ; i < NUMBER_OF_LIST_INDEXES ; i++){
    myList.add(myWords[i%6]);
}

long time = System.currentTimeMillis();

// BOTH TESTS WERE RUN SEPARATELY OF COURSE
// PUT THE UNUSED ONE IN COMMENTS WHEN THE OTHER WAS WORKING

// 390 milliseconds for the Lambda Expression
myList.removeIf(x -> x.contains("s"));

// 32854 milliseconds for the traditional Loop
for (int i = NUMBER_OF_LIST_INDEXES - 1 ; i >= 0 ; i--){ 
    if (myList.get(i).contains("s")) myList.remove(i);
}
System.out.println(System.currentTimeMillis() - time + " milliseconds");

为了使事情更简单，以下是结果：

|        |  10.000 | 1.000.000 |
| LAMBDA |  250ms  |   390ms   | 156% evolution
|FORLOOP |   16ms  |  32854ms  | 205000+% evolution

我有以下问题：

  • 这背后的魔力是什么？当使用的索引是* 100时，我们如何为数组而不是lambda带来如此大的差异。

  • 在性能方面，我们如何知道何时使用Lambdas以及何时坚持使用传统方式处理数据？

  • 这是List方法的特定行为吗？其他Lambda表达式是否也会产生像这样的随机性能？

3 个答案:

答案 0 :(得分：13)

因为public class MainActivityFragment extends Fragment { private ArrayAdapter<String> mForecastAdapter; public MainActivityFragment() { } @Override public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) { View rootView = inflater.inflate(R.layout.fragment_main, container, false); String[] forecastArray = { "Today - Foggy - 12/3", "Tomorrow - Rainy - 7/4", "Wed - Sunny - 12/3", "Thurs - Cloudy - 10/4", "Fri - Rainy - 12/8", "Sat - Heavy Rain- 10/5", "Sun - Sunny - 32/23" }; List<String> weekforecast = new ArrayList<String>( Arrays.asList(forecastArray)); mForecastAdapter = new ArrayAdapter<String>( getActivity(), R.layout.list_item_forecast, R.id.list_item_forecast_textview, weekforecast); ListView listView = (ListView) rootView.findViewById(R.id.list_item_forecast); listView.setAdapter(mForecastAdapter); return rootView; } public class FetchWeatherTask extends AsyncTask<Void, Void, Void> { private final String LOG_TAG = FetchWeatherTask.class.getSimpleName(); @Override protected Void doInBackground(Void... params) { HttpURLConnection urlConnection = null; BufferedReader reader = null; // Will contain the raw JSON response as a string. String forecastJsonStr = null; try { // Construct the URL for the OpenWeatherMap query // Possible parameters are avaiable at OWM's forecast API page, at // http://openweathermap.org/API#forecast URL url = new URL("api.openweathermap.org/data/2.5/find?q=7000&mode=json&units=metric&ch+7&appid=bd82977b86bf27fb59a04b61b657fb6f"); // Create the request to OpenWeatherMap, and open the connection urlConnection = (HttpURLConnection) url.openConnection(); urlConnection.setRequestMethod("GET"); urlConnection.connect(); // Read the input stream into a String InputStream inputStream = urlConnection.getInputStream(); StringBuffer buffer = new StringBuffer(); if (inputStream == null) { // Nothing to do. return null; } reader = new BufferedReader(new InputStreamReader(inputStream)); String line; while ((line = reader.readLine()) != null) { // Since it's JSON, adding a newline isn't necessary (it won't affect parsing) // But it does make debugging a *lot* easier if you print out the completed // buffer for debugging. buffer.append(line + "\n"); } if (buffer.length() == 0) { // Stream was empty. No point in parsing. return null; } forecastJsonStr = buffer.toString(); } catch (IOException e) { Log.e("PlaceholderFragment", "Error ", e); // If the code didn't successfully get the weather data, there's no point in attemping // to parse it. return null; } finally { if (urlConnection != null) { urlConnection.disconnect(); } if (reader != null) { try { reader.close(); } catch (final IOException e) { Log.e("PlaceholderFragment", "Error closing stream", e); } } } return null; } } 非常昂贵:)它需要复制和移动其余的元素，这在你的情况下会多次完成。

虽然remove(index)不需要这样做。它可以扫描一次并标记要删除的所有元素;然后最后阶段将幸存者复制到列表头部一次。

答案 1 :(得分：8)

我写了一个JMH基准来测量它。它有4种方法：

  • removeIf ArrayList。
  • removeIf LinkedList。
    • 在iterator.remove()上使用ArrayList的
  • 迭代器。
    • 在iterator.remove()上使用LinkedList的
  • 迭代器。

基准测试的目的是表明removeIf和迭代器应该提供相同的性能，但ArrayList不是这种情况。

默认情况下，removeIf在内部使用迭代器来删除元素，因此我们应该期望removeIf和iterator具有相同的效果。

现在考虑一个ArrayList，它在内部使用一个数组来保存元素。每当我们调用remove时，索引后面的其余元素必须移动一个;所以每次都要复制很多元素。当迭代器用于遍历ArrayList并且我们需要删除元素时，这种复制需要一次又一次地发生，这使得它非常慢。对于LinkedList，情况并非如此：删除元素时，唯一的变化是指向下一个元素的指针。

那么为什么removeIf ArrayList与LinkedList一样快？因为它实际上被覆盖并且它不使用迭代器：代码实际上标记了要在第一遍中删除的元素，然后在第二遍中删除它们（移动其余元素）。在这种情况下可以进行优化：每次需要移除剩余元素时，我们只会在知道需要删除的所有元素时执行一次。

结论：

    当需要删除与谓词匹配的每个元素时，应使用
  • removeIf。
  • remove应该用于删除单个已知元素。

基准测试结果：

Benchmark                            Mode  Cnt      Score      Error  Units
RemoveTest.removeIfArrayList         avgt   30      4,478 ±   0,194  ms/op
RemoveTest.removeIfLinkedList        avgt   30      3,634 ±   0,184  ms/op
RemoveTest.removeIteratorArrayList   avgt   30  27197,046 ± 536,584  ms/op
RemoveTest.removeIteratorLinkedList  avgt   30      3,601 ±   0,195  ms/op

基准：

@Warmup(iterations = 5, time = 1000, timeUnit = TimeUnit.MILLISECONDS)
@Measurement(iterations = 10, time = 1000, timeUnit = TimeUnit.MILLISECONDS)
@BenchmarkMode(Mode.AverageTime)
@OutputTimeUnit(TimeUnit.MILLISECONDS)
@Fork(3)
@State(Scope.Benchmark)
public class RemoveTest {

    private static final int NUMBER_OF_LIST_INDEXES = 1_000_000;
    private static final String[] words = "Testing Lamba expressions with this String array".split(" ");

    private ArrayList<String> arrayList;
    private LinkedList<String> linkedList;

    @Setup(Level.Iteration)
    public void setUp() {
        arrayList = new ArrayList<>();
        linkedList = new LinkedList<>();
        for (int i = 0 ; i < NUMBER_OF_LIST_INDEXES ; i++){
            arrayList.add(words[i%6]);
            linkedList.add(words[i%6]);
        }
    }

    @Benchmark
    public void removeIfArrayList() {
        arrayList.removeIf(x -> x.contains("s"));
    }

    @Benchmark
    public void removeIfLinkedList() {
        linkedList.removeIf(x -> x.contains("s"));
    }

    @Benchmark
    public void removeIteratorArrayList() {
        for (ListIterator<String> it = arrayList.listIterator(arrayList.size()); it.hasPrevious();){
            if (it.previous().contains("s")) it.remove();
        }
    }

    @Benchmark
    public void removeIteratorLinkedList() {
        for (ListIterator<String> it = linkedList.listIterator(linkedList.size()); it.hasPrevious();){
            if (it.previous().contains("s")) it.remove();
        }
    }

    public static void main(String[] args) throws Exception {
         Main.main(args);
    }

}

答案 2 :(得分：1)

我认为您所看到的性能差异可能更多是由于squirrel <- read.table("F:/Stats ass 5/squirrel.data", header=TRUE, quote="\"") View(squirrel) squirrel$age=as.factor(squirrel$age) squirrel$sex=as.factor(squirrel$sex) squirrel$weight=as.factor(squirrel$weight) squirrel$survive=as.numeric(squirrel$survive) squirrel$number=as.numeric(squirrel$number) 在内部使用迭代器而在for循环中使用get和remove。这个PAQ中的答案有一些关于迭代器好处的好信息。

bayou.io的回答是正确的，您可以看到code for removeIf here它会进行两次通过以避免一遍又一遍地移动剩余的元素。

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