我有一个库，它返回一些二进制数据作为二进制数组列表。这些byte[]需要合并到InputStream中。
这是我当前的实现:
public static InputStream foo(List<byte[]> binary) {
byte[] streamArray = null;
binary.forEach(bin -> {
org.apache.commons.lang.ArrayUtils.addAll(streamArray, bin);
});
return new ByteArrayInputStream(streamArray);
}
，但这是相当密集的CPU。有没有更好的办法?
谢谢你的回答。我做了一个性能测试。这些是我的结果:
- Function: 'NicolasFilotto' => 68,04 ms平均100次呼叫
- Function: 'NicolasFilottoEstSize' => 65,100次调用平均24毫秒
- 函数:'NicolasFilottoSequenceInputStream' => 63,09 ms平均100次调用
- Function: 'Saka1029_1' => 63,06 ms平均100次调用
- Function: 'Saka1029_2' => 100次调用平均0.79 ms
- 函数:'Coco' => 541, 10次调用平均60毫秒
我不确定'Saka1029_2'是否测量正确…
这是执行函数:
private static double execute(Callable<InputStream> funct, int times) throws Exception {
List<Long> executions = new ArrayList<>(times);
for (int idx = 0; idx < times; idx++) {
BufferedReader br = null;
long startTime = System.currentTimeMillis();
InputStream is = funct.call();
br = new BufferedReader(new InputStreamReader(is));
String line = null;
while ((line = br.readLine()) != null) {}
executions.add(System.currentTimeMillis() - startTime);
}
return calculateAverage(executions);
}
注意我读取了每个输入流
这些是使用的实现:
public static class NicolasFilotto implements Callable<InputStream> {
private final List<byte[]> binary;
public NicolasFilotto(List<byte[]> binary) {
this.binary = binary;
}
@Override
public InputStream call() throws Exception {
ByteArrayOutputStream baos = new ByteArrayOutputStream();
for (byte[] bytes : binary) {
baos.write(bytes, 0, bytes.length);
}
return new ByteArrayInputStream(baos.toByteArray());
}
}
public static class NicolasFilottoSequenceInputStream implements Callable<InputStream> {
private final List<byte[]> binary;
public NicolasFilottoSequenceInputStream(List<byte[]> binary) {
this.binary = binary;
}
@Override
public InputStream call() throws Exception {
return new SequenceInputStream(
Collections.enumeration(
binary.stream().map(ByteArrayInputStream::new).collect(Collectors.toList())));
}
}
public static class NicolasFilottoEstSize implements Callable<InputStream> {
private final List<byte[]> binary;
private final int lineSize;
public NicolasFilottoEstSize(List<byte[]> binary, int lineSize) {
this.binary = binary;
this.lineSize = lineSize;
}
@Override
public InputStream call() throws Exception {
ByteArrayOutputStream baos = new ByteArrayOutputStream(binary.size() * lineSize);
for (byte[] bytes : binary) {
baos.write(bytes, 0, bytes.length);
}
return new ByteArrayInputStream(baos.toByteArray());
}
}
public static class Saka1029_1 implements Callable<InputStream> {
private final List<byte[]> binary;
public Saka1029_1(List<byte[]> binary) {
this.binary = binary;
}
@Override
public InputStream call() throws Exception {
byte[] all = new byte[binary.stream().mapToInt(a -> a.length).sum()];
int pos = 0;
for (byte[] bin : binary) {
int length = bin.length;
System.arraycopy(bin, 0, all, pos, length);
pos += length;
}
return new ByteArrayInputStream(all);
}
}
public static class Saka1029_2 implements Callable<InputStream> {
private final List<byte[]> binary;
public Saka1029_2(List<byte[]> binary) {
this.binary = binary;
}
@Override
public InputStream call() throws Exception {
int size = binary.size();
return new InputStream() {
int i = 0, j = 0;
@Override
public int read() throws IOException {
if (i >= size) return -1;
if (j >= binary.get(i).length) {
++i;
j = 0;
}
if (i >= size) return -1;
return binary.get(i)[j++];
}
};
}
}
public static class Coco implements Callable<InputStream> {
private final List<byte[]> binary;
public Coco(List<byte[]> binary) {
this.binary = binary;
}
@Override
public InputStream call() throws Exception {
byte[] streamArray = new byte[0];
for (byte[] bin : binary) {
streamArray = org.apache.commons.lang.ArrayUtils.addAll(streamArray, bin);
}
return new ByteArrayInputStream(streamArray);
}
}
您可以使用ByteArrayOutputStream来存储列表的每个字节数组的内容，但为了使其高效，我们需要创建ByteArrayOutputStream 的实例，其初始大小尽可能与目标大小相匹配，因此，如果您知道字节数组的大小或至少平均大小，您应该使用它，代码将是:
public static InputStream foo(List<byte[]> binary) {
ByteArrayOutputStream baos = new ByteArrayOutputStream(ARRAY_SIZE * binary.size());
for (byte[] bytes : binary) {
baos.write(bytes, 0, bytes.length);
}
return new ByteArrayInputStream(baos.toByteArray());
}
另一种方法是使用SequenceInputStream，以便在逻辑上连接代表列表中一个元素的所有ByteArrayInputStream实例，如下所示:
public static InputStream foo(List<byte[]> binary) {
return new SequenceInputStream(
Collections.enumeration(
binary.stream().map(ByteArrayInputStream::new).collect(Collectors.toList())
)
);
}
这种方法的有趣之处在于，您不需要复制任何东西，您只需创建ByteArrayInputStream的实例，该实例将使用字节数组的原样。
为了避免将结果收集为具有成本的List，特别是如果您的初始List很大，您可以直接调用iterator()，如所建议的@Holger，然后我们只需要将iterator转换为enumeration，这可以通过Apache Commons Collection中的IteratorUtils.asEnumeration(iterator)完成，最终代码将是:
public static InputStream foo(List<byte[]> binary) {
return new SequenceInputStream(
IteratorUtils.asEnumeration(
binary.stream().map(ByteArrayInputStream::new).iterator()
)
);
}
试试这个
public static InputStream foo(List<byte[]> binary) {
byte[] all = new byte[binary.stream().mapToInt(a -> a.length).sum()];
int pos = 0;
for (byte[] bin : binary) {
int length = bin.length;
System.arraycopy(bin, 0, all, pos, length);
pos += length;
}
return new ByteArrayInputStream(all);
}
或
public static InputStream foo(List<byte[]> binary) {
int size = binary.size();
return new InputStream() {
int i = 0, j = 0;
@Override
public int read() throws IOException {
if (i >= size) return -1;
if (j >= binary.get(i).length) {
++i;
j = 0;
}
if (i >= size) return -1;
return binary.get(i)[j++];
}
};
}