Epa*_*aga 9 java algorithm performance
我想从表示十六进制值(大写或小写)的字符转换为字节,如
'0'->0, '1' -> 1, 'A' -> 10, 'a' -> 10, 'f' -> 15 etc...
我会经常调用这种方法,所以性能很重要.有没有比使用预先初始化HashMap<Character,Byte>从中获取值更快的方法？
回答
看起来这是使用switch-case和Jon Skeet的直接计算解决方案之间的一个折腾 - 尽管如此,交换机案例解决方案似乎有点微不足道.Greg的阵列方法胜出.以下是各种方法的200,000,000次运行的性能结果(以ms为单位):
Character.getNumericValue:
8360
Character.digit:
8453
HashMap<Character,Byte>:
15109
Greg's Array Method:
6656
JonSkeet's Direct Method:
7344
Switch:
7281
多谢你们!
基准方法代码
你好,JonSkeet,你是老竞争对手.;-)
public class ScratchPad {
private static final int NUMBER_OF_RUNS = 200000000;
static byte res;
static HashMap<Character, Byte> map = new HashMap<Character, Byte>() {{
put( Character.valueOf( '0' ), Byte.valueOf( (byte )0 ));
put( Character.valueOf( '1' ), Byte.valueOf( (byte )1 ));
put( Character.valueOf( '2' ), Byte.valueOf( (byte )2 ));
put( Character.valueOf( '3' ), Byte.valueOf( (byte )3 ));
put( Character.valueOf( '4' ), Byte.valueOf( (byte )4 ));
put( Character.valueOf( '5' ), Byte.valueOf( (byte )5 ));
put( Character.valueOf( '6' ), Byte.valueOf( (byte )6 ));
put( Character.valueOf( '7' ), Byte.valueOf( (byte )7 ));
put( Character.valueOf( '8' ), Byte.valueOf( (byte )8 ));
put( Character.valueOf( '9' ), Byte.valueOf( (byte )9 ));
put( Character.valueOf( 'a' ), Byte.valueOf( (byte )10 ));
put( Character.valueOf( 'b' ), Byte.valueOf( (byte )11 ));
put( Character.valueOf( 'c' ), Byte.valueOf( (byte )12 ));
put( Character.valueOf( 'd' ), Byte.valueOf( (byte )13 ));
put( Character.valueOf( 'e' ), Byte.valueOf( (byte )14 ));
put( Character.valueOf( 'f' ), Byte.valueOf( (byte )15 ));
put( Character.valueOf( 'A' ), Byte.valueOf( (byte )10 ));
put( Character.valueOf( 'B' ), Byte.valueOf( (byte )11 ));
put( Character.valueOf( 'C' ), Byte.valueOf( (byte )12 ));
put( Character.valueOf( 'D' ), Byte.valueOf( (byte )13 ));
put( Character.valueOf( 'E' ), Byte.valueOf( (byte )14 ));
put( Character.valueOf( 'F' ), Byte.valueOf( (byte )15 ));
}};
static int[] charValues = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -1, -1, -1, -1, -1, -1, -1, 10, 11, 12, 13,14,15,-1,-1,-1,-1,-1,-1,-1,-1,-1,
-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,10, 11, 12, 13,14,15};
static char[] cs = new char[]{'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f','A','B','C','D','E','F'};
public static void main(String args[]) throws Exception {
long time = System.currentTimeMillis();
for( int i = 0; i < NUMBER_OF_RUNS; i++ ) {
res = getNumericValue( i );
}
System.out.println( "Character.getNumericValue:" );
System.out.println( System.currentTimeMillis()-time );
time = System.currentTimeMillis();
for( int i = 0; i < NUMBER_OF_RUNS; i++ ) {
res = getDigit( i );
}
System.out.println( "Character.digit:" );
System.out.println( System.currentTimeMillis()-time );
time = System.currentTimeMillis();
for( int i = 0; i < NUMBER_OF_RUNS; i++ ) {
try {
res = getValueFromArray( i );
} catch (IllegalArgumentException e) {
}
}
System.out.println( "Array:" );
System.out.println( System.currentTimeMillis()-time );
time = System.currentTimeMillis();
for( int i = 0; i < NUMBER_OF_RUNS; i++ ) {
res = getValueFromHashMap( i );
}
System.out.println( "HashMap<Character,Byte>:" );
System.out.println( System.currentTimeMillis()-time );
time = System.currentTimeMillis();
for( int i = 0; i < NUMBER_OF_RUNS; i++ ) {
char c = cs[i%cs.length];
res = getValueFromComputeMethod( c );
}
System.out.println( "JonSkeet's Direct Method:" );
System.out.println( System.currentTimeMillis()-time );
time = System.currentTimeMillis();
for( int i = 0; i < NUMBER_OF_RUNS; i++ ) {
res = getValueFromSwitch( i );
}
System.out.println( "Switch:" );
System.out.println( System.currentTimeMillis()-time );
}
private static byte getValueFromSwitch( int i ) {
byte res;
char ch = cs[i%cs.length];
switch( ch ) {
case '0':
res = 0;
break;
case '1':
res = 1;
break;
case '2':
res = 2;
break;
case '3':
res = 3;
break;
case '4':
res = 4;
break;
case '5':
res = 5;
break;
case '6':
res = 6;
break;
case '7':
res = 7;
break;
case '8':
res = 8;
break;
case '9':
res = 9;
break;
case 'a':
case 'A':
res = 10;
break;
case 'b':
case 'B':
res = 11;
break;
case 'c':
case 'C':
res = 12;
break;
case 'd':
case 'D':
res = 13;
break;
case 'e':
case 'E':
res = 14;
break;
case 'f':
case 'F':
res = 15;
break;
default:
throw new RuntimeException("unknown hex character: " + ch );
}
return res;
}
private static byte getValueFromComputeMethod( char c ) {
byte result = 0;
if (c >= '0' && c <= '9')
{
result = (byte)(c - '0');
}
if (c >= 'a' && c <= 'f')
{
result = (byte)(c - 'a' + 10);
}
if (c >= 'A' && c <= 'F')
{
result = (byte)(c - 'A' + 10);
}
return result;
}
private static byte getValueFromHashMap( int i ) {
return map.get( Character.valueOf( cs[i%cs.length] ) ).byteValue();
}
private static byte getValueFromArray( int i ) {
char c = cs[i%cs.length];
if (c < '0' || c > 'f') {
throw new IllegalArgumentException();
}
byte result = (byte)charValues[c-'0'];
if (res < 0) {
throw new IllegalArgumentException();
}
return result;
}
private static byte getDigit( int i ) {
return (byte)Character.digit( cs[i%cs.length], 16 );
}
private static byte getNumericValue( int i ) {
return (byte)Character.getNumericValue( cs[i%cs.length] );
}
}
Gre*_*ill 16
预初始化的数组比HashMap快.像这样的东西:
int CharValues['f'-'0'+1] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -1, -1, ... -1, 10, 11, 12, ...};
if (c < '0' || c > 'f') {
throw new IllegalArgumentException();
}
int n = CharValues[c-'0'];
if (n < 0) {
throw new IllegalArgumentException();
}
// n contains the digit value
您应该将此方法与其他方法(例如Jon Skeet的直接方法)进行基准测试,以确定哪种方法对您的应用程序来说最快.
Jon*_*eet 14
哈希表会相对较慢.这很快:
if (c >= '0' && c <= '9')
{
return c - '0';
}
if (c >= 'a' && c <= 'f')
{
return c - 'a' + 10;
}
if (c >= 'A' && c <= 'F')
{
return c - 'A' + 10;
}
throw new IllegalArgumentException();
另一种选择是尝试切换/ case语句.如果数组在缓存中,则数组可能没问题,但是错过可能很昂贵.
我不记得以前见过这种方法，但 Mikko Rantanen 在对问题Code Golf - hex to (raw) binary conversion的评论中指出了这个等式
(char | 32) % 39 - 9
我不知道它的基准是什么（也许有人可以将它添加到上面的基准并运行它，但我猜 % 会降低性能） - 但它是一个整洁、简单的单字符十六进制单行十进制转换。处理 0-9、AF、AF。
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