从java字符串中有效地提取数字(已经尝试过番石榴和正则表达式)

Question

试图从字符串中有效地提取一些数字并尝试过

• java.util.regex.Matcher中
• com.google.common.base.Splitter

结果是:

• 通过正则表达式:24417 ms
• 通过Google Splitter:17730毫秒

还有另一种更快的推荐方式吗？

我知道之前提出的类似问题,例如如何从Java中的String中提取多个整数？但我的重点在于快速(但可维护/简单),因为它发生了很多.

---

编辑:以下是我的最终结果,与下面的Andrea Ligios相关:

• 正则表达式(无括号):18857
• Google Splitter(没有superflous trimResults()方法):15329
• Martijn Courteaux回答如下:4073

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import org.junit.Test;

import com.google.common.base.CharMatcher;
import com.google.common.base.Splitter;
import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Sample {

    final static int COUNT = 50000000;
    public static final String INPUT = "FOO-1-9-BAR1"; // I want 1, 9, 1

    @Test
    public void extractNumbers() {
        long startTime = System.currentTimeMillis();
        for (int i = 0; i < COUNT; i++) {
            // Output is list of 1, 9, 1
            Demo.extractNumbersViaGoogleSplitter(INPUT);
        }
        System.out.println("Total execution time (ms) via Google Splitter: " + (System.currentTimeMillis() - startTime));


        startTime = System.currentTimeMillis();
        for (int i = 0; i < COUNT; i++) {
            // Output is list of 1, 9, 1
            Demo.extractNumbersViaRegEx(INPUT);
        }
        System.out.println("Total execution time (ms) Regular Expression: " + (System.currentTimeMillis() - startTime));

    }
}

class Demo {

    static List<Integer> extractNumbersViaGoogleSplitter(final String text) {

        Iterator<String> iter = Splitter.on(CharMatcher.JAVA_DIGIT.negate()).trimResults().omitEmptyStrings().split(text).iterator();
        final List<Integer> result = new ArrayList<Integer>();
        while (iter.hasNext()) {
            result.add(Integer.parseInt(iter.next()));

        }
        return result;
    }
    /**
     * Matches all the numbers in a string, as individual groups. e.g.
     * FOO-1-BAR1-1-12 matches 1,1,1,12.
     */
    private static final Pattern NUMBERS = Pattern.compile("(\\d+)");

    static List<Integer> extractNumbersViaRegEx(final String source) {
        final Matcher matcher = NUMBERS.matcher(source);
        final List<Integer> result = new ArrayList<Integer>();

        if (matcher.find()) {
            do {
                result.add(Integer.parseInt(matcher.group(0)));
            } while (matcher.find());
            return result;
        }
        return result;
    }
}

Answer 1

这是一个非常快速的算法:

public List<Integer> extractIntegers(String input)
{
    List<Integer> result = new ArrayList<Integer>();
    int index = 0;
    int v = 0;
    int l = 0;
    while (index < input.length())
    {
        char c = input.charAt(index);
        if (Character.isDigit(c))
        {
            v *= 10;
            v += c - '0';
            l++;
        } else if (l > 0)
        {
            result.add(v);
            l = 0;
            v = 0;
        }
        index++;
    }
    if (l > 0)
    {
        result.add(v);
    }
    return result;
}

这段代码在我的机器上花了3672毫秒,用于"FOO-1-9-BAR1"和50000000次运行.我使用的是2.3 GHz核心.