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如何从{a1 | a2 | a3}格式的字符串中获得N个随机字符串?

algorithm java

Pentium10 · 技术社区 · 14 年前

string s1="planets {Sun|Mercury|Venus|Earth|Mars|Jupiter|Saturn|Uranus|Neptune}{?|!|.}"
string s2="some text {morning,night,evening} some text{?|!|.}"

我如何从集合中随机选择N,然后用逗号连接它们。集合在{}之间定义,选项用| pipe分隔维持秩序。字符串可以有多个集合{}。

string output1="planets Sun, Venus.";
string output2="planets Neptune!";
string output3="planets Earth, Saturn, Uranus, Neptune.";
string output4="planets Uranus, Saturn.";// bad example, order is not correct
string output5="some text morning!";

5 回复 | 直到 12 年前

Marcelo Cantos 14 年前

将行星排列成一个阵列,移除随机元素直到你有足够的空间。从算法上讲,有更可爱的解决方案,但对于处理的数据量来说,这既快速又简单。

Itay Karo 14 年前

我不确定java的语法,但应该是这样的。

string[] split = s.split("|");
Random r = new Random();
int first = r.nextInt(split.length);
string planets = "planets " + split[first++];
for (; first < split.length; first++)
{
    if (r.nextInt(2) == 1)
         planets += " " + split[first];
}

tucuxi 14 年前

这是另一个选择。这个只能从中选择一个元素 {|||} 设置,并在中的1到N个元素之间进行选择 {,,,}

public static String generateVariant(String s) {
    Pattern p = Pattern.compile("[{]([^}]+)[}]");
    Matcher m = p.matcher(s);
    StringBuilder output = new StringBuilder();

    int offset = 0;
    while (m.find()) {
        output.append(s.substring(offset, m.start()));
        String[] choices = m.group(1).split("[|,]");

        // if '|' used as separator, only echo 1 random choice
        int n = m.group(1).contains("|") ? 1
                : (int) (Math.random() * (choices.length - 1)) + 1;

        // permutation with n random elements
        int[] permutation = new int[choices.length];
        for (int i = 0; i < choices.length; i++) {
            permutation[i] = i;
        }
        for (int i=0; i<n; i++) {
            int r = (int)(Math.random() * (choices.length - i)) + i;
            int aux = permutation[r];
            permutation[r] = permutation[i];
            permutation[i] = aux;
        }

        // sort and echo first n
        Arrays.sort(permutation, 0, n);
        for (int i=0; i<n; i++) {
            output.append((i == 0 ? "" : ", ") + choices[permutation[i]]);
        }
        offset = m.end();
    }
    output.append(s.substring(offset, s.length()));
    return output.toString();
}

public static void main(String[] args) {
    String s1 = "planets {Sun,Mercury,Venus,Earth,Mars,Jupiter,Saturn,Uranus,Neptune}{?|!|.}";
    for (int i = 0; i < 10; i++) {
        System.err.println(generateVariant(s1));
    }
}

{|||} 套。下面是一个简短的版本,其中选择数的分布不再是统一的(灵感来自丹尼尔·特雷宾的答案):

public static String generateVariant(String s) {
    Pattern p = Pattern.compile("[{]([^}]+)[}]");
    Matcher m = p.matcher(s);
    StringBuilder output = new StringBuilder();
    Random r = new Random();

    int offset = 0;
    while (m.find()) {
        output.append(s.substring(offset, m.start()));
        String[] choices = m.group(1).split("[|,]");
        // if '|' used as separator, only echo 1 random choice
        if (m.group(1).contains("|")) {
            output.append(choices[r.nextInt(choices.length)]);
        } else {
            boolean first = true;
            for (int i=0; i<choices.length; i++) {
                if (r.nextBoolean()) {
                    output.append((first ? "" : ", ") + choices[i]);
                    first = false;
                }
            }                
        }
        offset = m.end();
    }
    output.append(s.substring(offset, s.length()));
    return output.toString();
}

Daniel Trebbien 14 年前

我测试了这个Java程序,它可以工作:

import java.util.Random;

/** @author Daniel Trebbien */
// License: Public Domain
public class SO2965185 {
    public static String randomFormat(final String templ) {
        int i = templ.indexOf('{');
        if (i < 0) {
            return templ;
        }
        else {
            Random r = new Random();

            int prevI = 0;
            StringBuilder sb = new StringBuilder();
            do {
                sb.append(templ, prevI, i);
                int j = templ.indexOf('}', i + 1);
                if (j < 0)
                    throw new java.util.MissingFormatArgumentException(templ.substring(i));
                int pipeCount = 0;
                for (int k = templ.indexOf('|', i + 1); i < k && k < j; k = templ.indexOf('|', k + 1))
                    ++pipeCount;
                if (pipeCount == 0) {
                    sb.append(templ, i + 1, j);
                }
                else {
                    String m0Selection;
                    final int m0 = r.nextInt(pipeCount + 1); // must pick one from each set
                    if (m0 >= pipeCount) {
                        m0Selection = templ.substring(templ.lastIndexOf('|', j - 1) + 1, j);
                    }
                    else {
                        int k = i + 1;
                        int m = m0;
                        for(; m > 0; --m)
                            k = templ.indexOf('|', k) + 1;
                        m0Selection = templ.substring(k, templ.indexOf('|', k + 1));
                    }

                    int selectionCount = 0;
                    for (int n = 0; n <= pipeCount; ++n) {
                        if (n == m0) {
                            if (selectionCount != 0)
                                sb.append(", ");
                            sb.append(m0Selection);
                            ++selectionCount;
                        }
                        else if (r.nextBoolean()) {
                            int m = n;
                            if (selectionCount != 0)
                                sb.append(", ");
                            if (m >= pipeCount) {
                                sb.append(templ, templ.lastIndexOf('|', j - 1) + 1, j);
                            }
                            else {
                                int k = i + 1;
                                for(; m > 0; --m)
                                    k = templ.indexOf('|', k) + 1;
                                sb.append(templ, k, templ.indexOf('|', k + 1));
                            }
                            ++selectionCount;
                        }
                    }
                }
                prevI = j + 1;
                i = templ.indexOf('{', j + 1);
            } while(i >= 0);
            return sb.toString();
        }
    }

    public static void main(String[] args) {
        System.out.println(randomFormat("test"));
        System.out.println(randomFormat("{oneOption}"));
        System.out.println(randomFormat("{first|second}"));
        String s1 = "planets {Sun|Mercury|Venus|Earth|Mars|Jupiter|Saturn|Uranus|Neptune}{?|!|.}";
        System.out.println(randomFormat(s1));
        //System.out.println(randomFormat("jjj{test"));
    }
}

test
oneOption
first, second
planets Sun, Mercury, Jupiter, Neptune?, !, .

和

test
oneOption
second
planets Sun, Jupiter, Saturn!, .

test
oneOption
first
planets Venus, Earth, Jupiter, Saturn, Uranus, Neptune.

代码从每个集合中选择至少有一个条目的条目的随机组合。所以,对于一套 N 条目,共有2个 ^N M 随机组合中的条目:

P(正是 M 生成的组合中的条目)=( 选择 M ^N

例子: N = 9 ( "{Sun|Mercury|Venus|Earth|Mars|Jupiter|Saturn|Uranus|Neptune}" ).

P(生成的组合中正好有2个条目)= 0.0704

dev13 12 年前

对于行星,只要写一个循环

result = empty list
For p in planets
  throw a dice
  If (dice > 3) // 50% probability, adjust as required
    append p to result
If result is empty, start over // do this if you don't want result to be empty