寻找最相似值的有效方法

Question

我有一个价值，如颜色，和字符串列表：{颜色，颜色，主题，品牌，主题……等等

我想得到最相似的字符串，除了搜索到的字符串本身。在本例中，希望得到颜色。(不是颜色)

我使用以下规则对列表进行排序，并对规则进行排序：

1. 过滤相同的值
2. 检查上下位病例
3. 删除空白空间。纵倾
4. 使用Levenshtein距离
5. 字符串顺序:主颜色=主色
6. 检查首字母缩写: HP -惠普

查看1000名相关候选人的名单需要很长时间。此外，我有很多候选人要检查。

还有其他有效的方法吗？

原始守则：

public static List findSimilarity(String word, List candidates) {
    List recommendations = new ArrayList();
    if (!word.equals("")) {
        for (String candidate : candidates) {
            if (!word.equals(candidate)) { //1. same token , lower/upper cases , ignore white spaces
                if (StringUtils.deleteWhitespace(word).equalsIgnoreCase(StringUtils.deleteWhitespace(candidate))) {
                    recommendations.add(candidate);
                }
                //2. same tokens diff order
                else if (candidate.split(" ").length == word.split("     ").length) {
                    String[] candidatearr = candidate.split(" ");
                    String[] wordarr = word.split(" ");
                    boolean status = true;
                    SortIgnoreCase icc = new SortIgnoreCase();
                    Arrays.sort(candidatearr, icc);
                    Arrays.sort(wordarr, icc);
                    for (int i = 0; i < candidatearr.length; i++) {
                        if (!(candidatearr[i] == null ? wordarr[i] == null : wordarr[i].equalsIgnoreCase(candidatearr[i])))
                            status = false;
                    }

                    if (status) {
                        recommendations.add(candidate);
                    }
                }
            }
        }
        //3. distance between words
        if (recommendations.size() == 0) {
            for (String candidate : candidates) {
                if (!word.equals(candidate)) {
                    String[] candidatearr = candidate.split(" ");
                    String[] wordarr = word.split(" ");
                    //check for acronym
                    if ((wordarr.length == 1 && candidatearr.length > 1) || (wordarr.length > 1 && candidatearr.length == 1)) {
                        String acronym = "";
                        if (wordarr.length > candidatearr.length) {
                            for (String tmp : wordarr) {
                                if (!tmp.equals("")) {
                                    acronym = acronym + tmp.substring(0, 1);
                                }
                            }

                            if (acronym.equalsIgnoreCase(candidatearr[0])) {
                                recommendations.add(candidate);
                            }
                        } else {
                            for (String tmp : candidatearr) {
                                if (!tmp.equals("")) {
                                    acronym = acronym + tmp.substring(0, 1);
                                }
                            }

                            if (acronym.equalsIgnoreCase(wordarr[0])) {
                                recommendations.add(candidate);
                            }
                        }
                    }
                }
            }
        }

        if (recommendations.size() == 0) {
            for (String candidate : candidates) {
                if (!word.equals(candidate)) {
                    int dist = 0;
                    String check = "";
                    if (word.length() > candidate.length()) {
                        check = candidate;
                    } else {
                        check = word;
                    }
                    if (check.length() <= 3) {
                        dist = 0;
                    } else if (check.length() > 3 && check.length() <= 5) {
                        dist = 1;
                    } else if (check.length() > 5) {
                        dist = 2;
                    }

                    if (StringUtils.getLevenshteinDistance(word, candidate) <= dist) {
                        //if(Levenshtein.distance(word,candidate) <= dist){
                        recommendations.add(candidate);
                    }
                }
            }
        }

        if (recommendations.size() == 0) {
            for (String candidate : candidates) {
                if (!word.equals(candidate)) {
                    String[] candidatearr = candidate.split(" ");
                    String[] wordarr = word.split(" ");

                    for (String cand : candidatearr) {
                        for (String wor : wordarr) {
                            if (cand.equals(wor) && cand.length() > 4) {
                                recommendations.add(candidate);

                            }
                        }
                    }
                }
            }//for
            if (recommendations.size() > 4) {
                recommendations.clear();
            }
        }

        //4. low priority - starts with
        if (recommendations.size() == 0) {
            for (String candidate : candidates) {
                if (!word.equals(candidate)) {
                    if (candidate.startsWith(word) || word.startsWith(candidate)) {
                        recommendations.add(candidate);
                    }
                }
            }
            if (recommendations.size() > 4) {
                recommendations.clear();
            }
        }

        //5. low priority - contain word
        if (recommendations.size() == 0) {
            for (String candidate : candidates) {
                if (!word.equals(candidate)) {
                    if (candidate.contains(word) || word.contains(candidate)) {
                        recommendations.add(candidate);
                    }
                }
            }
            if (recommendations.size() > 4) {
                recommendations.clear();
            }
        }
    }
    return recommendations;
}

谢谢你，M。

Answer 1

编辑的

为了让您更好地理解，我将波希米亚给出的答案包装在您的原始代码的上下文中。

行.map(term -> Arrays.stream(term.split(" ")).sorted().collect(Collectors.joining(" ")))拆分多个单词的术语、排序和再次连接，以消除相同单词的排列。这是对像“主颜色”和“颜色主”这样的术语的置换平等挑战的回答。

但是，在这个问题的上下文中捕获任务的所有业务需求是没有意义的。通过这个答案，你得到了解决方案的大纲。讨论了效率问题。您可能需要更多的阶段，但这是一个不同的故事。该方法的优点是所有阶段都是独立的，因此您可以独立地提出问题并寻求帮助。

public static String findSimilarity(String word, List<String> candidatesList) {

    // Populating the set with distinct values of the input terms
    Set<String> candidates = candidatesList.stream()
            .map(String::toLowerCase)
            .map(term -> Arrays.stream(term.split(" ")).sorted().collect(Collectors.joining(" "))) // eliminates permutations
            .collect(Collectors.toSet());

    Map<String, Integer> cache = new ConcurrentHashMap<>();

    return candidates.parallelStream()
            .map(String::trim)
                    // add more mappers if needed
            .filter(s -> !s.equalsIgnoreCase(word))
                    // add more filters if needed
            .min((a, b) -> Integer.compare(
                    cache.computeIfAbsent(a, k -> getLevenshteinDistance(word, k)),
                    cache.computeIfAbsent(b, k -> getLevenshteinDistance(word, k))))
            .get(); // get the closest match
}