从文本中查找前 10 个单词：95% 的工作代码 [关闭]答案

【问题标题】：Find TOP 10 words from text: 95%-working code [closed]从文本中查找前 10 个单词：95% 的工作代码 [关闭]
【发布时间】：2013-06-16 18:38:29
【问题描述】：

我几个月前开始学习 Java。对于我的家庭作业，我编写了这段代码 - 从文本文件中查找前 10 个单词。对我来说这是非常有趣的任务。而且很难。第三天我无法完成这个软件——我发现我的代码很难导航。我认为，这项任务可以用更少的代码和更高的性能来完成。但我的一般问题 - 我无法修复下面的代码块之一。

总而言之：我读取文件，剪切一些无意义的单词，将文本中的所有单词添加到数组中（不重复），并且并行地我有第二个数组来计算单词的频率。还有我的问题 - 查找十个最流行单词的代码块。我所需要的 - 从 freq[] 中找到最大的数字，而不会丢失这些数字的索引，因为 words[9] == freq[9] - 九个位置的单词在 freq[] 的九个位置有频率数。

package CountWords21;

import java.io.IOException;
import java.nio.charset.StandardCharsets;

public class CountWords21 {
    public static void main(String[] args) throws IOException {

    String textSource = ReadFromFileToString.readFile("/home/vitaly/Desktop/text", StandardCharsets.UTF_8);
    int start = 0;
    int end = 0;
    int w = 0; // counter for array of words - position for add new word in array
    StringBuilder textStringBuilder = new StringBuilder(textSource);
    for (int i = 0; i < textSource.length(); i++) {
        if (textSource.charAt(i) == ','
                || textSource.charAt(i) == '.'
                || textSource.charAt(i) == '!'
                || textSource.charAt(i) == ';'
                || textSource.charAt(i) == '\''
                || textSource.charAt(i) == '-'
                || textSource.charAt(i) == '+'
                || textSource.charAt(i) == '*'
                || textSource.charAt(i) == '/'
                || textSource.charAt(i) == '('
                || textSource.charAt(i) == '>'
                || textSource.charAt(i) == ':'
                || textSource.charAt(i) == '\n' // new line
                || (textSource.charAt(i) == 'a'
                && textSource.charAt(i - 1) == ' ' 
                && textSource.charAt(i + 1) == ' '))
            textStringBuilder.setCharAt(i, ' ');
        if (textSource.charAt(i) == '\'' 
                && textSource.charAt(i + 1) == 's') {
            textStringBuilder.setCharAt(i, ' '); // 's
            textStringBuilder.setCharAt(i + 1, ' ');
        }
        if (textSource.charAt(i) == 'a' && textSource.charAt(i + 1) == 's'
                && textSource.charAt(i + 2) == ' '
                && textSource.charAt(i - 1) == ' ' && i > 2) {
            textStringBuilder.setCharAt(i, ' '); // as
            textStringBuilder.setCharAt(i + 1, ' ');
        }
        if ((textSource.charAt(i) == 't' || textSource.charAt(i) == 'T')&& textSource.charAt(i + 1) == 'h'
                && textSource.charAt(i + 2) == 'e'
                && textSource.charAt(i - 1) == ' '
                && textSource.charAt(i + 3) == ' ' && i > 3) {
            textStringBuilder.setCharAt(i, ' '); // the
            textStringBuilder.setCharAt(i + 1, ' ');
            textStringBuilder.setCharAt(i + 2, ' ');
        }
        if (           textSource.charAt(i) == 'a' // and
                && textSource.charAt(i + 1) == 'n'
                && textSource.charAt(i + 2) == 'd'
                && (textSource.charAt(i - 1) == ' ' || textSource.charAt(i - 1) == '\n' 
                && textSource.charAt(i + 3) == ' ' && i > 3)) {
            textStringBuilder.setCharAt(i, ' '); 
            textStringBuilder.setCharAt(i + 1, ' ');
            textStringBuilder.setCharAt(i + 2, ' ');
        }
        if ((textSource.charAt(i) == 'f' || i < (textSource.length() - 4)) // for
                && textSource.charAt(i + 1) == 'o'
                && textSource.charAt(i + 2) == 'r'
                && (i == 0 || textSource.charAt(i - 1) == ' ' || textSource.charAt(i - 1) == '\n')
                && textSource.charAt(i + 3) == ' ') {
            textStringBuilder.setCharAt(i, ' '); 
            textStringBuilder.setCharAt(i + 1, ' ');
            textStringBuilder.setCharAt(i + 2, ' ');
        }
        if (textSource.charAt(i) == 'o' && textSource.charAt(i + 1) == 'f'
                && textSource.charAt(i - 1) == ' '
                && textSource.charAt(i + 2) == ' ' && i > 2) {
            textStringBuilder.setCharAt(i, ' '); // of
            textStringBuilder.setCharAt(i + 1, ' ');
            textStringBuilder.setCharAt(i + 2, ' ');
        }
        if (textSource.charAt(i) == 'i' && textSource.charAt(i + 1) == 's'
                && textSource.charAt(i + 2) == ' '
                && textSource.charAt(i - 1) == ' ' && i > 2) {
            textStringBuilder.setCharAt(i, ' '); // is
            textStringBuilder.setCharAt(i + 1, ' ');
        }
        if (textSource.charAt(i) == 't' && textSource.charAt(i + 1) == 'o'
                && textSource.charAt(i + 2) == ' '
                && textSource.charAt(i - 1) == ' ' && i > 2) {
            textStringBuilder.setCharAt(i, ' '); // to
            textStringBuilder.setCharAt(i + 1, ' ');
        }
        if (textSource.charAt(i) == 'i' && textSource.charAt(i + 1) == 'n'
                && textSource.charAt(i + 2) == ' '
                && textSource.charAt(i - 1) == ' ' && i > 2) {
            textStringBuilder.setCharAt(i, ' '); // in
            textStringBuilder.setCharAt(i + 1, ' ');
        }
        if ((textSource.charAt(i) == 't' || textSource.charAt(i) == 'T')
                && textSource.charAt(i + 1) == 'h'
                && textSource.charAt(i + 2) == 'i'
                && textSource.charAt(i + 3) == 's'
                && textSource.charAt(i + 4) == ' '
                && (i == 0 || textSource.charAt(i - 1) == ' ')) {
            textStringBuilder.setCharAt(i, ' '); // this
            textStringBuilder.setCharAt(i + 1, ' ');
            textStringBuilder.setCharAt(i + 2, ' ');
            textStringBuilder.setCharAt(i + 3, ' ');
        }
        if ((textSource.charAt(i) == 't' || textSource.charAt(i) == 'T')
                && textSource.charAt(i + 1) == 'h'
                && textSource.charAt(i + 2) == 'a'
                && textSource.charAt(i + 3) == 't'
                && textSource.charAt(i + 4) == ' '
                && (i == 0 || textSource.charAt(i - 1) == ' ')) {
            textStringBuilder.setCharAt(i, ' '); // that
            textStringBuilder.setCharAt(i + 1, ' ');
            textStringBuilder.setCharAt(i + 2, ' ');
            textStringBuilder.setCharAt(i + 3, ' ');
        }
        if ((textSource.charAt(i) == 'm' || textSource.charAt(i) == 'M')
                && textSource.charAt(i + 1) == 'o'
                && textSource.charAt(i + 2) == 's'
                && textSource.charAt(i + 3) == 't'
                && textSource.charAt(i + 4) == ' '
                && (i == 0 || textSource.charAt(i - 1) == ' ')) {
            textStringBuilder.setCharAt(i, ' '); // this
            textStringBuilder.setCharAt(i + 1, ' ');
            textStringBuilder.setCharAt(i + 2, ' ');
            textStringBuilder.setCharAt(i + 3, ' ');
        }
    }
    textSource = textStringBuilder.toString();
    textSource = textSource.trim();

    String[] words = new String[textSource.length() / 2];
    int[] freq = new int[textSource.length() / 2]; 
    // number of usage - index the same for words[]

    while (end < textSource.length()) {
        // global search for next word
        while (end != -1) { // -1 will be if string finish
            end = textSource.indexOf(' ', end);
            // search for space, start from end 
            // (last founded space position)
            if (end != -1) // when space
                break; // because we already have end position
            else { // (end == -1) -- finish of sourceText - without space on end
                end = textSource.length();
                break;
            }
        }
        if (words[w] != null) w++; // or last word in array will be replaced
        if (end != start) { 
            words[w] = (textSource.substring(start, end)).trim(); // add word to array
            freq[w] = 1;
        }
        if (end == start) w--;

        if (w == 0) { // first word
            freq[0] = 1;
            w = 1; // next word will be added in next position
        }

        if (words[1] != null) // if it not for the first word
            searchDuplicate: for (int i = 0; i <= w - 1; i++) {
                // search word in array for duplicate
                if (words[w].compareToIgnoreCase(words[i]) == 0) {
                    // if new word equal to one of word from array
                    freq[i] += 1; 
                    // increment count of this word or set 1 if first added
                    words[w] = null;
                    freq[w] = 0;
                    // clear duplicated new (last added) word
                    break searchDuplicate; // exit from for-search-iterator
                } // if we can't find duplicate and this is new word
            }
        if (w != 1 && words[w] != null && words[w].compareToIgnoreCase(words[w - 1]) != 0 && freq[w] == 0) {
            freq[w] += 1; // new word achieve first 1.
            w += 1; // next time add new word in next position
        }
        start = end + 1; 
        // start next word extract from first symbol after space
        end += 1; 
        // without this line end == current space end 
        // indexOf will find this current space
    }

    int[] PopularWordsIndexes = new int[10]; // indexes of max used words
    for (int index = 0, minFreqIndexTemp = 1; freq[index] != 0; index++) { // max frequency
        if (index < PopularWordsIndexes.length) 
            PopularWordsIndexes[index] = index;
        // add first words in empty TOP

        else { // TODO error somewhere here
            /* if we don't have empty position in PopularWordsIndexes
             find the lowest freq index and replace with higher index */
            for (int top10IndexIterator = 0; top10IndexIterator < PopularWordsIndexes.length; top10IndexIterator++)
                if (freq[PopularWordsIndexes[minFreqIndexTemp]] < freq[PopularWordsIndexes[top10IndexIterator]])
                    minFreqIndexTemp = top10IndexIterator;
            if (freq[PopularWordsIndexes[minFreqIndexTemp]] < freq[index])
                PopularWordsIndexes[minFreqIndexTemp] = index;

        }
    }
        // output
        for (int i2 = 0; i2 < PopularWordsIndexes.length; i2++) {
            System.out.println(words[PopularWordsIndexes[i2]] + " : "
                    + freq[PopularWordsIndexes[i2]]);
        }

}
}

【问题讨论】：

听说过 HashMaps 吗？
哇。即使不阅读代码，我也可以告诉您走错了路！如果您不想查找新单词（或从文件中加载感兴趣的单词列表）会怎样？
@tejas 不，今天我不熟悉 HashMaps。去阅读这个。

标签： java arrays sorting text statistics

【解决方案1】：

尝试使用 HashMap，其中键作为文本，值作为频率计数。与使用数组相比，将帮助您减少运行时间。

【讨论】：

【解决方案2】：

你可能想看看这个Java Strings，对你特别有用的是方法

substring()   // get a string of x length rather than so many charAt(i)
toLowerCase() // to ignore capitals
contains()    // to see if the string you have contains the string you dont want
equals()      // to see if the string you have is the same the string you dont want
split()       // split into an array based on a expression, in you case I assume a space

此外，如果您使用 Map 来存储字符串，则无需担心重复。

【讨论】：