Java8流，以确定文本文件中的最大计数

Question

String filename = "SophieSallyJack.txt";
if (args.length == 1) {
    filename = args[0];
}
Map<String, Integer> wordFrequency = new TreeMap<>();

List<String> incoming = Utilities.readAFile(filename);

wordFrequency = incoming.stream()
    .map(String::toLowerCase)
    .filter(word -> !word.trim().isEmpty())
    .collect(Collectors.toMap(word -> word, word -> 1, (a, b) -> a + b, TreeMap::new));                

int maxCnt = 0;

// TODO add a single statement that uses streams to determine maxCnt
for (String word : incoming) {
    Integer cnt = wordFrequency.get(word);
    if (cnt != null) {
        if (cnt > maxCnt) {
            maxCnt = cnt;
        }
    }
}
System.out.print("Words that appear " + maxCnt + " times:");

wordFrequency = incoming.parallelStream().
    collect(Collectors.toConcurrentMap(w -> w, w -> 1, Integer::sum));

但这是不对的，我不确定如何将maxCnt合并到流中。

Answer 1

假设您已经从列表 中的文件提取了所有单词，可以使用这种方法计算每个单词的单词计数，

Map<String, Long> wordToCountMap = words.stream()
                .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

然后可以使用上面的映射计算最自由的单词，如下所示，

Entry<String, Long> mostFreequentWord = wordToCountMap.entrySet().stream()
    .max(Map.Entry.comparingByValue())
    .orElse(new AbstractMap.SimpleEntry<>("Invalid", 0l));

如果您愿意，您可以将上述两个管道一起更改，

Entry<String, Long> mostFreequentWord = words.stream()
    .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()))
    .entrySet().stream()
    .max(Map.Entry.comparingByValue())
    .orElse(new AbstractMap.SimpleEntry<>("Invalid", 0l));

Optional<Entry<String, Long>> mostFreequentWord = words.stream()
    .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()))
    .entrySet().stream()
    .max(Map.Entry.comparingByValue());