print(frequency); // instead of print(list);
}
public static List<String> sort(List<String> list){
/** We use string compare to sort the list by length first then Collections.sort will sort it alphabetically **/
Collections.sort(list, new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
if (o1.length() < o2.length()) { /**Second string length is greater than first string length**/
return -1;
} else if (o1.length() > o2.length()) {/**Second string length is less than first string length**/
return 1;
} else { /** Equal **/
return o1.compareTo(o2);
}
}
});
return list;
}
/** Simply prints the length and sorted words. Sorted by their length first then alphapetically. **/
public static void print(Map<String, Integer> frequency) {
frequency.forEach((word, freq) -> System.out.printf("%d - %w: %d%n", word.length(), word, freq));
}
}
}
/** Counts how many words **/
public static void count(SimpleCharacterReader stream) {
try {
while (true) {
a = getReader(stream);
/** When a space, next line or tab is met, we assume a word was met. **/
if ((a == ' ') || (a == '\n') || (a == '\t')) {
count++;
}
}
} catch (EOFException eof) {
stream.Dispose();
}
arr = new String[count];
}
/** Gets the characters **/
public static char getReader(ICharacterReader reader) throws EOFException {
return reader.GetNextChar();
}
/** Store unsorted words in array **/
public static void store(ICharacterReader info) throws EOFException {
int i = 0;
while (i < count) {
s = info.GetNextChar();
if (Character.isLetterOrDigit(s)) {
word += Character.toString(s);
} else if (s == ' ' || s == '\n' || s == '\t') {
arr[i++] = word;
word = "";
}
}
}
}
我的任务是创建一个字符阅读器，它从项目文件中读取并创建一个按长度排序的词频图表，如果它们的长度相同，则按字母顺序排列。这是我到目前为止的代码，但我的输出有重复的单词并且没有频率？怎么了？谢谢！我不确定我哪里出错了。 Isimplecharacter 和 Character 是我得到的文件。
我的代码是最后一个
import java.io.EOFException;
import java.util.Random;
public class SimpleCharacterReader implements ICharacterReader {
private int m_Pos = 0;
public static final char lf = '\n';
private String m_Content = "It was the best of times, it was the worst of times," +
lf +
"it was the age of wisdom, it was the age of foolishness," +
lf +
"it was the epoch of belief, it was the epoch of incredulity," +
lf +
"it was the season of Light, it was the season of Darkness," +
lf +
"it was the spring of hope, it was the winter of despair," +
lf +
"we had everything before us, we had nothing before us," +
lf +
"we were all going direct to Heaven, we were all going direct" +
lf +
"the other way--in short, the period was so far like the present" +
lf +
"period, that some of its noisiest authorities insisted on its" +
lf +
"being received, for good or for evil, in the superlative degree" +
lf +
"of comparison only." +
lf +
"There were a king with a large jaw and a queen with a plain face," +
lf +
"on the throne of England; there were a king with a large jaw and" +
lf +
"a queen with a fair face, on the throne of France. In both" +
lf +
"countries it was clearer than crystal to the lords of the State" +
lf +
"preserves of loaves and fishes, that things in general were" +
lf +
"settled for ever";
Random m_Rnd = new Random();
public char GetNextChar() throws EOFException {
if (m_Pos >= m_Content.length()) {
throw new EOFException();
}
return m_Content.charAt(m_Pos++);
}
public void Dispose() {
// Do nothing
}
}
import java.io.EOFException;
public interface ICharacterReader {
char GetNextChar() throws EOFException;
void Dispose();
}
import java.lang.Character;
import java.io.EOFException;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
public class Analyse {
static String arr[]; /** Store unsorted words in array**/
public static int count = 0;
static char a, s;
static String word = "";
public static void main(String[] args) throws EOFException {
SimpleCharacterReader stream = new SimpleCharacterReader();
List<String> list = new ArrayList<>();
String str = "";
/** Need to know how may words we have**/
count(stream);
SimpleCharacterReader st = new SimpleCharacterReader();
store(st);
/**Store valid words in list to sort later**/
for (int i = 0; i < arr.length; i++) {
if (arr[i] != " " && arr[i] != null) {
str = arr[i];
}
list.add(str);
}
sort(list); /**Sort list**/
print(list);/**Print list**/
}
public static List<String> sort(List<String> list){
/** We use string compare to sort the list by length first then Collections.sort will sort it alphabetically **/
Collections.sort(list, new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
if (o1.length() < o2.length()) { /**Second string length is greater than first string length**/
return -1;
} else if (o1.length() > o2.length()) {/**Second string length is less than first string length**/
return 1;
} else { /** Equal **/
return o1.compareTo(o2);
}
}
});
return list;
}
/** Simply prints the length and sorted words. Sorted by their length first then alphapetically. **/
public static void print(List<String> list) {
for (int i = 0; i < arr.length; i++) {
System.out.println(list.get(i).length() + " - " + list.get(i));
}
}
/** Counts how many words **/
public static void count(SimpleCharacterReader stream) {
try {
while (true) {
a = getReader(stream);
/** When a space, next line or tab is met, we assume a word was met. **/
if ((a == ' ') || (a == '\n') || (a == '\t')) {
count++;
}
}
} catch (EOFException eof) {
stream.Dispose();
}
arr = new String[count];
}
/** Gets the characters **/
public static char getReader(ICharacterReader reader) throws EOFException {
return reader.GetNextChar();
}
/** Store unsorted words in array **/
public static void store(ICharacterReader info) throws EOFException {
int i = 0;
while (i < count) {
s = info.GetNextChar();
if (Character.isLetterOrDigit(s)) {
word += Character.toString(s);
} else if (s == ' ' || s == '\n' || s == '\t') {
arr[i++] = word;
word = "";
}
}
}
}
答案 0 :(得分：1)
您需要将单词收集到一个 TreeMap 中，该 arr 支持有序键并且可能会提供一个 custom comparator via constructor: public TreeMap(Comparator<? super K> comparator)。
每个单词的频率将存储为一个值。可以使用 Map::merge 函数累积。
假设所有的词都被方法store读入数组// class Analyse, method main
store(st);
/**Store valid words in a sorted map and count word frequency **/
// create TreeMap with custom Comparator as a lambda
Map<String, Integer> frequency = new TreeMap<>((s1, s2) -> Integer.compare(s1.length(), s2.length()) == 0 ? s1.compareTo(s2) : Integer.compare(s1.length(), s2.length()));
for (String word : arr) {
if (null != word && !" ".equals(word)) {
frequency.merge(word, 1, (acc, one) -> acc + one);
}
}
print(frequency); // instead of print(list);
，代码可以更新如下：
print接下来，需要更新方法 public static void print(Map<String, Integer> frequency) {
frequency.forEach((word, freq) -> System.out.printf("%d - %w: %d%n", word.length(), word, freq));
}
以处理地图而不是列表：
Comparator.comparing(String::length).thenComparing(String::compareTo)此外，可以使用 Stream API 使用构建为 Collectors.toMap 链的自定义比较器对单词进行排序，然后使用 LinkedHashMap 将频率收集到保持插入的 // class Analyse, method main
store(st);
Map<String, Integer> frequency = Arrays
.stream(arr) // Stream<String>
.filter(word -> null != word && !" ".equals(word))
.sorted( // sort
Comparator.comparingInt(String::length)
.thenComparing(String::compareTo) // or compareToIgnoreCase if needed
)
.collect(Collectors.toMap(
word -> word, // use word as a key
word -> 1, // 1 as initial value
Integer::sum, // merge function to count frequency
LinkedHashMap::new // maintain insertion order (by sorted keys)
));
print(frequency); // instead of print(list);
中顺序：
{{1}}