Java 如何从.txt文件中读取字符串,并根据出现的次数将其排序到ArrayList中?
我有一个程序可以读取一个.txt文件,创建一个包含每个唯一字符串及其出现次数的HashMap,我想创建一个ArrayList,以降序显示这些唯一字符串的出现次数 目前,我的程序从字母顺序的角度按降序排序(我假设使用ASCII值) 我如何根据它们的出现次数按降序排列 以下是代码的相关部分:Java 如何从.txt文件中读取字符串,并根据出现的次数将其排序到ArrayList中?,java,sorting,arraylist,hashmap,counter,Java,Sorting,Arraylist,Hashmap,Counter,我有一个程序可以读取一个.txt文件,创建一个包含每个唯一字符串及其出现次数的HashMap,我想创建一个ArrayList,以降序显示这些唯一字符串的出现次数 目前,我的程序从字母顺序的角度按降序排序(我假设使用ASCII值) 我如何根据它们的出现次数按降序排列 以下是代码的相关部分: Scanner in = new Scanner(new File("C:/Users/ahz9187/Desktop/counter.txt")); while
Scanner in = new Scanner(new File("C:/Users/ahz9187/Desktop/counter.txt"));
while(in.hasNext()){
String string = in.next();
//makes sure unique strings are not repeated - adds a new unit if new, updates the count if repeated
if(map.containsKey(string)){
Integer count = (Integer)map.get(string);
map.put(string, new Integer(count.intValue()+1));
} else{
map.put(string, new Integer(1));
}
}
System.out.println(map);
//places units of map into an arrayList which is then sorted
//Using ArrayList because length does not need to be designated - can take in the units of HashMap 'map' regardless of length
ArrayList arraylist = new ArrayList(map.keySet());
Collections.sort(arraylist); //this method sorts in ascending order
//Outputs the list in reverse alphabetical (or descending) order, case sensitive
for(int i = arraylist.size()-1; i >= 0; i--){
String key = (String)arraylist.get(i);
Integer count = (Integer)map.get(key);
System.out.println(key + " --> " + count);
}
您尚未显示地图的声明,但为了回答此问题,我假设您的地图声明如下:
Map<String,Integer> map = new HashMap<String,Integer>();
在Java 8中:
public static void main(final String[] args) throws IOException {
final Path path = Paths.get("C:", "Users", "ahz9187", "Desktop", "counter.txt");
try (final Stream<String> lines = Files.lines(path)) {
final Map<String, Integer> count = lines.
collect(HashMap::new, (m, v) -> m.merge(v, 1, Integer::sum), Map::putAll);
final List<String> ordered = count.entrySet().stream().
sorted((l, r) -> Integer.compare(l.getValue(), r.getValue())).
map(Entry::getKey).
collect(Collectors.toList());
ordered.forEach(System.out::println);
}
}
使用自定义的
比较器
。从HashMap
中获取出现次数并进行比较…并使用entrySet()而不是keySet(),这样您就拥有了唯一的字符串和出现次数。使用比较器将完成的HashMap中的所有项目推到一个列表中,就像Boris所说的那样。您需要的是一个最大堆,这样,一旦您将所有内容放入priqueue(按出现次数排序),您就可以poll()
,直到提取所有元素。旁注:Integer
支持+
运算符:map.put(string,map.get(string)+1)代码>@Boristespider你为什么不回答而不是评论?@sp00m,如果你坚持的话。这太完美了。非常感谢。
public static void main(final String[] args) throws IOException {
final Path path = Paths.get("C:", "Users", "ahz9187", "Desktop", "counter.txt");
try (final Stream<String> lines = Files.lines(path)) {
final Map<String, Integer> count = lines.
collect(HashMap::new, (m, v) -> m.merge(v, 1, Integer::sum), Map::putAll);
final List<String> ordered = count.entrySet().stream().
sorted((l, r) -> Integer.compare(l.getValue(), r.getValue())).
map(Entry::getKey).
collect(Collectors.toList());
ordered.forEach(System.out::println);
}
}
final Map<String, Integer> counts = /*from somewhere*/
final List<String> sorted = new ArrayList<>(counts.keySet());
Collections.sort(sorted, new Comparator<String>() {
@Override
public int compare(final String o1, final String o2) {
return counts.get(o1).compareTo(counts.get(o2));
}
});