我已经编写了Java代码来对发生的总数进行计数。它使用2个.txt
文件作为输入,并给出单词和频率作为输出。
我也想打印,哪个文件包含一个给定的单词多少次。你有什么想法吗?
public class JavaApplication2
{
public static void main(String[] args) throws IOException
{
Path filePath1 = Paths.get("test.txt");
Path filePath2 = Paths.get("test2.txt");
Scanner readerL = new Scanner(filePath1);
Scanner readerR = new Scanner(filePath2);
String line1 = readerL.nextLine();
String line2 = readerR.nextLine();
String text = new String();
text=text.concat(line1).concat(line2);
String[] keys = text.split("[!.?:;\\s]");
String[] uniqueKeys;
int count = 0;
System.out.println(text);
uniqueKeys = getUniqueKeys(keys);
for(String key: uniqueKeys)
{
if(null == key)
{
break;
}
for(String s : keys)
{
if(key.equals(s))
{
count++;
}
}
System.out.println("["+key+"] frequency : "+count);
count=0;
}
}
private static String[] getUniqueKeys(String[] keys)
{
String[] uniqueKeys = new String[keys.length];
uniqueKeys[0] = keys[0];
int uniqueKeyIndex = 1;
boolean keyAlreadyExists = false;
for(int i=1; i<keys.length ; i++)
{
for(int j=0; j<=uniqueKeyIndex; j++)
{
if(keys[i].equals(uniqueKeys[j]))
{
keyAlreadyExists = true;
}
}
if(!keyAlreadyExists)
{
uniqueKeys[uniqueKeyIndex] = keys[i];
uniqueKeyIndex++;
}
keyAlreadyExists = false;
}
return uniqueKeys;
}
首先,不要将数组用于唯一键,而应使用HashMap<String, Integer>
。效率更高。
最好的选择是分别对每个行/文件运行处理,并分别存储这些计数。然后合并两个计数以获得总频率。
更多详情:
String[] keys = text.split("[!.?:;\\s]");
HashMap<String,Integer> uniqueKeys = new HashMap<>();
for(String key : keys){
if(uniqueKeys.containsKey(key)){
// if your keys is already in map, increment count of it
uniqueKeys.put(key, uniqueKeys.get(map) + 1);
}else{
// if it isn't in it, add it
uniqueKeys.put(key, 1);
}
}
// You now have the count of all unique keys in a given text
// To print them to console
for(Entry<String, Integer> keyCount : uniqueKeys.getEntrySet()){
System.out.println(keyCount.getKey() + ": " + keyCount.getValue());
}
// To merge, if you're using Java 8
for(Entry<String, Integer> keyEntry : uniqueKeys1.getEntrySet()){
uniqueKeys2.merge(keyEntry.getKey(), keyEntry.getValue(), Integer::add);
}
// To merge, otherwise
for(Entry<String, Integer> keyEntry : uniqueKeys1.getEntrySet()){
if(uniqueKeys2.containsKey()){
uniqueKeys2.put(keyEntry.getKey(),
uniqueKeys2.get(keyEntry.getKey()) + keyEntry.getValue());
}else{
uniqueKeys2.put(keyEntry.getKey(), keyEntry.getValue());
}
}
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