编码行数：850
编码时长：13h
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所学知识点：
统计文章中各单词出现的频率，并去掉常用单词。问题解决基本思路：先将所要读取的文件打开，先定义一个fileReader对象，用来初始化BufferedReader，new一个BufferedReader对象，将文件内容读取到缓存，定义一个字符串缓存，将字符串存放缓存中。然后按行读取，每一行读取的内容用append方法放在字符串缓存后边，再用toString()将其转化为字符串，根据".",",","?","!",":","‘","’","“","”","—",";","-"等将他们替换成空格，在根据空格分割成一个个单词放进字符串数组中，这里的方法用到Map方法，Map<String, Integer> map = new TreeMap<String, Integer>();两个参数分别代表出现的单词和其出现的次数。用for循环遍历单词数组，先统一转化成小写，在调用自定义方法judgeNouse（）去掉常见单词，用map.get方法看其出现的次数，如果为0则不重复出现则其count记为1，否则为重复出现，其count++；并
class entity11{//定义单词的实体类，便于list对象集合操作
String danci;
int cishu;
public entity11(String zimu,int cishu) {
this.danci = zimu;
this.cishu = cishu;
}
public String getDanci() {
return danci;
}
public int getCishu() {
return cishu;
}
}
public class DanciCollect {
public static boolean judgeNouse(String str) throws IOException {//去掉常用单词的方法
boolean flag = true;
FileReader fr = new FileReader("judge.txt");
BufferedReader bf = new BufferedReader(fr);
String str1; // 按行读取字符串
while ((str1 = bf.readLine()) != null) {
if(str.equals(str1)) {
flag = false;
}
}
bf.close();
fr.close();
return flag;
}
public static String toLowerCase(String str) {//将字符串转化成小写
char []StringArr = str.toCharArray();
for (int i = 0; i < StringArr.length; i++) {
StringArr[i] = Character.toLowerCase(StringArr[i]);
}
StringBuffer sb = new StringBuffer();
for(int i = 0;i < StringArr.length;i++) {
sb.append(StringArr[i]);
}
String str1 = sb.toString();
return str1;
}
public static void collect1() throws IOException {
try {
File file1 = new File("piao.txt");//定义一个file对象，用来初始化FileReader
FileReader reader1 = new FileReader(file1);//定义一个fileReader对象，用来初始化BufferedReader
BufferedReader bReader1 = new BufferedReader(reader1);//new一个BufferedReader对象，将文件内容读取到缓存
StringBuilder sb1 = new StringBuilder();//定义一个字符串缓存，将字符串存放缓存中
String s1 = "";
while ((s1 =bReader1.readLine()) != null) {//逐行读取文件内容，不读取换行符和末尾的空格
sb1.append(s1);//将读取的字符串添加换行符后累加p存放在缓存中
}
bReader1.close();
String text = sb1.toString();
int i=0;
String[] array = {".",",","?","!",":","‘","’","“","”","—",";","-"};
for (int j = 0; j < array.length; j++) {
text = text.replace(array[j]," "); //将text中的array数组中包含的特殊字符用空格代替
}
String[] textArray = text.split(" "); //根据空格将text分割并存放在textArray中
Map<String, Integer> map = new TreeMap<String, Integer>();
Integer count = 0;//每个字母的次数
Integer total = 0;//总共多少个字母
while(i < textArray.length) {
String str = toLowerCase(textArray[i]);
if(!judgeNouse(str)) {
total = total + 1;
count = map.get(str+"");
if (count == null) {//单词没有出现重复;
count = 1;
}else {//单词出现重复，count+1;
count++;
}
map.put(str+"", count);
i++;
}
else {
i++;
}
}
List<entity11> result = new ArrayList<>();
entity11 e = null;
for (String str : map.keySet()) {
e = new entity11(str,map.get(str));
result.add(e);
}
result.sort((entity11 e1,entity11 e2)->{
return e2.getCishu()-e1.getCishu();});
System.out.println("文章共计"+total+"个单词");
for(int ii = 0 ; ii < result.size();ii++) {
System.out.println(result.get(ii).getDanci()+"在文章中出现"+result.get(ii).getCishu()+"次,其频率为"+String.format("%.2f",result.get(ii).getCishu()*1.0/total));
}
// for(entity11 ee : result) {
// System.out.println("单词"+ee.getDanci()+"在文章中出现"+ee.getCishu()+"次,其频率为"+String.format("%.2f",ee.getCishu()*1.0/total));
// }
// for (String str : map.keySet()) {
// System.out.println("字母"+str+"出现"+map.get(str)+"次,其频率为:"+String.format("%.2f",map.get(str)*1.0/total));
// }
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
public static void main(String args[]) throws IOException {
try {
DanciCollect dancicollect = new DanciCollect();
DanciCollect.collect1();
}catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
且将其单词本身字符串和出现次数count赋值给map对象map.put（）；最后将其转化成List集合，根据出现频率排序输出。源代码如下：