Java问题 怎么将将一个英文句子中的每一个单词的首字母转化为大写字母？急~~~~~！！！

Posted 2023-04-22

完全不在状态，这是我编的，一堆错误。。。。财富值没了，一共就6个。。。
Capitalized.java
public class Capitalized
char a = new char[13];
char[13] a = "T","o","d","a","y"," ","i","s"," ","s","u","n","n","y","!";

//构造方法
public Capitalized()



//get和set方法
public void setCapitalized(String[] a)

this.a = a;


public String getCapitalized()
return a;


public char printCapitalized()
int i;
for(i=0;i<=13;i++)
boolean Character.isSpaceChar(char[i] a);
if (char[i] a == " ")
i=i+1;
char Character.toUpperCase(char[i] a);





CapitalizedTest.java
import java.util.*;

public class CapitalizedTest
public static void main(String[] args)

char[13] a = "T","o","d","a","y"," ","i","s"," ","s","u","n","n","y","!";
String s = new String(a);
Capitalized c = new Capitalized();

System.out.println("Enter a string > Today is sunny!")

c.setCapitalized(s);
s = c.getCapitalized();
System.out.println(s);

//方法有很多种，这是你方法
//方法一
public class Capitalized
char[] cs = 'T', 'o', 'd', 'a', 'y', ' ', 'i', 's', ' ', 's', 'u', 'n',
'n', 'y', '!' ;

// 构造方法
public Capitalized()


public void printCapitalized()
int i;
for (i = 0; i <cs.length; i++)
if (Character.isSpaceChar(cs[i]))
i = i + 1;
cs[i]=Character.toUpperCase(cs[i]);



for (i = 0; i< cs.length; i++)
System.out.print(cs[i]);



public static void main(String[] args)
new Capitalized().printCapitalized();




//方法二

public class Capitalized

public static void main(String[] args)
String s="today is sunny!";
String arr[]=s.trim().split("\\s+");
if(s.length()>0)
for (int i = 0; i < arr.length; i++)
arr[i]=Character.toUpperCase(arr[i].charAt(0))+arr[i].substring(1);
System.out.print(arr[i]+" ");

参考资料：还有其他问题的话，给我发百度消息

参考技术A String s="today is sunny!";
public String toUppString(String str)

String arg[]=str.split(" ");
String s="";
for(int i=0;i<arg.length;i++)

s+=arg[i].substring(0,1).toUpperCase()+arg[i].substring(1);


return s;
参考技术B 根据要求写了一段小程序，希望对你有帮助

import java.util.*;

public class Capitalized

/*
* Capital one word
*/
public String capitalOneWord(String word)
if (word == null)
return null;

//get the upper case for first character, then add the rest
return word.substring(0, 1).toUpperCase() + word.substring(1);


/*
* Capital one sentence
*/
public String capitalOneSentence(String sentence)
if (sentence == null)
return null;

StringTokenizer st = new StringTokenizer(sentence);
StringBuffer result = new StringBuffer();

//devide sentence by word, process with capitalOneWord() once a time
while (st.hasMoreTokens())

result.append(capitalOneWord(st.nextToken())+" ");


return result.toString();


public static void main(String[] args)

String[] test = "Today is Sunndy, how aboout tomorrow?",
"How are you, darling",
"hi, sweat heart~",
"",
null
;

Capitalized tool = new Capitalized();

for(int i=0;i<test.length;i++)
System.out.println(tool.capitalOneSentence(test[i]));




参考技术C 你可以试试把这句话存到一个stringbuffer对象里，stringbuffer提供很多对字符串操作的方法

如果非要用char属于重复发明轮子 参考技术D printCapitalized()函数有问题啊。。。。
public char printCapitalized(String s)
char[] a=s.toCharArray();
for(int i=0;i<=13;i++)
if (a[i] == ' ')
i=i+1;
char Character.toUpperCase(a[i]);


刚开始的这两句也有问题，应该删掉：
char a = new char[13];
char[13] a = "T","o","d","a","y"," ","i","s"," ","s","u","n","n","y","!";
还有主函数里的这两句也对，可是多余的呀
char[13] a = "T","o","d","a","y"," ","i","s"," ","s","u","n","n","y","!";
String s = new String(a);
直接String s=“Today is sunny！”；

句子中的单词共现

【中文标题】句子中的单词共现

【英文标题】：Word cooccurence in sentences

【发布时间】：2016-06-26 05:52:37

【问题描述】：

我在一个文件中有大量句子（10,000 个）。该文件每个文件包含一个句子。在整个集合中，我想找出哪些单词在一个句子中一起出现以及它们的频率。

例句：

"Proposal 201 has been accepted by the Chief today.", 
"Proposal 214 and 221 are accepted, as per recent Chief decision",     
"This proposal has been accepted by the Chief.",
"Both proposal 3 MazerNo and patch 4 have been accepted by the Chief.",     
"Proposal 214, ValueMania, has been accepted by the Chief.";

我想对以下输出进行编码。我应该能够提供三个起始词作为程序的参数：“Chief, accepted, Proposal”

Chief accepted Proposal            5
Chief accepted Proposal has        3
Chief accepted Proposal has been   3

... 
...
for all combinations.

我知道组合可能很大。

我在网上搜索过，但没有找到。我已经编写了一些代码，但无法理解它。也许知道该域的人可能知道。

ReadFileLinesIntoArray rf = new ReadFileLinesIntoArray();

            try 
                String[] tmp = rf.readFromFile("c:/scripts/SelectedSentences.txt");
                for (String t : tmp)
                      String[] keys = t.split(" ");
                      String[] uniqueKeys;
                      int count = 0;
                      System.out.println(t);
                      uniqueKeys = getUniqueKeys(keys);
                        for(String key: uniqueKeys)
                        
                            if(null == key)
                            
                                break;
                                       
                            for(String s : keys)
                            
                                if(key.equals(s))
                                
                                    count++;
                                               
                            
                            System.out.println("Count of ["+key+"] is : "+count);
                            count=0;
                        
                
             catch (IOException e) 
                // TODO Auto-generated catch block
                e.printStackTrace();
            

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;
    

有人可以帮忙编码吗？

【问题讨论】：

是的，你会有非常大的排列集。您可以使用 Map（如 TreeMap）将地图中的键存储为唯一字符串，并将地图的值存储为计数。或者，您可以创建自己的小型数据结构来存储名称/值信息。

3 的输出对于 Chief、accepted、Proposal 意味着什么？这是否意味着有 3 个句子在句子中出现这 3 个单词？大小写重要吗？

抱歉“首席接受提案”应为 5，“已接受首席提案”应为 3...将编辑

@JonathanGrey：为什么“首席接受提案”的值为 5？

可能是您应该考虑的一个问题，您打算如何处理这些排列？取决于此，您是否真的需要生成所有这些？给定一组维度，您能否构建一个带有显式接收器/顶部节点的图并对其进行操作，边表示出现次数

【参考方案1】：

您可以应用标准信息检索数据结构，尤其是倒排索引。这是你的做法。

考虑您的原始句子。用一些整数标识符对它们进行编号，如下所示：

“201 号提案今天已被酋长接受。”, “根据最近的首席决定，接受提案 214 和 221”， “此提议已被酋长接受。”, “提案 3 MazerNo 和补丁 4 均已被酋长接受。”, “提议 214，ValueMania，已被酋长接受。”

对于您在句子中遇到的每一对单词，将其添加到倒排索引中，该倒排索引会将这对单词映射到一组句子标识符（一组唯一项）。对于一个长度为 N 的句子，有 N-choose-2 对。

适当的 Java 数据结构将是 Map<String, Map<String, Set<Integer>>。按字母顺序排列这些对，这样“has”和“Proposal”对将仅作为 ("has", "Proposal") 而不是 ("Proposal", "has") 出现。

此地图将包含以下内容：

"has", "Proposal" --> Set(1, 5)
"accepted", "Proposal" --> Set(1, 2, 5)
"accepted", "has" --> Set(1, 3, 5)
etc.

例如，单词对“has”和“Proposal”有一组 (1, 5)，表示它们出现在句子 1 和 5 中。

现在假设您要查找“accepted”、“has”和“Proposal”列表中单词的共现次数。从此列表中生成所有对并与它们各自的列表相交（使用 Java 的 Set.retainAll()）。这里的结果将最终设置为 (1, 5)。它的大小是2，意思是有两个句子包含“accepted”、“has”和“Proposal”。

要生成所有对，只需根据需要遍历您的地图。要生成大小为 N 的所有单词元组，您需要迭代并根据需要使用递归。