java - 如何找到每个单词的第一个字母？

Question

我在哪里编辑这段代码以从我的输入文件中找到每个单词的第一个字母，保持频率和百分比，而不是每个字符？例如我在哪里可以实现 charAt(0) 或者我需要更改/添加什么？

import java.util.Scanner;
import java.io.File;
import java.io.FileNotFoundException;

public class FirstWordLetters
{
    public static void main(String[] args) throws FileNotFoundException
    {
        char[] capital = { 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J','K', 'L', 'M', 'N',
                'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};

        char[] small = { 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n',
                'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z' };


        //Input Scanner into the system        
        Scanner scan; 
        //2. Locate file using the scanner class - search to computer data location
        try {
            scan = new Scanner(new File("F:/programming principles/Programming Principles - PART B/enciphered.txt"));
        } catch (Exception e) { //throw exception e (meaning prevent any runtime errors)
            System.out.println("File not found");
            return;
        }
        //3. Set up int's (to count, and for the complete count.)
        //the aplhabet has 26 characters so the new int will be 26. The mem. space of the array.
        int[] count = new int[26];
        int completeTotal = 0;
        //4. Start scanning the system
        //Scan every line and notify user that each line has been read properly. 
        //each time line has been read store value in array and increment by one
        while(scan.hasNextLine()) {
            String line = scan.nextLine();
            System.out.println("Line read: " + line);
            char[] digit = line.toCharArray();
            for(int i = 0; i < digit.length; i++) {
                for(int j = 0; j < 26; j++) {

                    if(digit[i] == capital[j] || digit[i] == small[j]) {
                        count[j]++;
                        completeTotal = completeTotal + 1;
                        break;
                    }
                }
            }
        }
        //5. Display results
        //Print the overall data - letter, frequency and percentage.
        System.out.println("");
        System.out.println("First Word Count"); //notify user of what has been counted? (full count)
        for (int i = 0; i < 26; i++) //increment each letter frequency by one each time
        {
            System.out.print(" " + small[i]);
            System.out.print(" " + count[i]);
            //calculate and display percentage for the full count
            if (count[i] > 0)
                System.out.println(" " + (((float) count[i]/completeTotal)*100) + "%");
            else
                System.out.println(" 0%");

        }
    }    
} //end of source code.

Answer 1

这段代码可以做到。 inputString 是将整个文件存储为单个字符串。另请注意，如果任何单词以非字母字符开头，它将崩溃。此外，它假定每个单词都用空格或换行符分隔。

    int[] charCounts = new int[26];

    //Separate the string into individual words
    //The string " /n" tells it to look for spaces and newline characters "/n"
    StringTokenizer st = new StringTokenizer(inputString, " /n");

    //Loop until all the words are processed
    while (st.hasMoreTokens()) {

        //Select the next word
        String word = st.nextToken();

        //Convert the string to upper case so that lower case and upper case characters are represented the same
        word = word.toUpperCase();

        //Get the first character from the word
        char firstChar = word.charAt(0);

        //Convert the character to an integer representing it as an ASCII code
        int charCode = (int) firstChar;

        //Increment the count for that character by 1 
        //(We subtract 65 from the ASCII code because the array starts at 0 but 'A' is at 65)
        charCounts[charCode - 65]++;
    }

    //Obviously replace this section with whatever you like. It's just to show you how to get the values out again.
    for (int i = 0; i < charCounts.length; i++){
        System.out.println((char)(i + 65) + ": " + charCounts[i]);
    }

对于字符串“asduhasa ioajsdu ijqwjfoscn kopeurfc eqiwdfjs/nijoasdij iohwefscd aosd8ch”，输出:

A: 2
B: 0
C: 0
D: 0
E: 1
F: 0
G: 0
H: 0
I: 4
J: 0
K: 1
L: 0
M: 0
N: 0
O: 0
P: 0
Q: 0
R: 0
S: 0
T: 0
U: 0
V: 0
W: 0
X: 0
Y: 0
Z: 0

如果有什么不明白的，我会尽力解释。