计算URL中字母的出现次数

Question

I'm trying to count the occurrence of each letter from a URL.

I found this code, which seems to do the trick, but there are a few things I would hope to get explained.

1) I'm using a Norwegian alphabet so I need to add three more letters. I changed the array to 29, but it did not work.

2) Could you please explain to me what %c%7d\\n means?

01  import java.io.FileReader;
02  import java.io.IOException;
03   
04   
05  public class FrequencyAnalysis {
06      public static void main(String[] args) throws IOException {
07      FileReader reader = new FileReader("PlainTextDocument.txt");
08   
09      System.out.println("Letter Frequency");
10   
11      int nextChar;
12      char ch;
13   
14      // Declare 26 char counting
15      int[] count = new int[26];
16   
17      //Loop through the file char
18      while ((nextChar = reader.read()) != -1) {
19          ch = Character.toLowerCase((char) nextChar);
20   
21          if (ch >= 'a' && ch <= 'z')
22          count[ch - 'a']++;
23      }
24   
25      // Print out
26      for (int i = 0; i < 26; i++) {
27          System.out.printf("%c%7d\n", i + 'A', count[i]);
28      }
29   
30      reader.close();
31      }
32  }

Answer 1

You havent said how you checked for the 3 additional letters. It not enough to increase the size of the count array. You will need to account for the new characters unicode point values here. Chances are that the values are no longer conveniently sequentially ordered. In that case, you can use a Map<Integer, Integer> to store the frequencies.

%c is the format specifier for a unicode character. %7d is the specifier for integer with leftmost space padding. \\n is a newline character

Documented in the Formatter javadoc

Answer 2

An important thing here is that when you want to increment the number of occurences in your array, you are implicitly using the ASCII code of the characters :

//Here, ch is a char.
ch = Character.toLowerCase((char) nextChar);

  //I hate *if statements* without curly brackets but this is off-topic :)
  if (ch >= 'a' && ch <= 'z')

    /*
     * but here, ch is implicitly cast to an integer.
     * The int value of a char is its ASCII code.
     * for example, the value of 'a' is 97.
     * So if ch is 'a', (ch - 'a') = (97 - 97) = 0.
     * That's why you are incrementing count[0] in this case.
     *
     * Now, what happens if ch ='ø'? What is the ASCII code of ø?
     * Probably something quite high so that ch-'a' is probably out of bounds
     * but the size of your array is 26+3 only.
     *
     * EDIT : after a quick test, 'ø' = 248.
     *
     * This would work if norvegian characters had ASCII code between 98 and 100.
     */
     count[ch - 'a']++;

You should rewrite the algorithm using a HashMap<Character, Integer> instead.

//HashMap<Character, nb occurences of this character>
HashMap<Character, Integer> map = new HashMap<Character, Integer>();

while ((nextChar = reader.read()) != -1) {
  if(!map.containsKey(nextChar)) {
    map.put(nextChar, 0);
  }
  map.put(nextChar, map.get(nextChar)+1);
}