Hi I need to calculate the entropy of order m of a file where m is the number of bit (m <= 16).
So:
H_m(X)=-sum_i=0 to i=2^m-1{(p_i,m)(log_2 (p_i,m))}
So, I thought to create an input stream to read the file and then calculate the probability of each sequence composed by m bit.
For m = 8 it's easy because I consider a byte. Since that m<=16 I tought to consider as primitive type short, save each short of the file in an array short[] and then manipulate bits using bitwise operators to obtain all the sequences of m bit in the file. Is this a good idea?
Anyway, I'm not able to create a stream of short. This is what I've done:
public static void main(String[] args) {
readFile(FILE_NAME_INPUT);
}
public static void readFile(String filename) {
short[] buffer = null;
File a_file = new File(filename);
try {
File file = new File(filename);
FileInputStream fis = new FileInputStream(filename);
DataInputStream dis = new DataInputStream(fis);
int length = (int)file.length() / 2;
buffer = new short[length];
int count = 0;
while(dis.available() > 0 && count < length) {
buffer[count] = dis.readShort();
count++;
}
System.out.println("length=" + length);
System.out.println("count=" + count);
for(int i = 0; i < buffer.length; i++) {
System.out.println("buffer[" + i + "]: " + buffer[i]);
}
fis.close();
}
catch(EOFException eof) {
System.out.println("EOFException: " + eof);
}
catch(FileNotFoundException fe) {
System.out.println("FileNotFoundException: " + fe);
}
catch(IOException ioe) {
System.out.println("IOException: " + ioe);
}
}
But I lose a byte and I don't think this is the best way to proced.
This is what I think to do using bitwise operator:
int[] list = new int[l];
foreach n in buffer {
for(int i = 16 - m; i > 0; i-m) {
list.add( (n >> i) & 2^m-1 );
}
}
I'm assuming in this case to use shorts. If I use bytes, how can I do a cycle like that for m > 8? That cycle doesn't work because I have to concatenate multiple bytes and each time varying the number of bits to be joined..
Any ideas? Thanks
I think you just need to have a byte array:
public static void readFile(String filename) {
ByteArrayOutputStream outputStream=new ByteArrayOutputStream();
try {
FileInputStream fis = new FileInputStream(filename);
byte b=0;
while((b=fis.read())!=-1) {
outputStream.write(b);
}
byte[] byteData=outputStream.toByteArray();
fis.close();
}
catch(IOException ioe) {
System.out.println("IOException: " + ioe);
}
Then you can manipulate byteData as per your bitwise operations.
--
If you want to work with shorts you can combine bytes read this way
short[] buffer=new short[(int)(byteData.length/2.)+1];
j=0;
for(i=0; i<byteData.length-1; i+=2) {
buffer[j]=(short)((byteData[i]<<8)|byteData[i+1]);
j++;
}
To check for odd bytes do this
if((byteData.length%2)==1) last=(short)((0x00<<8)|byteData[byteData.length-1]]);
last is a short so it could be placed in buffer[buffer.length-1]; I'm not sure if that last position in buffer is available or occupied; I think it is but you need to check j after exiting the loop; if j's value is buffer.length-1 then it is available; otherwise might be some problem.
Then manipulate buffer.
The second approach with working with bytes is more involved. It's a question of its own. So try this above.