Using an array, reading from a file

Array Tutorials

Accessing Arrays

Print part of the array

Create the array

Use the array

Reading from a file

In these tutorials you are passed a token stream containing numbers.

• When the tokens run out then token.nextToken() returns the value StreamTokenizer.TT_EOF
• You can read the current token as a double using token.nval
• to change this to an integer you can cast as an int using (int)token.nval

There are three files used in these examples:

array01.txt: 1 1 1 5 9 9
array02.txt: 3 1 4 1 5
array03.txt: 2 4 6 1 2 2 6

Print each digit in the file

Add the following line inside the while loop, after the declaration of n.

    System.out.println(n);

to see each number in the file.

.

import java.io.FileReader; import java.io.StreamTokenizer; class P{ static void doStuff(StreamTokenizer token)throws Exception{ while (token.nextToken() != StreamTokenizer.TT_EOF){ int n = (int)token.nval; } } public static void main(String [] argv)throws Exception{ doStuff(new StreamTokenizer(new FileReader("array01.txt"))); } }

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The following questions are the same as the previous tutorial - it's just that you are reading data from a file.

Show the frequency of each digit

You can create an array with ten items - one for each digit (you may be sure that all numbers in x are between 0 and 9).

if x is the list [1,1,1,5,9,9]
then the result will be

freq
0      0
1      3
2      0
3      0
4      0
5      1
6      0
7      0
8      0
9      2

Print the frequency of each digit

Hint

Replace

freq[n] = 999;

with

freq[n]++;

.

import java.io.FileReader; import java.io.StreamTokenizer; class P{ static void doStuff(StreamTokenizer token)throws Exception{ int [] freq = new int[10]; while (token.nextToken() != StreamTokenizer.TT_EOF){ int n = (int)token.nval; freq[n]=999; } for (int i=0;i<freq.length;i++) System.out.println(i + " " + freq[i]); } public static void main(String [] argv)throws Exception{ for (String file:new String[] {"array01.txt","array02.txt","array03.txt"}){ System.out.println("Processing file: "+file); doStuff(new StreamTokenizer(new FileReader(file))); System.out.println(); } } }

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Show the number of different digits

If x is the list [1,1,1,5,9,9] then the result will be 3 because there are three different number (namely 1,5 and 9)

You can use the freq array created as above - then just count the number of non-zero elements using an accumulator:

 int acc = 0;
 for (int i=0;i<freq.length;i++)
   if (freq[i]>0)
     acc++;
 System.out.println(acc);

Print the number of different digits

.

import java.io.FileReader; import java.io.StreamTokenizer; class P{ static void doStuff(StreamTokenizer token)throws Exception{ int [] freq = new int[10]; while (token.nextToken() != StreamTokenizer.TT_EOF){ int n = (int)token.nval; freq[n]++; } for (int i=0;i<freq.length;i++) System.out.println(i + " " + freq[i]); } public static void main(String [] argv)throws Exception{ for (String file:new String[] {"array01.txt","array02.txt","array03.txt"}){ System.out.println("Processing file: "+file); doStuff(new StreamTokenizer(new FileReader(file))); System.out.println(); } } }

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Show the repeated digits

If x is the list [1,1,1,5,9,9] then the result will be 1 and 9 because 1 occurs more than once and 9 occurs twice.

You can use the freq array created as above - then just show the index when the freq is more than 1:

 for (int i=0;i<freq.length;i++)
   if (freq[i]>1)
     System.out.println(i);

Print the repeated digits

.

import java.io.FileReader; import java.io.StreamTokenizer; class P{ static void doStuff(StreamTokenizer token)throws Exception{ int [] freq = new int[10]; while (token.nextToken()!=StreamTokenizer.TT_EOF){ int n = (int)token.nval; freq[n]++; } for (int i=0;i<freq.length;i++) System.out.println(i + " " + freq[i]); } public static void main(String [] argv)throws Exception{ for (String file:new String[] {"array01.txt","array02.txt","array03.txt"}){ System.out.println("Processing file: "+file); doStuff(new StreamTokenizer(new FileReader(file))); System.out.println(); } } }

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test text

 

 

 

 

Show the missing digits

If x is the list [1,1,1,5,9,9] then the result will be 2 3 4 6 7 8 because these digits are not in the input list.

You can use the freq array created as above - then just print the indexes when the freq is 0:

  for (int i=0;i<freq.length;i++)
    if (freq[i]==0)
      System.out.println(i);

Print the missing digits

.

import java.io.FileReader; import java.io.StreamTokenizer; class P{ static void doStuff(StreamTokenizer token)throws Exception{ int [] freq = new int[10]; while (token.nextToken()!=StreamTokenizer.TT_EOF){ int n = (int)token.nval; freq[n]++; } for (int i=0;i<freq.length;i++) System.out.println(i + " " + freq[i]); } public static void main(String [] argv)throws Exception{ for (String file:new String[] {"array01.txt","array02.txt","array03.txt"}){ System.out.println("Processing file: "+file); doStuff(new StreamTokenizer(new FileReader(file))); System.out.println(); } } }

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