So, I am trying to create 2 randomly generated arrays,(a, and b, each with 10 unique whole numbers from 0 to 20), and then creating 2 arrays with the info of the last two. One containing the numbers that appear in both a and b, and another with the numbers that are unique to a and to b. The arrays must be listed in a "a -> [1, 2, 3,...]" format. At the moment I only know how to generate the 2 arrays, and am currently at the Intersection part. The problem is, that I can create a array with the correct list of numbers, but it will have the same length of the other two, and the spaces where it shouldn't have anything, it will be filled with 0s when its supposed to create a smaller array with only the right numbers.
package tps.tp1.pack2Arrays;
public class P02ArraysExtractUniqsAndReps {
public static void main(String[] args) {
int nbr = 10;
int min = 0;
int max = 20;
generateArray(nbr, min, max);
System.out.println();
}
public static int[] generateArray(int nbr, int min, int max) {
int[] a = new int[nbr];
int[] b = new int[nbr];
int[] s = new int[nbr];
s[0] = 0;
for (int i = 0; i < a.length; i++) {
a[i] = (int) (Math.random() * (max - min));
b[i] = (int) (Math.random() * (max - min));
for (int j = 0; j < i; j++) {
if (a[i] == a[j]) {
i--;
}
if (b[i] == b[j]) {
i--;
}
}
}
System.out.println("a - > " + Arrays.toString(a));
System.out.println("b - > " + Arrays.toString(b));
for (int k = 0; k < a.length; k++) {
for (int l = 0; l < b.length; l++) {
if (a[k] == b[l]) {
s[l] = b[l];
}else {
}
}
}
System.out.println("(a ∪ (b/(a ∩ b)) - > " + Arrays.toString(s));
return null;
}
public static boolean hasValue(int[] array, int value) {
for (int i = 0; i < array.length; i++) {
if (array[i] == value) {
return true;
}
}
return false;
}
}
Is there any way to create the array without the incorrect 0s? (I say incorrect because it is possible to have 0 in both a and b).
Any help/clarification is appreciated.
First, allocate an array large enough to hold the intersection. It needs to be no bigger that the smaller of the source arrays.
When you add a value to the intersection array, always add it starting at the beginning of the array. Use a counter to update the next position. This also allows the value 0 to be a valid value.
Then when finished. use Array.copyOf() to copy only the first part of the array to itself, thus removing the empty (unfilled 0 value) spaces. This works as follow assuming count is the index you have been using to add to the array: Assume count = 3
int[] inter = {1,2,3,0,0,0,0};
inter = Arrays.copyOf(inter, count);
System.out.println(Arrays.toString(inter);
prints
[1,2,3]
Here is an approach using a List
int[] b = {4,3,1,2,5,0,2};
int [] a = {3,5,2,3,7,8,2,0,9,10};
Add one of the arrays to the list.
List<Integer> list = new ArrayList<>();
for(int i : a) {
list.add(i);
}
Allocate the intersection array with count used as the next location. It doesn't matter which array's length you use.
int count = 0;
int [] intersection = new int[a.length];
Now simply iterate thru the other array.
if the list contains the value, add it to the intersection array.
then remove it from the list and increment count. NOTE - The removed value must be converted to an Integer object, otherwise, if a simple int value, it would be interpreted as an index and the value at that index would be removed and not the actual value itself (or an Exception might be thrown).
once finished the intersection array will have the values and probably unseen zeroes at the end.
for(int i = 0; i < b.length; i++) {
int val = b[i];
if (list.contains(val)) {
intersection[count++] = val;
list.remove(Integer.valueOf(val));
}
}
To shorten the array, use the copy method mentioned above.
intersection = Arrays.copyOf(intersection, count);
System.out.println(Arrays.toString(intersection));
prints
[3, 2, 5, 0, 2]
Note that it does not matter which array is which. If you reverse the arrays for a and b above, the same intersection will result, albeit in a different order.
The first thing I notice is that you are declaring your intersection array at the top of the method.
int[] s = new int[nbr];
You are declaring the same amount of space for the array regardless of the amount you actually use.
Method Arrays.toString(int []) will print any uninitialized slots in the array as "0"
There are several different approaches you can take here:
You can delay initializing the array until you have determined the size of the set you are dealing with.
You can transfer your content into another well sized array after figuring out your result set.
You could forego using Array.toString, and build the string up yourself.
I have a string of values separated by commas that I converted into an array, which I was then going to use to create a 2D array. When creating a loop to add the data from the first array to the 2D array it is repeating the data.
The output I'm getting is:
4428,40,401,610,2016,3821,31,347,572,2015,4381,38,341,520,2014,2536,17,193,290,2013,4295,39,371,552,2012,4643,45,343,502,2011,3922,28,312,475,2010,4434,30,350,541,2009,4038,28,341,536,2008,218,1,20,28,2007,46,0,6,15,2006,65,0,9,16,2005,4428,40,401,610,2016,3821,31,347,572,2015,4381,38,341,520,2014,2536,17,193,290,2013,4295,39,371,552,2012,4643,45,343,502,2011,3922,28,312,475,2010,4434,30,350,541,2009,4038,28,341,536,2008,218,1,20,28,2007,46,0,6,15,2006,65,0,9,16,2005,4428,40,401,610,2016,3821,31,347,572,2015,4381,38,341,520,2014,2536,17,193,290,2013,4295,39,371,552,2012,4643,45,343,502,2011,3922,28,312,475,2010,4434,30,350,541,2009,4038,28,341,536,2008,218,1,20,28,2007,46,0,6,15,2006,65,0,9,16,2005,4428,40,401,610,2016,3821,31,347,572,2015,4381,38,341,520,2014,2536,17,193,290,2013,4295,39,371,552,2012,4643,45,343,502,2011,3922,28,312,475,2010,4434,30,350,541,2009,4038,28,341,536,2008,218,1,20,28,2007,46,0,6,15,2006,65,0,9,16,2005,4428,40,401,610,2016,3821,31,347,572,2015,4381,38,341,520,2014,2536,17,193,290,2013,4295,39,371,552,2012,4643,45,343,502,2011,3922,28,312,475,2010,4434,30,350,541,2009,4038,28,341,536,2008,218,1,20,28,2007,46,0,6,15,2006,65,0,9,16,2005
and the correct output should be:
4428,40,401,610,2016,3821,31,347,572,2015,4381,38,341,520,2014,2536,17,193,290,2013,4295,39,371,552,2012,4643,45,343,502,2011,3922,28,312,475,2010,4434,30,350,541,2009,4038,28,341,536,2008,218,1,20,28,2007,46,0,6,15,2006,65,0,9,16,2005
Expected results:
{{4428,40,401,610,2016}
{3821,31,347,572,2015}
{4381,38,341,520,2014}
...} and so on, every 5
My code for adding the array to the 2D array is below:
{String[] columns = {"Yards","Touchdowns","Attempts","Incompletions","Year"};
String[] data1 = results1.split(",");
Object [][] data11 = new Object[columns.length][data1.length];
for(int i = 0; i<columns.length;i++){
for(int j = 0; j<data1.length;j++){
data11[i][j] = data1[j];
//System.out.print(data11[i][j]+",");
}
}}
EDIT: Solution!
Object [][] data11 = new Object[data1.length/columns.length][columns.length];
int column = -1;
for(int j = 0; j<data1.length;j++){
if(j % columns.length == 0) column = column+1;
data11[column][j % 5] = data1[j];
}
Maybe this work for you:
Object [][] data11 = new Object[columns.length][data1.length / 5];
int column = -1;
for(int j = 0; j<data1.length;j++){
if(j % 5 == 0) column = column + 1
data11[j % 5][column] = data1[j];
}
Note the matrix size changed and the assignation too. Haven't tried, probably you can make a prettier version. Hope it helps!
You also loop the first array (i), so he will repeat it. Just write this:
data11[0][j] = data1[j];
Or even don't do the first loop at all, depending on what you need.
Assuming your output, you only want to fill in the first position (0) of the array.
Right, so I have a 2 part sorting algorithm. It's all based on an array of 14 random integers. For example:
int[] a = {9,2,4,8,9,4,3,2,8,1,2,7,2,5};
Now, the first thing I'm trying to figure out how to do is to count how many a certain number exists in the original array. So, we know that 1 exists once, and 2 exists four times in the original array. But as easy as it is to visually see this, what if we don't have access to the original array. So I need to craft a method that will count how many of each number 1-9 exists and put this in a new array called count. So that index 0 in count would represent the integer 1 and would have a value of 1. Index 1 will represent the integer 2 and have a value of 4. And so on and so forth. Here is what I've got so far but I'm stuck. Sorting is pretty challenging for me.
public static void main(String[] args)
{
// int[] countFinal = {1,4,1,2,1,0,1,2,2}; // The number of times a number 1-9 appears in a[].
// int[] sortedFinal = {1,2,2,2,2,3,4,4,5,7,8,8,9,9}; // What we need as a final product.
int[] a = {9,2,4,8,9,4,3,2,8,1,2,7,2,5};
//int[] count = {};
int[] sorted = {};
countHowMany(a, 1);
countHowMany(a, 2);
countHowMany(a, 3);
countHowMany(a, 4);
countHowMany(a, 5);
countHowMany(a, 6);
countHowMany(a, 7);
countHowMany(a, 8);
countHowMany(a, 9);
}
public static int countHowMany(int[] array, int value)
{
// Gathering a count for how many times a number 1-9 exists and adding it to count[];
int howManyCount = 0;
for (int i = 0; i < array.length; i++)
{
if (array[i] == value)
{
howManyCount++;
}
}
System.out.println(howManyCount);
count = new int[9];
count[howManyCount];
System.out.println(Arrays.toString(count); // Testing the input
return howManyCount;
}
It appears to count the number of times an item in the array exists properly. Now I just gotta figure out how I can add that value into a new array count[] and do it for each countHowMany(). This is the part I'm stuck on.
Once I have figured out count[] I can use it to create sorted[]. Now what sorted is supposed to do is take the data from the original array and count[] and create a new array that sorts it in ascending order and allows duplicates. So, since 1 occurs once and 2 occurs four times, the new array would be sorted[] = {1, 2, 2, 2, 2, ...}
It's a relatively small program and a small amount of integers, so it's ok that I create array's as necessary. The key being that I'm limited to using arrays and cannot use say ArrayLists for this.
You don't need to count each value individually. You can just iterate through the entire array and increment your counters for each element as you encounter it.
int counts = new int[20]; // Choose a value that's bigger than anything in your array.
int[] a = {9,2,4,8,9,4,3,2,8,1,2,7,2,5};
for (int value : a) {
counts[value]++;
}
If you don't know what the largest value in your array is likely to be, you're better to use either a Map to store the counts, or some kind of List that you increase the size of as needed.
You're better off just going through the array once and incrementing a counter for each value that might appear:
int counts[] = new int[10];
for (int n: array)
counts[n]++;
That's enough to put the count for each n in counts[n]. You can then read the values out of your count[] array.
You might not have come across this syntax for a for loop over an array, by the way. It's equivalent to
int counts[] = new int[10];
for (int i=0; i<array.length; i++) {
int n = array[i];
counts[n]++;
}
but it's less verbose.
Your method may as well be void, since you're not doing anything with the returned values of your countHowMany function. This will accomplish what you want:
public static void main(String[] args)
{
int[] a = {9,2,4,8,9,4,3,2,8,1,2,7,2,5};
//count the instances of each number in the array
int[] count = new int[9];
for(int i = 0; i < count.length; i++)
count[i] = countHowMany(a, i+1);
//put the values in the sorted array
int[] sorted = new int[a.length];
int position = 0; // stores the place in the array to put the new digit
for(int digit = 0; digit < 9; digit++)
{
for(int inst = 0; inst < count[digit]; inst++)
{
sorted[position] = digit + 1;
position++;
}
}
System.out.println(Arrays.toString(sorted));
}
The issue with your code is that you were trying to create the count array in each call of the countHowMany method, but this array is destroyed once the method finishes. The method calls should just return the counts, and then those returns should be put into the count array from outside the method. Note, however, that there are other ways to count the number of instances of each value, as noted by other answers.
How do I get the second dimension of an array if I don't know it? array.length gives only the first dimension.
For example, in
public class B {
public static void main(String [] main){
int [] [] nir = new int [2] [3];
System.out.println(nir.length);
}
}
See that code run live at Ideone.com.
2
How would I get the value of the second dimension of nir, which is 3?
which 3?
You've created a multi-dimentional array. nir is an array of int arrays; you've got two arrays of length three.
System.out.println(nir[0].length);
would give you the length of your first array.
Also worth noting is that you don't have to initialize a multi-dimensional array as you did, which means all the arrays don't have to be the same length (or exist at all).
int nir[][] = new int[5][];
nir[0] = new int[5];
nir[1] = new int[3];
System.out.println(nir[0].length); // 5
System.out.println(nir[1].length); // 3
System.out.println(nir[2].length); // Null pointer exception
In the latest version of JAVA this is how you do it:
nir.length //is the first dimension
nir[0].length //is the second dimension
You can do :
System.out.println(nir[0].length);
But be aware that there's no real two-dimensional array in Java. Each "first level" array contains another array. Each of these arrays can be of different sizes. nir[0].length isn't necessarily the same size as nir[1].length.
use
System.out.print( nir[0].length);
look at this for loop which print the content of the 2 dimension array
the second loop iterate over the column in each row
for(int row =0 ; row < ntr.length; ++row)
for(int column =0; column<ntr[row].length;++column)
System.out.print(ntr[row][column]);
int secondDimensionSize = nir[0].length;
Each element of the first dimension is actually another array with the length of the second dimension.
Here's a complete solution to how to enumerate elements in a jagged two-dimensional array (with 3 rows and 3 to 5 columns):
String row = "";
int[][] myArray = {{11, 12, 13}, {14, 15, 16, 17}, {18, 19, 20, 21, 22}};
for (int i=0; i<myArray.length; i++) {
row+="\n";
for (int j = 0; j<myArray[i].length; j++) {
row += myArray[i][j] + " ";
}
}
JOptionPane.showMessageDialog(null, "myArray contains:" + row);
nir[0].length
Note 0: You have to have minimum one array in your array.
Note 1: Not all sub-arrays are not necessary the same length.
Assuming that the length is same for each array in the second dimension, you can use
public class B {
public static void main(String [] main){
int [] [] nir= new int [2] [3];
System.out.println(nir[0].length);
}
}
Remember, 2D array is not a 2D array in real sense.Every element of an array in itself is an array, not necessarily of the same size.
so, nir[0].length may or may not be equal to nir[1].length or nir[2]length.
Hope that helps..:)
Expansion for multi-dimension array total length,
Generally for your case, since the shape of the 2D array is "squared".
int length = nir.length * nir[0].length;
However, for 2D array, each row may not have the exact same number of elements.
Therefore we need to traverse through each row, add number of elements up.
int length = 0;
for ( int lvl = 0; lvl < _levels.length; lvl++ )
{
length += _levels[ lvl ].length;
}
If N-D array, which means we need N-1 for loop to get each row's size.
//initializing few values
int[][] tab = new int[][]{
{1,0,1,0,1,0,1,0},
{0,1,0,1,0,1,0,1},
{1,0,1,0,1,0,1,0},
{0,1,0,1,0,1,0,1},
{1,0,1,0,1,0,1,0},
{0,1,0,1,0,1,0,1},
{1,0,1,0,1,0,1,0},
{0,1,0,1,0,1,0,1}
};
//tab.length in first loop
for (int row = 0; row < tab.length; row++)
{
//tab[0].length in second loop
for (int column = 0; column < tab[0].length; column++)
{
//printing one value from array with space
System.out.print(tab[row][column]+ " ");
}
System.out.println(); // new row = new enter
}