You can convert your arrays to a List and then use the containsAll method to see if a particular list contains all elements described in another list.
You would get better performance out of it if they were Sets instead.
Example:
List<String> firstList = Arrays.asList("chery", "chery", "unid", ...);
List<String> secondList = Arrays.asList("chery", "unid", ...);
System.out.println(secondList.containsAll(firstList));
If the performance of this method in particular is getting a bit dodgy, then consider converting the lists into Sets instead:
Set<String> firstSet = new HashSet<>(Arrays.asList("chery", "chery", "unid", ...));
Answer from Makoto on Stack OverflowYou can convert your arrays to a List and then use the containsAll method to see if a particular list contains all elements described in another list.
You would get better performance out of it if they were Sets instead.
Example:
List<String> firstList = Arrays.asList("chery", "chery", "unid", ...);
List<String> secondList = Arrays.asList("chery", "unid", ...);
System.out.println(secondList.containsAll(firstList));
If the performance of this method in particular is getting a bit dodgy, then consider converting the lists into Sets instead:
Set<String> firstSet = new HashSet<>(Arrays.asList("chery", "chery", "unid", ...));
In the example I am using integers but can be used for other types also with slight modifications. First put a loop on array A elements.
for(int i =0; i<A.length(); i++)
{
//this loop will transverse with all elements in array A.
}
Now inside this for loop make another for loop which transverse through elements of loop B.
for(int i =0; i<A.length(); i++)
{
for(int j=0; j<B.length();j++)
{
if(A[i] == B[j])
{ System.out.println("this element is in array A and B"); }
}
}
Now if you want to check if all elements of A are in B you can make a boolean. this boolean is true as long each element in A is found at least once in B. as soon as you find one element which is not present on both arrays you can exit.
Judging by the picture, you are not placing the values in the correct index in the second array, or you are not matching the correct ones.
If you want it to look exactly like in the picture, the second array should be declared:
double[] newArray = new double[array.length - 2];
And the loop to fill it should be changed to:
for(j = 2; j < array.length - 2; j++)
{
for(int k = -2; k <= 2; k++)
{
newArray[j] += array[j+k] / 5;
}
}
This will put the averages in the third, fourth, fifth... elements in newArray. And now you can compare them directly:
for(int i = 2; i < array.length - 2; i++)
{
if(array[i] > (newArray[i] + 0.999))
{
count++;
tempVal = count;
}
System.out.println(tempVal);
}
If you want to save the two unused spaces, as you originally did, rather than responding exactly to the picture, then you should calculate the values as you originally did. But remember to compare each element to the one two places before it and stop 2 places before the end.
Instead of
for(int i = 2; i < array.length; i++)
use
for(int i = 2; i < array.length - 2; i++)
To print the location, your construct with the count and tempVal is unnecessary. You just need to print i+1. Also note that you have a ; after your if. This means it's an empty if, and the block after it is always performed. Never have a ; after an if, for, while etc.
Not clear with what you are asking for in your question but without questioning what's the logic, by just looking at your code:
for(int i = 2; i < array.length; i++) { if(array[i] > (newArray[i-2] + 0.999)); { count++; tempVal = count; } System.out.println(tempVal); } }
if you relocate the system.out line as follows, I think you will get what you expect as follows:
for(int i = 2; i < array.length - 2; i++)
{
if(array[i] > (newArray[i-2] + 0.999));
{
System.out.println(tempVal);
// count++;
// tempVal = count;
}
}
}
PS: Please note that I have also changed the boundary for the loop to stop iteration on 13th member of the array, instead of 15.
From what I see you just try to see if they are equal, if this is true, just go with something like this:
boolean areEqual = Arrays.equals(arr1, arr2);
This is the standard way of doing it.
Please note that the arrays must be also sorted to be considered equal, from the JavaDoc:
Two arrays are considered equal if both arrays contain the same number of elements, and all corresponding pairs of elements in the two arrays are equal. In other words, two arrays are equal if they contain the same elements in the same order.
Sorry for missing that.
use Arrays.equals(ary1,ary2); // returns boolean value
EDIT
you can use Arrays.deepEquals(ary1,ary2) to compare 2D arrays as well
also check this link for comparision comparision between Arrays.equls(ar1,ar2) and Arrays.deepEquals(ar1,ar2)
Java Arrays.equals() returns false for two dimensional arrays
EDIT 2
if you dont want to use these library methods then you can easily implement your method like this:
public static boolean ArrayCompare(int[] a, int[] a2) {
if (a==a2) // checks for same array reference
return true;
if (a==null || a2==null) // checks for null arrays
return false;
int length = a.length;
if (a2.length != length) // arrays should be of equal length
return false;
for (int i=0; i<length; i++) // compare array values
if (a[i] != a2[i])
return false;
return true;
}
Here is how you will compare your two double arrays for equality:
public static boolean compArray (double[] a, double[] b, int n) {
//if the arrays dont have the same length, then they are not equals
if (a.length != b.length) {
return false;
}
for (int i = 0; i < a.length; i++) {
//if one element at array a is different than the other at array b,
//at the same index, then the arrays are not equals
if (a[i] != b[i]) {
return false;
}
}
}
//otherwise, the two arrays are equals
return true;
}
Don't reinvent the wheel. There's a library function for that:
public static boolean compArray(double[] a, double[] b, int n) {
return Arrays.equals(a, 0, n, b, 0, n);
}
See Arrays.equals(double[] a, int aFromIndex, int aToIndex, double[] b, int bFromIndex, int bToIndex)