Hi everyone,

My data is stored in a two-dimensional array, which rows correspond to different products, and row elements correspond to product components. Product values are stored in a separate one-dimensional array.

I would like to retain only the products with highest value that consist of distinct and unique components, i.e. identify and retain the rows of the 2d array that differ on all elements. Here is an example:

// 2d array with product IDs

int [][] duplicate =

 {{0, 1, 2, 3}, // retain

  {0, 5, 6, 11}, // remove

  {4, 1, 6, 3}, // remove

  {8, 9, 10, 11}}; // retain

// 1d array with product values

 int [] value = {10,4,9,15}; 

 int rows =4;// number of components (a constant equal to 4)

// My objective is to filter out distinct products:

Int [][] distProd =

{{0, 1, 2, 3},

  {8, 9, 10, 11}};

My idea was (1) to identify the product with the highest value, (2) check if other products overlap with the best one; (3) if products do not have overlapping components with the best one, they are stored in a new array. Here is the code:

int [][] duplicate =

1.  {{0, 1, 2, 3}, // retain

     {0, 5, 6, 11}, // remove

     {4, 1, 6, 3}, // remove

     {8, 9, 10, 11}}; // retain

    

   // 1d array with product values

    int [] value = {10,4,9,15};

    

   int rows =4;// number of components (a constant equal to 4)

   int [] maxval = new int [rows]; // product id with the max value of the fixed length

   int max = max(value); // 15

   for (int i=0;i<duplicate.length; i++){

     if (value[i]==max) {

       maxval = duplicate[i];

   }

   }

   println(maxval); // 8, 9, 10, 11

   // check if there are duplicates by substracting each array from an array with max value:

   // if products consist of distinct cells neither of the cells in test array should be equal to zero

   int [][] dup_test = new int [duplicate.length][rows];

    for (int i=0;i<duplicate.length; i++){ 

      for (int j=0; j<duplicate[i].length;j++){

        dup_test[i][j]=maxval[j]-duplicate[i][j];

      }

    }

    

    for (int i=0;i<duplicate.length; i++){

      println(i);

      println(dup_test[i]);

    }

    

   //initialise boolean array

   boolean [][] isZero = new boolean [duplicate.length][rows];

    for (int i = 0; i<duplicate.length; i++){

       for (int j = 0; j<duplicate[i].length; j++){

         isZero[i][j] = false;}

    }

   // check which positions are zeros

   for (int i = 0; i<duplicate.length; i++){

       for (int j = 0; j<duplicate[i].length; j++){    

         if (dup_test[i][j]==0) {isZero[i][j] = true;};       

         }

       }

      

   for (int i = 0; i<duplicate.length; i++){

     println(i);

     println(isZero[i]);

   } 

   // count how many zeros in each product

   int [] zeros = new int [isZero.length];

       int count_null =0;

      for (int i = 0; i<isZero.length; i++){ 

       for (int j = 0; j<isZero[i].length;j++){

        if ( isZero [i][j] == true) {count_null++;

      }

       } zeros [i] = count_null; count_null =0; // reset count for each row

      }

    

   for (int i = 0; i<zeros.length; i++){

     println(zeros[i]);

   }

    

   // filter out distinct products (zeros [i]=0) and the product with max value (zeros[i]=4)

   int [][] distinct = new int [duplicate.length][rows]; // temp array with excessive length

       int count_dist =0;

    for (int i = 0; i<duplicate.length; i++){  

         if (zeros[i] ==rows || zeros [i] == 0) {distinct[count_dist]=duplicate[i]; count_dist++;}

       }

      

    for (int i = 0; i<zeros.length; i++){

     println(i);

     println(distinct[i]);

   }

    

   // count how many distinct products (to input as a length of an array)

   int count_cutoff =0;

      for (int i =0; i<zeros.length; i++){

        if (zeros[i] ==rows || zeros [i] == 0) {count_cutoff++;}

      } 

    

   int [][] distProd = Arrays.copyOf(distinct,count_cutoff);

   // prints: {{0, 1, 2, 3},

   // {4, 1, 6, 3},

   // {8, 9, 10, 11}}; 

Question 1: I was wondering if there is a shorter and more elegant way to arrive at the desired solution. Moreover, the current version uses Arrays.copyOf which is no longer available in more recent versions of Processing.

Question 2: The code only partially fulfills its purpose because although products may not have overlapping cells with the max value product (8, 9, 10, 11), they can still have common components between themselves  (like in case of 0, 1, 2, 3 and 4, 1, 6, 3). I could just repeat the same algorithm comparing elements to the second best product in distProd, but maybe there is an easier way..

Thank you very much for your consideration!

Hi everyone,

Hi everyone,

Hello everyone!

I need to compare two 2D arrays and append equal ones to a new empty array but this time I would need to two arrays to result equal even if the same elements are in different order, i.e. so {1,2,3,4} and {2,1,3,4} would be considered equal rows... Arrays.equal no longer applicable in this case.

I tried to solve this looping through each element but unfortunately this attempt failed:

Hello everyone,

Hi all!

I am new to Processing and I need help with the code. I am looking for a way to get all possible combinations of elements which are separated in two groups, for example for 4 elements it would look like this:

(AC) = (AD)(AE)(AF)
(AD) = (AC)(AE)(AF)
(AE) = (AC)(AD)(AF)
(AF) = (AC)(AD)(AE)
(AC)(AD) = (AE)(AF)
(AC)(AE) = (AD)(AF)
(AC)(AF) = (AD)(AE)
(AC)(AD)(AE)(AF)

Below I attach code in C++ which gives an idea of an algorithm, but I was just wondering if there is a way to create the same thing in processing. Maybe it would help.. thank you ver much in advance for any suggestions!

void

getCombination( int k, int n) //generation of combinations

{

int* A=new int[k]; //array for group 1

int * B= new int [n-k]; //array for group 2

char Companies[20][3]={"AB","AC","AD","AE","AF","BB","BC","BD","BE","BF","CB","CC","CD","CE","CF","DB","DC","DD","DE","DF"}; // companies names

if (k<n)

{

for ( int i=1;i<=k;i++) //initialis array

A[i]=i;

// algorithm execution

int p=k;

while(p>=1 && res>0)

{

res--;

int z=1;

for(int ii=1; ii<=n;ii++)

{

bool ok=1;

for(int jj=1; jj<=k;jj++)

if (ii==A[jj])

ok=0;

if(ok)

B[z++]=ii;

}

cnt++;

for(int j=1;j<=k;j++)

{

cout << "(" << Companies[A[j]] << ")" ;

myfile << "(" << Companies[A[j]] << ")" ;

}

cout << " = ";

myfile << " = ";

for(int ii=1; ii<=(n-k);ii++)

{

cout << "(" << Companies[B[ii]] << ")";

myfile << "(" << Companies[B[ii]] << ")";

}

cout << "\n";

myfile << "\n";

if(A[k]==n)

p--;

else

p=k;

if (p>=1)

{

for ( int ii=k; ii>=p;ii--)

A[ii]=A[p]+ii-p+1;

}

}

}

else //last combination (if all companies are in the same group)

{

cnt++;

for(int j=1;j<=k;j++)

{

cout << "(" << Companies[j] << ")" ; //to screen

myfile << "(" << Companies[j] << ")" ; //to file

}

cout << "\n";

myfile << "\n";

}

}

　

int

_tmain( int argc, _TCHAR* argv[])

{

int keyCode=0;

while(keyCode!=113 && keyCode!=81) //113 and 81 code for Q and q

{

cnt=0;

int maxval=0;

while (maxval<2 || maxval>20) //limitations - no more than 20 and no less than 2 companies

{

cout << "Number of companies[2-20]? \n";

cout << "===========================\n";

char tmp[100];

cin >> tmp;

maxval = atoi(tmp);

}

res=pow(2,double(maxval-1))-1; //verification Steerling number

cout << "Summary: " << res + 1 << "\n"; // add 1 to account for the last combination when all companies are in the same group

myfile << "Summary: " << res + 1 << "\n";

myfile << "============================================\n";

for(int i=1;i<=maxval;i++)

{

getCombination(i,maxval);

}

　

　

cout << "----------\n";

cout << "Total: " << cnt << "\n";

myfile << "----------\n";

myfile << "Total: " << cnt << "\n";

cout << "Press any key to repeat or 'Q' for quit \n";

keyCode=_getch();

}

myfile.close();

return 0;

}