CODEWITHSUNDEEP
javaarraysnatural-join
Lisa
Lisa

Reputation: 3181

Natural Join of two dimensional Arrays

I'm trying to implement the same idea of Natural Join in Database but on Two dimensional Arrays, so What I'm trying to do is if I have

A={[a,b],[a',b']} and B={[b,c], [b',c],[b,c']}

The result of NaturalJoin(A,B) should be:

Result = {[a,b,c], [a,b,c'],[a',b',c]}

So after I find the shared column B and compare it in both arrays, how can I combine the rows? Can you please give me some hints on how to create the joinedTableau as I don't know the number of rows from the beginning of the join, how can I create it dynamically?

This is my pseudo code:

int[][] array1;
int[][] array2;
int shared = prtiallyEqualColumn(array1,array2);
           for(int i = 0; i <array1.length; i++)
                        {
                            for(int j = 0 ; j < array2.length; j++)
                            {
                                if(array1[i][shared] == array2[j][shared]) 
                                {
                                        for(int s = 0 ; s < shared; s++)
                                        {
                                            joinedTableau[joinedCountRow][s] = array1[i][s];
                                        }
                                        for(int y=shared+1; y<array2.length;y++)
                                        {
                                            joinedTableau[joinedCountRow][y] = array2[j][y];
                                        }
                                    }
                                }
                            }

Upvotes: 0

Views: 664

Answers (2)

Manuel Hernandez
Manuel Hernandez

Reputation: 524

This yields the right answer. is it the most efficient? I posted the results of my timing. definitely exponential growth in times

count avg time
10 1.931190E-05
100 4.734993E-04
1000 2.540604E-02
10000 1.400114E+00
100000 9.096992E+01 
#!python3
from time import time


def natural_join( S, T):
    theJoin = []
    for i in T:
        for j in S:
            if i[0] == j[1]:
                theJoin.append((i[1], j[0], j[1]))
                break
        
    return theJoin


for n in range(1, 6):
    A = []
    B = []
    for i in range(10 ** n):
        A.append(('x' + str(i), 'y' + str(i)))

    for i in range(10 ** n):
        B.append(('y' + str(i), 'z' + str(i)))

    start = time()
    joined = natural_join(A ,B)
    end = time()
    print('{:d}, {:E}'.format(10 ** n, (end-start)/n))

Upvotes: 0

Am_I_Helpful
Am_I_Helpful

Reputation: 19168

I don't know what you've done in the code as you have hidden several implementations from the code presented here in the question. I am giving you the algo :-

Each column value of array1 must be compared with each row value of array2 to produce a natural join,only in case if they are equal, else not.

a1 = array1.countRow();
a2 = array1.countColumn();
b1 = array2.countRow();
b2 = array2.countColumn();
i = j = 1;
while(i<=a1)
while(j<=b1)
if(array1[i][a2]==array2[j][1]) // I've made array-indices start from 1
// perform natural-join operation between array1[i]+array2[j]-common_element
// Similarly iterate the array for next passes.

If there is some mistake or something which is unclear to you,please notify me. Good luck for your code.

Upvotes: 1

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