mathematical operations using php scripting

Question

I am implementing k- nearest neighbor alog using euclidian distance formula for 2-D data in my sql table. Things look ok below even it shows strange behavious.

Code:

    $result1 = mysqli_query($con,"SELECT pcount, ncount from test");
    $result2 = mysqli_query($con,"SELECT pcount, ncount from test");
    $i = 0;
    $min = 0;
    while ($row1 = @mysqli_fetch_array($result1))
    {
        $pcount = $row1['pcount'];
        $ncount = $row1['ncount'];
        echo "pcount is $pcount<br/>";
        echo "ncount is $ncount<br/></br>";
        $a[$i] = $pcount ;
        $b[$i] = $pcount ;
        $j = 0;
        while ($row2 = @mysqli_fetch_array($result2))
        {
            echo "j is $j <br/>";
            $a[$j] = $row2['pcount'];   
            $b[$j] = $row2['ncount'];
            $diff = sqrt(($a[$i] - $a[$j])^2 + ($b[$i] - $b[$j])^2 );
        $j= $j + 1;
        echo "$diff <br>";
        }               
        echo "i is $i <br/>";
                $i = $i + 1;    
    }

it gives this result:

j is 0 
Difference : 0 
j is 1 
Difference : 1.4142135623731 
j is 2 
Difference : 0 
j is 3 
Difference : 0 
j is 4 
Difference : 1.4142135623731 
j is 5 
Difference : 1.4142135623731 
j is 6 
Difference : 1.4142135623731 
i is 0

In table I have total 8 rows, but above result is for 7 rows only. Why?

And First while loop gets executed for $i = 0 on, for rest i=1 to 7 no operation. Can some one tell me what is bug?

data in table is

pcount  ncount
20  80
21  79
20  80
21  79
19  81
20  80
20  80
20  80

result after fallens answer:

j is 0 
Difference : 0 
j is 1 
Difference : NAN 
j is 2 
Difference : 0 
j is 3 
Difference : NAN 
j is 4 
Difference : 1.4142135623731 
j is 5 
Difference : 0 
j is 6 
Difference : 0 
j is 7 
Difference : 0 
i is 0 
pcount is 21
ncount is 79

i is 1 
pcount is 20
ncount is 80

i is 2 
pcount is 21
ncount is 79

i is 3 
pcount is 19
ncount is 81

i is 4 
pcount is 20
ncount is 80

i is 5 
pcount is 20
ncount is 80

i is 6 
pcount is 20
ncount is 80

i is 7

Could it fail because you used the same variable in both while-loops? Try replacing the $row in the second loop with something like $row2. — mnme
– mnme, Commented Feb 20, 2014 at 13:09

Fallen · Accepted Answer · 2014-02-20 13:17:03Z

1

You are using same variable in inner and outer loop. Try different variable names for rows and results in these two loops. Like result1, row1 for outer and result2, row2 for inner loop

It shows 7 rows instead of 8, because first row from result is picked by outer loop already

Try this piece of code:

$result1 = mysqli_query($con,"SELECT pcount, ncount from test");
$i = 0;
$min = 0;
while ($row1 = @mysqli_fetch_array($result1))
{
    $pcount = $row1['pcount'];
    $ncount = $row1['ncount'];
    echo "pcount is $pcount<br/>";
    echo "ncount is $ncount<br/></br>";
    $a[$i] = $pcount ;
    $b[$i] = $pcount ;
    $j = 0;

    $result2 = mysqli_query($con,"SELECT pcount, ncount from test");

    while ($row2 = @mysqli_fetch_array($result2))
    {
        echo "j is $j <br/>";
        $a[$j] = $row2['pcount'];   
        $b[$j] = $row2['ncount'];
        $diff = ($a[$i] - $a[$j])^2 + ($b[$i] - $b[$j])^2;
        if( $diff != 0)
        $diff = sqrt( $diff );
    $j= $j + 1;
    echo "$diff <br>";
    }               
    echo "i is $i <br/>";
            $i = $i + 1;    
}

Note To avoid NaN, make sure ($a[$i] - $a[$j])^2 + ($b[$i] - $b[$j])^2 doesn't yield 0.

edited Feb 20, 2014 at 13:17

answered Feb 20, 2014 at 13:12

Fallen

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3 Comments

Fallen Over a year ago

@Programming_crazy: I've added code. Please note that instead of initializing $result2 once, this updated code initialize it for every possible outer loop

user2129623 Over a year ago

can you please suggest me to encount ($a[$i] - $a[$j])^2 + ($b[$i] - $b[$j])^2, bcoz my data has frequent value where it yields 0

Fallen Over a year ago

If you check the updated code, you can see this part $diff = ($a[$i] - $a[$j])^2 + ($b[$i] - $b[$j])^2; if( $diff != 0) $diff = sqrt( $diff ); filters $diff

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