Group values from a 3d array into an associative 2d array and sum values with a shared path

Question

I have already seen this stackoverflow page but it is not helping me.

I want to group by two columns and sum the values of a third column.

If the discount_id and dis_percent are the same then add the discount_value.

Here is my array:

$dis = [
    [['Dis_id' => 'Dl-Dis1'], ['Dis_per' => '7.500'], ['Dis_val' => '192.75']],
    [['Dis_id' => 'Dl-Dis2'], ['Dis_per' => '2.500'], ['Dis_val' => '97.88']],
    [['Dis_id' => 'Dl-Dis1'], ['Dis_per' => '5.000'], ['Dis_val' => '39.90']],
    [['Dis_id' => 'Dl-Dis2'], ['Dis_per' => '2.500'], ['Dis_val' => '99.90']]
];

The output that I need is:

D1-Dis1->7.5->192.75
D1-Dis1->5.0->39.9
D1-Dis2->2.5->197.78

My code looks like this:

$newarr = array();
$reverse_map = array();

foreach($dis as $idx => $entry) {
    if (isset($reverse_map[$entry['Dis_id']])) {
         // have we seen this name before? retrieve its original index value
         $idx = $reverse_map[$entry['Dis_id']]; 
    } else {
         // nope, new name, so store its index value
         $reverse_map[$entry['Dis_id']] = $idx;
    }

    // copy the 'constant' values
    $newarr[$idx]['Dis_id'] = $entry['Dis_id'];
    $newarr[$idx]['Dis_per'] = $entry['Dis_per'];

    // sum the qtd_post values to whatever we previously stored.        
    foreach($entry['Dis_val'] as $x => $y) {
        $newarr[$idx]['Dis_val'][$x] += $y;
    }
}

If the dis_id and dis_per are what? Also, have you got any examples of code where you've tried to solve this so we can see where you are having issues? — Pudge601
– Pudge601, Commented Jun 4, 2014 at 8:51
dis_id is the name and dis_per is the percentage for the dis_id. i gave the amt and the result after deduct from the dicount percent is dis_val.. please see the code i tried — rAm
– rAm, Commented Jun 4, 2014 at 9:16
"a condition if the discount_id and dis_percent are then add the discount_value" <- if they are what? You haven't said what the condition is — Pudge601
– Pudge601, Commented Jun 4, 2014 at 9:26
Right.. So I'm getting the idea that you want to sum up the Dis_val for all entries, grouped by Dis_id and Dis_per. I've written a solution, but also noticed that your array structure is quite odd, and doesn't match how your attempted solution works. Is it intentional that each entry contains a subarray, in which there is just a single element for either Dis_id, Dis_per or Dis_val? Hopefully you'll get what I mean by this; pastebin.com/FDjVWy6z — Pudge601
– Pudge601, Commented Jun 4, 2014 at 10:57

Pudge601 · Accepted Answer · 2014-06-04 11:07:10Z

2

This is the solution I've come up with based off of the understanding that your intended array structure was as so;

$dis = array(
    array(
        'Dis_id'  => 'Dl-Dis1',
        'Dis_per' => 7.500,
        'Dis_val' => 192.75
    ),
    ...
);

It determines the solution by creating a multidimensional array where the first dimension is the Dis_id, and the second dimension is the Dis_per, and the value becomes the sum of the Dis_val;

$sums = array();

foreach ($dis as $entry) {
    if (!isset($sums[$entry['Dis_id']])) {
        $sums[$entry['Dis_id']] = array();
    }
    if (!isset($sums[$entry['Dis_id']]["{$entry['Dis_per']}"])) {
        $sums[$entry['Dis_id']]["{$entry['Dis_per']}"] = 0;
    }
    $sums[$entry['Dis_id']]["{$entry['Dis_per']}"] += $entry['Dis_val'];
}

See this working example; https://eval.in/158661

answered Jun 4, 2014 at 11:07

Pudge601

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Comments

mickmackusa · Accepted Answer · 2024-01-16 12:35:34Z

0

As you iterate your input array, use array destructuring to isolate the Dis_id and Dis_per values and use them as keys when storing your Dis_val.

If the key path hasn't been encountered before, default the stored amount to 0 and add the current amount to it.

Code: (Demo)

$result = [];
foreach ($dis as [['Dis_id' => $id], ['Dis_per' => $per], ['Dis_val' => $val]]) {
    $result[$id][$per] = ($result[$id][$per] ?? 0) + $val;
}
var_export($result);

edited Jan 16, 2024 at 12:35

answered Aug 24, 2018 at 5:29

mickmackusa♦

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