Combine columnar values in each row of a 2d array to form dynamically keyed elements

Question

I have this array:

$params = [
    ['name'  => 'xxx', 'value' => 'yyy'],
    ['name'  => 'uuu', 'value' => 'vvv']
];

and I want to achieve this:

$params = [
    ['xxx' => 'yyy'],
    ['uuu' => 'vvv']
];

I can do it this way:

foreach ($params as &$param) {
    $param[$param['name']] = $param['value'];
    unset($param['name']);
    unset($param['value']);
    unset($param);
}

But I am wondering if this could be done on more elegant way?

codereview.stackexchange.com may be more appropriate -- elegance is extremely subjective — Mm-Art-In
– Mm-Art-In, Commented May 24, 2018 at 12:23
If you're looking for a built in way to accomplish this, there isn't one. You're code is short, accomplishes the task and is readable. I would use it. — Questionable
– Questionable, Commented May 24, 2018 at 12:24

deceze · Accepted Answer · 2018-05-24 12:25:08Z

6

This calls for an array mapping of values:

$params = array_map(function ($i) { return [$i['name'] => $i['value']]; }, $params);

answered May 24, 2018 at 12:25

deceze♦

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4 Comments

deceze Over a year ago

That's in the eye of the beholder. A mapping is a defined action, and all you need to know is that 1) this is a mapping, 2) what the callback returns. Then it's very precise and readable. Too bad PHP's verbose syntax makes it somewhat ungainly.

deceze Over a year ago

E.g. in Javascript this would be params = params.map(i => ({ [i.name]: i.value })). It hardly gets any more concise than this, and the concept translates to most other languages.

Mm-Art-In Over a year ago

Yes, I was making the point that OP was asking for elegance and NOT for the smallest possible codeblock. I believe it to be a poor question and this is only a subjectively beautiful answer :-p

deceze Over a year ago

Well, I consider references and superfluous variables inelegant. IMO this is about as elegant as it gets in PHP.

Eddie · Accepted Answer · 2018-05-24 12:29:17Z

1

You can use array_map to reiterate the array.

$params = [
  [
    'name'  => "xxx",
    'value' => "yyy"
  ],
  [
    'name'  => "uuu",
    'value' => "vvv"
  ]
];

$params = array_map(function($o){
    return [ $o['name'] => $o['value'] ];
}, $params);


echo "<pre>";
print_r( $params );
echo "</pre>";

This will result to:

Array
(
    [0] => Array
        (
            [xxx] => yyy
        )

    [1] => Array
        (
            [uuu] => vvv
        )

)

Doc: array_map()

answered May 24, 2018 at 12:29

Eddie

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2 Comments

The fourth bird Over a year ago

Just curious, what is the difference with the answer of @deceze?

Eddie Over a year ago

@Thefourthbird I didn't realize that someone already answered this approach. Until just now. Haha

Mm-Art-In · Accepted Answer · 2018-05-24 12:27:19Z

0

This is a bit shorter. By not using foreach you can directly replace the elements.

for ($i = 0; $i < count ($params); $i++)
    $params[$i] = array ($params[$i]['name'] => $params[$i]['value']);

edited May 24, 2018 at 12:27

Mm-Art-In

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answered May 24, 2018 at 12:26

wayneOS

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Comments

The fourth bird · Accepted Answer · 2018-05-24 12:40:44Z

0

To make your current code a little bit more elegant, you could pass multiple arguments to unset

foreach ($params as &$param) {
    $param[$param['name']] = $param['value'];
    unset($param['name'], $param['value'], $param);
}

answered May 24, 2018 at 12:40

The fourth bird

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3 Comments

The fourth bird Over a year ago

@Martin You don't have to do it to be able to use the new key, but as the OP wanst to remove the old keys and values in the example you could use it.

Mm-Art-In Over a year ago

Yes that's a fair point. I had missed that; I was looking at it purely as a case of tidying up the foreach system.

Andreas Over a year ago

No it's not, if you don't unset the values you will have three items in each subarray. But if you build a new array, then yes.

Meloman · Accepted Answer · 2019-10-03 13:08:13Z

0

Try this :

foreach ($params as $key => $sub_array) {

    // unset original array data
    unset($params[$key]);

    // construct the new values
    $params[$key][$sub_array['name']] = $sub_array['value'];
}

EDIT:

Another approach more elegant is the following :

foreach($params as $k=>$s){
    $params[$k]=array($s['name']=>$s['value']);
}

edited Oct 3, 2019 at 13:08

answered May 24, 2018 at 12:24

Meloman

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2 Comments

Mm-Art-In Over a year ago

why, what does this do? explain your answer

Meloman Over a year ago

It simply put the name value in the key of parent array as you want. And unset the current name and value values. That's what you want. You only forgot to use $key in foreach to help you construct the array.

mickmackusa · Accepted Answer · 2024-10-04 21:04:13Z

All of the following snippets will produce the same desired result from the provided input array. Demo

Make a new array with a body-less loop.

$result = [];
foreach ($params as ['name' => $k, 'value' => $result[][$k]]);
var_export($result);

Make a new array using iterated chunk&combining.

var_export(
    array_map(
        fn($row) => array_combine(...array_chunk($row, 1)),
        $params
    )
);

Make a new array using iterated hardcoded declarations.

var_export(
    array_map(
        fn($row) => [$row['name'] => $row['value']],
        $params
    )
);

Make a new array by reducing the columns to a flat associative array, then re-expand to a 2d array.

var_export(
    array_chunk(
        array_column($params, 'value', 'name'),
        1,
        true
    )
);

Modify the original array by reference without using first level keys.

foreach ($params as &$row) {
    $row = [$row['name'] => $row['value']];
}
var_export($params);

treyBake · Accepted Answer · 2018-05-25 06:45:43Z

-3

Please try this:

foreach ($params as $k => $v)
{
    $nary[$v[name]] = $v[value];
}

$params = $nary;

edited May 25, 2018 at 6:45

treyBake

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answered May 24, 2018 at 12:32

Anant Ram Verma

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3 Comments

treyBake Over a year ago

this won't work without quotes around name and value. Also OP already says they can do it this way and want something more elegant

deceze Over a year ago

@ThisGuy Depressingly "$v[name]" will work just fine and $v[value] will in fact work for the time being (albeit generating a notice)…

treyBake Over a year ago

@deceze ah I edited question for easier reading and forgot no one else can see it XD agreed that it is depressing it still works how it is

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