Grouping intersections of arrays

Question

I have an indefinite of texts which needs to be translated into different languages,
e.g. $text1 needs to be translated in French and Hungarian.

$text1 = array ('fr', 'hr');
$text2 = array ('bg', 'el', 'hr');
$text3 = array ('bg', 'el', 'en', 'es');
$text4 = array ('bg', 'el', 'en', 'es');
$text5 = array ('bg', 'el', 'en', 'es', 'fr', 'hr');

Now I am looking for a way to combine texts and languages in order to get a minimum of language combinations, to get something like this:

$order1 = array('languages' => array('bg', 'el'), 'texts' => array ('text2', 'text3', 'text4', 'text5'));
$order2 = array('languages' => array('en', 'es'), 'texts' => array ('text3', 'text4', 'text5'));
$order3 = array('languages' => array('hr'),       'texts' => array ('text1', 'text2', 'text5'));
$order4 = array('languages' => array('fr'),       'texts' => array ('text1', 'text5'));

I have absolutly no idea how to start. Could anybody please give me a hint? Thank you very much.

Can we add the var name to the end of each array for making it easier to code? For example : $text1 = array ('fr', 'hr','text1'); — Mobin F.R.G
– Mobin F.R.G, Commented Jul 19, 2018 at 12:48
I'm not sure if this question so hard or am I missing some important easy notes in PHP! I'll try to solve this even for days! But It will be awesome that someone comes and give a short good answer! — Mobin F.R.G
– Mobin F.R.G, Commented Jul 19, 2018 at 14:01

Decent Dabbler · Accepted Answer · 2018-07-19 15:15:44Z

2

This was a pretty tricky problem, indeed.

Here's my solution:

// first collect all texts per individual language
for( $i = 1, $data = []; ( $key = 'text' . $i++ ) && isset( ${$key} ); ) {
  $data = array_merge_recursive( $data, array_fill_keys( ${$key}, $key ) );
}

// then walk through them and find other languages that match the same texts
for( $i = 1, $orders = []; count( $data ) > 0; $i++ ) {
  $texts = reset( $data );
  $order = [ 'languages' => [], 'texts' => $texts ];
  while( false !== ( $lang = array_search( $texts, $data, true ) ) ) {
    $order[ 'languages' ][] = $lang;
    unset( $data[ $lang ] );
  }
  $orders[ 'order' . $i ] = $order;
}

^{view sample on eval.in}

Notes

The orders are not in the order of your example. Let me know if that's a prerequisite.
If you insist on having individual order variables ($order1, $order2, etc.), in stead of an orders array, you could do extract( $orders ); to extract them into the current scope, but I would advice against that, for you could accidentally overwrite pre-existing variables, if you are not careful.

edited Jul 19, 2018 at 15:15

answered Jul 19, 2018 at 15:05

Decent Dabbler

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2 Comments

Mobin F.R.G Over a year ago

Interesting answer!

ondrums Over a year ago

Thank you very much for your help!

Pbras · Accepted Answer · 2018-07-19 14:44:49Z

1

-->Link for 3v4l.org full code<--

You have to do a while to check all the array positions

while($aux< count($order1)

And you also need a $containPrev1 that will turn false if the $text don't have all the 'languages' in $order1.

You use the function "in_array" to check if you element of the array is in the $text. You will check all the elements cause you are in a while cicle.

in_array($order1[$aux], $text1)

If its not in the array you set the $containPrev1 to false to point the that atleast one of elements of $order1 are not in $text1 and soo you will ignor that $text1 for the rest of the while.

Im adding elements into a string and then explode() the string to create a array, you can directaly creat an array soo you should check it and adapt the code to that (I assumed you added the 'text1' as the last element of the array).

$finalstr = $finalstr . $text1[$lastpos].' ';
$arrayfinal = explode(" ", $finalstr);

The code is too complex and you can simplify the code repetions with functions, use it as a base and improve it.

Sorry any bad english and the lack of proper text format, I'm still new to stackoverflow

answered Jul 19, 2018 at 14:44

Pbras

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1 Comment

ondrums Over a year ago

Thank you very much for your effort!

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