Checking if an array contains all elements of another array

Question

I'm designing an electrical engineering application. However, i'm stuck on this: I have the following array

<?php 
// Static Array
$GroupOfEight = array (
                      array(0,1,3,2,4,5,7,6),
                      array(4,5,6,7,16,12,13,14),
                      array(12,13,15,14,8,9,11,10),
                      array(2,6,14,10,3,7,15,11),
                      array(1,3,5,7,13,15,9,11),
                      array(0,4,12,8,1,5,13,9),
                      array(0,1,3,2,8,9,11,10)
                      );
?>

And I have another array, but this one is one dimensional.

<?php
$myStack = array(0,1,3,2,4,5,7,6); //Dynamic, gets value by POST method.
?>

What I want to do is to check if $myStack is equal to any sub array of $GroupOfEight array. ( Number ordering is not important. The script should just check if every elements contained. It's not important if their order is same or not. )

Here is what I've done to solve the issue so far:

<?php
//Check if stackArray contains 8group
for($i=0; $i<count($GroupOfEight);$i++)
for($j=0; $j<count($GroupOfEight[$i]); $j++){
    //$containsSearch = count(array_intersect($search_this,$all)) == count($search_this);
    $containsSearch = count(array_intersect($stackArray,$GroupOfEight[$j])) == count($stackArray);
    echo $containsSearch;
}
?>

Please help me correct my code or introduce me the solution of this issue, Thanks.

EDIT: It should give only 1 index number. for example stackArray is 0,1,3,2,4,1,2,3 and it should find GroupOfEight[N] that matches the same numbers, regardless of the order of the numbers. I should get the N if there is a matching case.

have you assigned to $stackarray somewhere? Also, you don't need two loops; one is fine. — user1726343
– user1726343, Commented Oct 12, 2012 at 21:29

score 3 · Accepted Answer · 2012-10-12 22:28:28Z

3

Given your sample arrays, the output of this will be:

> 0

In case you HAD to have only one number output, this should do that:

<?php
//Check if stackArray contains 8group
$check=false;
for($i=0; $i<count($GroupOfEight);$i++){
    //$containsSearch = count(array_intersect($search_this,$all)) == count($search_this);
    $containsSearch = (count(array_intersect($stackArray,$GroupOfEight[$i])) == count($stackArray) && count(array_intersect($stackArray,$GroupOfEight[$i])) == count($GroupOfEight[$i]));
    if($containsSearch && !$check){
        echo $i; //This specifies which index in GroupOfEight contains a matching array
        $check=true;
    }
}
?>

EDIT: Made a function. Returns first matched index or -1 for no matches:

function searcheight($stackArray,$GroupOfEight){
    for($i=0; $i<count($GroupOfEight);$i++){
        $containsSearch = (count(array_intersect($stackArray,$GroupOfEight[$i])) == count($stackArray) && count(array_intersect($stackArray,$GroupOfEight[$i])) == count($GroupOfEight[$i]));
        if($containsSearch){
            return $i; //This specifies which index in GroupOfEight contains a matching array
        }
    }
    return -1;
}
echo searcheight($stackArray,$GroupOfEight);

edited Oct 12, 2012 at 22:28

answered Oct 12, 2012 at 21:31

user1726343

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13 Comments

mozcelikors Over a year ago

seems like the error has disappeared, but let me ask you something. How do i check the $stackArray for each indexes in $GroupOfEight, and return the index number if there is any. That will totally fix my problem and will save me some time. Thanks.

user1726343 Over a year ago

when you run array_intersect, that returns an array filled with the values present in both.

user1726343 Over a year ago

I have edited it to also echo the number. So if you get true0true1false2true3... that means stackArray matches 0,1,3 etc. Obviously you want to do something better than just echo the index.

mozcelikors Over a year ago

Its not what I ask actually, it should give only 1 index number. for example stackArray is 0,1,3,2,4,1,2,3 and it should find GroupOfEight[N] that matches the same numbers, regardless of the order of the numbers. I should get the N if there is a matching case. Please help. Thanks

user1726343 Over a year ago

Yeah there was a problem, I had to check intersections both ways. Fixed now, here is the fiddle

|

Baba · Accepted Answer · 2012-10-12 21:40:18Z

1

You can try :

$searchKeys = array();
foreach ( $GroupOfEight as $key => $values ) {
    (count(array_intersect($values, $myStack)) == count($myStack)) and $searchKeys[] = $key;
}

#Output all keys it found same match
var_dump($searchKeys);

#OR Output Each Array it found a match
foreach($searchKeys as $key)
{
    var_dump($GroupOfEight[$key]);
}

edited Oct 12, 2012 at 21:40

answered Oct 12, 2012 at 21:30

Baba

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8 Comments

mozcelikors Over a year ago

(count(array_intersect($values, $myStack)) == count($myStack)) and $searchKeys[] = $key; This line encounters a problem

Baba Over a year ago

It works fine ... codepad.viper-7.com/M718f0 also see codepad.org/JUZdy1zF .. what verion of PHP are you using ???

mozcelikors Over a year ago

Warning: array_intersect() [function.array-intersect]: Argument #2 is not an array in \scripts\script.php on line 1771 array(0) { } This is the error i got

Baba Over a year ago

Please note that i used $myStack ???? You are having both $stackArray and $myStack ... am sure that is where thing are missed up

mozcelikors Over a year ago

Yes I mistyped myStack, i see. but this is what i seek :it should give only 1 index number. for example stackArray is 0,1,3,2,4,1,2,3 and it should find GroupOfEight[N] that matches the same numbers, regardless of the order of the numbers. I should get only 1 N number if there is a matching case. Please help. Thanks

|

Community · Accepted Answer · 2017-05-23 12:30:55Z

The thing that is wrong with your original approach is that you loop through GroupOfEight twice. You have two for-loops.
First you select every array within GroupOfEight and in the second for loop you go through each value of the array.

If you would like to use your original approach, get rid of the extra for loop:

echo "Hello, World!";
$GroupOfEight = array (
              array(0,1,3,2,4,5,7,6),
              array(4,5,6,7,16,12,13,14),
              array(12,13,15,14,8,9,11,10),
              array(2,6,14,10,3,7,15,11),
              array(1,3,5,7,13,15,9,11),
              array(0,4,12,8,1,5,13,9),
              array(0,1,3,2,8,9,11,10)
              );

$myStack = array(0,1,3,2,4,5,7,6); //Dynamic, gets value by POST method.


for($i=0; $i<count($GroupOfEight);$i++) {       
    $containsSearch = count(array_intersect($myStack,$GroupOfEight[$i])) == count($myStack);
    if($containsSearch===true) {
        echo "Woo! GroupOfEight[$i], <br/>" . print_r($GroupOfEight[$i], true) . "<br/>==<br/>" . print_r($myStack, true);
    }
}

Demo: http://codepad.viper-7.com/0hRNHz

You could accomplish the same with array_diff:

for($i=0; $i<count($GroupOfEight);$i++) {               
    if(count(array_diff($myStack,$GroupOfEight[$i]))==0) {
        echo "Woo! GroupOfEight[$i], <br/>" . print_r($GroupOfEight[$i], true) . "<br/>==<br/>" . print_r($myStack, true);
    }
}

Demo: http://codepad.viper-7.com/6uLd9L

Update
A related SO post is: Check whether two array values are equal (ignoring order)

Riz · Accepted Answer · 2012-10-12 21:33:27Z

0

You are comparing count which is not enough, as numbers could be changed. Try this:

// Static Array
$GroupOfEight = array (
    array(0,1,3,2,4,5,7,6),
    array(4,5,6,7,16,12,13,14),
    array(12,13,15,14,8,9,11,10),
    array(2,6,14,10,3,7,15,11),
    array(1,3,5,7,13,15,9,11),
    array(0,4,12,8,1,5,13,9),
    array(0,1,3,2,8,9,11,10)
    );

$myStack = array(0,1,3,2,4,5,7,6); //Dynamic, gets value by POST method.

$containsSearch = false;
foreach($GroupOfEight as $key => $value){
    if ($myStack == $value) {
        $containsSearch = true;
    }

}

var_dump($containsSearch);

answered Oct 12, 2012 at 21:33

Riz

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Comments

rinchik · Accepted Answer · 2012-10-12 21:34:54Z

0

Count the total of $GroupOfEight[$i]
Count the total of $myStack
if totals are equal:
In a loop -
if $myStack[$c] is in_array($GroupOfEight[$i]): $equal = 1
else $equal = 0; exit;

if $equal == 1 -> arrays are identical

answered Oct 12, 2012 at 21:34

rinchik

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1 Comment

user1386320 Over a year ago

please format your code better and use Code Sample around your code snippets.

hemnath mouli · Accepted Answer · 2015-10-23 13:03:21Z

0

We do not need any loops. Try this

<?php 
$GroupOfEight = array (
                      array(0,1,3,2,4,5,7,6),
                      array(4,5,6,7,16,12,13,14),
                      array(12,13,15,14,8,9,11,10),
                      array(2,6,14,10,3,7,15,11),
                      array(1,3,5,7,13,15,9,11),
                      array(0,4,12,8,1,5,13,9),
                      array(0,1,3,2,8,9,11,10)
                      );
$myStack = array(0,1,3,2,4,5,7,6);

$key = '';
$key = array_search($myStack,$GroupOfEight);
echo $key;
?>

Output

0

Note: Output $key is location of the array in $GroupOfEight i.e ( $GroupOfEight[0] )

answered Oct 23, 2015 at 13:03

hemnath mouli

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