How do I count occurrence of duplicate items in array

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I would like to count the occurrence of each duplicate item in an array and end up with an array of only unique/non duplicate items with their respective occurrences.

Here is my code; BUT I don't where am going wrong!

<?php
$array = array(12,43,66,21,56,43,43,78,78,100,43,43,43,21);

//$previous[value][Occurrence]

for($arr = 0; $arr < count($array); $arr++){

    $current = $array[$arr];
    for($n = 0; $n < count($previous); $n++){
        if($current != $previous[$n][0]){// 12 is not 43 -----> TRUE
            if($current != $previous[count($previous)][0]){
                $previous[$n++][0] = $current;
                $previous[$n++][1] = $counter++;
            }
        }else{  
            $previous[$n][1] = $counter++;
            unset($previous[count($previous)-1][0]);
            unset($previous[count($previous)-1][1]);
        }   
    }
}
//EXPECTED VALUES
echo 'No. of NON Duplicate Items: '.count($previous).'<br><br>';// 7
print_r($previous);// array( {12,1} , {21,2} , {43,6} , {66,1} , {56,1} , {78,2} , {100,1})
?>

Answers

array_count_values, enjoy :-)

$array = array(12,43,66,21,56,43,43,78,78,100,43,43,43,21);
$vals = array_count_values($array);
echo 'No. of NON Duplicate Items: '.count($vals).'<br><br>';
print_r($vals);

Result:

No. of NON Duplicate Items: 7
Array
(
    [12] => 1
    [43] => 6
    [66] => 1
    [21] => 2
    [56] => 1
    [78] => 2
    [100] => 1
)

Thursday, September 1, 2022

emeksedarc1-85339b449b37

You are assigning $count to 0 in your conditional statement

Instead of...

if ($count = 0)

Do this

if ($count === 0)

Saturday, November 19, 2022

mike_fitzpatrick

You can do this:

echo count($food['fruits']);
echo count($food['veggie']);

If you want a more general solution, you can use a foreach loop:

foreach ($food as $type => $list) {
    echo $type." has ".count($list). " elementsn";
}

Wednesday, December 14, 2022

the_name_of_my_cat

You could use a MultiSet from Google Collections/Guava or a Bag from Apache Commons.

If you have a collection instead of an array, you can use addAll() to add the entire contents to the above data structure, and then apply the count() method to each value. A SortedMultiSet or SortedBag would give you the items in a defined order.

Google Collections actually has very convenient ways of going from arrays to a SortedMultiset.

Sunday, October 2, 2022

endyourif

You could use the bitwise AND in python and compare them by converting the list of lists to a list of sets

>>> set(['cat','dog']) & set(['cat','dog','monkey','horse','fish'])
set(['dog', 'cat'])

You could use this property and achieve the count you've wanted.

def listOccurences(item, names):
    # item is the list that you want to check, eg. ['cat','fish']
    # names contain the list of list you have.
    set_of_items = set(item) # set(['cat','fish'])
    count = 0
    for value in names:
        if set_of_items & set(value) == set_of_items:
            count+=1
    return count

names =  [['cat', 'fish'], ['cat'], ['fish', 'dog', 'cat'],['cat', 'bird', 'fish'], ['fish', 'bird']]
# Now for each of your possibilities which you can generate
# Chain flattens the list, set removes duplicates, and combinations generates all possible pairs.
permuted_values = list(itertools.combinations(set(itertools.chain.from_iterable(names)), 2))
d = {}
for v in permuted_values:
    d[str(v)] = listOccurences(v, names)
# The key in the dict being a list cannot be possible unless it's converted to a string.
print(d)
# {"['fish', 'dog']": 1, "['cat', 'dog']": 1, "['cat', 'fish']": 3, "['cat', 'bird']": 1, "['fish', 'bird']": 2}

Friday, December 9, 2022

phil_mccullick

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