How to Count the Consecutive Duplicate Values in an Array

Arraylist find the count of consecutive duplicate elements

A simple approach here would be to just iterate the arraylist once, and then keep tallies as we go along:

List<Integer> list = new ArrayList<>();
list.add(20);
list.add(20);
list.add(30);
list.add(40);
list.add(40);
list.add(20);
list.add(20);
list.add(20);

Integer curr = null;
int count = 0;
System.out.print("{");
for (int val : list) {
    if (curr == null) {
        curr = val;
        count = 1;
    }
    else if (curr != val) {
        System.out.print("(" + curr + ", " + count + ")");
        curr = val;
        count = 1;
    }
    else {
        ++count;
    }
}
System.out.print("(" + curr + ", " + count + ")");
System.out.print("}");

{(20, 2)(30, 1)(40, 2)(20, 3)}

Count consecutive repeated elements in array in the same sequence as they are

You can iterate over the array, counting values while they are the same, and pushing a value to the result array when they are different:

$result = [];
$count = 0;
$current = $array[0];
for ($i = 0; $i < count($array); $i++) {
    if ($array[$i] == $current) {
        $count++;
    }
    else {
        $result[] = array($current, $count);
        $count = 1;
    }
    $current = $array[$i];
}
$result[] = array($current, $count);
print_r($result);

Output:

Array
(
    [0] => Array
        (
            [0] => a
            [1] => 1
        )
    [1] => Array
        (
            [0] => b
            [1] => 1
        )
    [2] => Array
        (
            [0] => a
            [1] => 3
        )
    [3] => Array
        (
            [0] => c
            [1] => 2
        )
    [4] => Array
        (
            [0] => b
            [1] => 2
        )
    [5] => Array
        (
            [0] => a
            [1] => 4
        )
)

Demo on 3v4l.org

Count consecutive duplicate values in ArrayList (Kotlin)

You could manually build the list, like this:

fun count(values: List<String>): List<Group> {
    val groups = mutableListOf<Group>()
    values.forEach {
        val last = groups.lastOrNull()
        if (last?.value == it) {
            last.count++
        } else {
            groups.add(Group(it, 1))
        }
    }
    return groups
}

data class Group(val value: String, var count: Int)

This results in:

[Group(value=training, count=2), Group(value=assessment, count=1), Group(value=training, count=1), Group(value=assessment, count=2)]

Count of consecutive duplicate integers in a list with frequency

I didn't try running this code on an IDE but I think this should suffice:

int count = 1;
int index = 0;

byte current = bytes[index];
while (index < bytes.length - 1) {
    index++;
    if (bytes[index] == current) {
        count++;
    } else {
        integers.add(current);
        frequencies.add(count);
        count = 1;
        current = bytes[index];
    }
}

integers.add(current);
frequencies.add(count);

how to count consecutive duplicates in a python list

You can use itertools.groupby:

from itertools import groupby
list1 = [-1, -1, 1, 1, 1, -1, 1]
count_dups = [sum(1 for _ in group) for _, group in groupby(list1)]
print(count_dups)

Output:

[2, 3, 1, 1]

How to count the consecutive duplicate values in an array?

It can be done simply manually:

$arr = array(1,1,1,2,2,3,3,1,1,2,2,3);

$result = array();
$prev_value = array('value' => null, 'amount' => null);

foreach ($arr as $val) {
    if ($prev_value['value'] != $val) {
        unset($prev_value);
        $prev_value = array('value' => $val, 'amount' => 0);
        $result[] =& $prev_value;
    }

    $prev_value['amount']++;
}

var_dump($result);

Counting consecutive duplicates of strings from a list

Use itertools.groupby:

from itertools import groupby
li = ['aaa','bbb','aaa','abb','abb','bbb','bbb','bbb','aaa','aaa']

a = [[i, sum(1 for i in group)] for i, group in groupby(li)]
print(a)
[['aaa', 1], ['bbb', 1], ['aaa', 1], ['abb', 2], ['bbb', 3], ['aaa', 2]]

Thank you @user3483203 for improvement:

a = [[i, len([*group])] for i, group in groupby(li)]

Count consecutive equal values in array

Here is one option adapted from this answer:

def count_consecutive(arr, n):
    # pad a with False at both sides for edge cases when array starts or ends with n
    d = np.diff(np.concatenate(([False], arr == n, [False])).astype(int))
    # subtract indices when value changes from False to True from indices where value changes from True to False
    return np.flatnonzero(d == -1) - np.flatnonzero(d == 1)

count_consecutive(a, 5)
# array([2, 1, 3])

How to count duplicate value in an array in javascript

function count() {    array_elements = ["a", "b", "c", "d", "e", "a", "b", "c", "f", "g", "h", "h", "h", "e", "a"];
    array_elements.sort();
    var current = null;    var cnt = 0;    for (var i = 0; i < array_elements.length; i++) {        if (array_elements[i] != current) {            if (cnt > 0) {                document.write(current + ' comes --> ' + cnt + ' times<br>');            }            current = array_elements[i];            cnt = 1;        } else {            cnt++;        }    }    if (cnt > 0) {        document.write(current + ' comes --> ' + cnt + ' times');    }
}
count();

Checking for duplicate strings in JavaScript array

The findDuplicates function (below) compares index of all items in array with index of first occurrence of same item. If indexes are not same returns it as duplicate.

let strArray = [ "q", "w", "w", "w", "e", "i", "u", "r"];
let findDuplicates = arr => arr.filter((item, index) => arr.indexOf(item) != index)

console.log(findDuplicates(strArray)) // All duplicates
console.log([...new Set(findDuplicates(strArray))]) // Unique duplicates

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