Grouping Consecutive Numbers in an Array

Javascript group the numbers from an array with series of consecutive numbers

Iterate with Array#reduce, and whenever the last number is not equal to the new number - 1, add another sub array. Add the current number to the last sub array:

const a = [0,1,2,5,6,9];
const result = a.reduce((r, n) => {  const lastSubArray = r[r.length - 1];    if(!lastSubArray || lastSubArray[lastSubArray.length - 1] !== n - 1) {    r.push([]);  }     r[r.length - 1].push(n);    return r;  }, []);
console.log(result);

group together consecutive numbers in a list

You could use negative indexing:

def group_by_missing(seq):
    if not seq:
        return seq
    grouped = [[seq[0]]]
    for x in seq[1:]:
        if x == grouped[-1][-1] + 1:
            grouped[-1].append(x)
        else:
            grouped.append([x])
    return grouped

Example Usage:

>>> lst = [1, 2, 3, 4, 5, 12, 13, 14, 15, 20, 21, 22, 23, 30, 35, 36, 37, 38, 39, 40]
>>> group_by_missing(lst)
[[1, 2, 3, 4, 5], [12, 13, 14, 15], [20, 21, 22, 23], [30], [35, 36, 37, 38, 39, 40]]

How to count group of consecutive numbers in array

In the future, try to include more info on what you've tried or what you've researched. As Azro said, we're here to help, not complete tasks for you.

int[] myArr = {1,2,3,4,9,12,17,23,34,54,55,56};
int consecutiveSums = 0;

// Iterate through the array
for (int i = 0; i < myArr.length; i++) {
    // Check if the next value is consecutive
    if (i + 1 < myArr.length && myArr[i] + 1 == myArr[i+1]) {
        consecutiveSums ++;
        // Skip any remaining consecutives until we reach a new set
        while (i + 1 < myArr.length && myArr[i] + 1 == myArr[i+1]) {
            i++;
        }
    }
}

System.out.println(consecutiveSums);

Group the consecutive numbers from an array with series of numbers which are greater than 2 into subarrays using javascript

You can use a global variable to keep track of the indexes so that it's possible to detect when exists a gap between two numbers, which allows you to add another array. In combination, using ternary operators can make the code cleaner to test for conditions.

const arr = [1, 2, 3, 5, 6, 7, 8, 10, 0, 1, 1, 2, 4, 10, 6, 7, 3];
let lastIdx = -1
const results = arr.reduce((acc, current, idx) => 
     (
        current > 2 ? (
            lastIdx + 1 === idx && acc.length ?
                (
                    lastIdx = idx,
                    acc[acc.length - 1].push(current), 
                    acc
                )
                :
                (
                    lastIdx = idx,
                    [...acc, [current]]
                )
        ) : acc
    )
, [])

console.log(results);

Best way to loop through array and group consecutive numbers in another array SWIFT 4?

My suggestion is IndexSet where consecutive items are stored as ranges.

Create the index set from the array.
Get the rangeView.
Map the ranges to arrays.

let myNumbersArray = [1,2,3,4,10,11,15,20,21,22,23]
let indexSet = IndexSet(myNumbersArray)
let rangeView = indexSet.rangeView
let newNumbersArray = rangeView.map { Array($0.indices) }

Grouping consecutive elements together using Javascript

You can use a counter variable which has to be incremented and the difference between the index and the consecutive elements are the same, group them in a temporary array. If the difference is varies for two consecutive array elements, the temporary element has to be moved to the result and the temporary array has to be assigned a new array object.

var array = [2, 3, 4, 5, 8, 9, 12, 13, 14, 15, 16, 17, 20];

var result = [], temp = [], difference;
for (var i = 0; i < array.length; i += 1) {
    if (difference !== (array[i] - i)) {
        if (difference !== undefined) {
            result.push(temp);
            temp = [];
        }
        difference = array[i] - i;
    }
    temp.push(array[i]);
}

if (temp.length) {
    result.push(temp);
}

console.log(result);
# [ [ 2, 3, 4, 5 ], [ 8, 9 ], [ 12, 13, 14, 15, 16, 17 ], [ 20 ] ]

Identify groups of continuous numbers in a list

more_itertools.consecutive_groups was added in version 4.0.

Demo

import more_itertools as mit

iterable = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
[list(group) for group in mit.consecutive_groups(iterable)]
# [[2, 3, 4, 5], [12, 13, 14, 15, 16, 17], [20]]

Code

Applying this tool, we make a generator function that finds ranges of consecutive numbers.

def find_ranges(iterable):
    """Yield range of consecutive numbers."""
    for group in mit.consecutive_groups(iterable):
        group = list(group)
        if len(group) == 1:
            yield group[0]
        else:
            yield group[0], group[-1]

iterable = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20]
list(find_ranges(iterable))
# [(2, 5), (12, 17), 20]

The source implementation emulates a classic recipe (as demonstrated by @Nadia Alramli).

Note: more_itertools is a third-party package installable via pip install more_itertools.

How to group consecutive data in 2d array in python

Here you go ...

Steps are:

Get a list xUnique of unique column 1 values with sort order preserved.
Build a list xRanges of items of the form [col1_value, [col2_min, col2_max]] holding the column 2 ranges for each column 1 value.
Build a list xGroups of items of the form [[col1_min, col1_max], [col2_min, col2_max]] where the [col1_min, col1_max] part is created by merging the col1_value part of consecutive items in xRanges if they differ by 1 and have identical [col2_min, col2_max] value ranges for column 2.
Turn the ranges in each item of xGroups into strings and print with the required row and column headings.
Also package and print as a numpy.array to match the form of the input.

import numpy as np
data = np.array([
    [1, 1],
    [1, 2],    
    [2, 1],    
    [2, 2],
    [3, 1],
    [5, 1],
    [5, 2]])
xUnique = list({pair[0] for pair in data})
xRanges = list(zip(xUnique, [[0, 0] for _ in range(len(xUnique))]))
rows, cols = data.shape
iRange = -1
for i in range(rows):
    if i == 0 or data[i, 0] > data[i - 1, 0]:
        iRange += 1
        xRanges[iRange][1][0] = data[i, 1]
    xRanges[iRange][1][1] = data[i, 1]
xGroups = []
for i in range(len(xRanges)):
    if i and xRanges[i][0] - xRanges[i - 1][0] == 1 and xRanges[i][1] == xRanges[i - 1][1]:
        xGroups[-1][0][1] = xRanges[i][0]
    else:
        xGroups += [[[xRanges[i][0], xRanges[i][0]], xRanges[i][1]]]

xGroupStrs = [ [f'{a}-{b}' for a, b in row] for row in xGroups]

groupArray = np.array(xGroupStrs)
print(groupArray)

print()
print(f'{"":<10}{"Col1":<8}{"Col2":<8}')
[print(f'{"group " + str(i) + ":":<10}{col1:<8}{col2:<8}') for i, (col1, col2) in enumerate(xGroupStrs)]

Output:

[['1-2' '1-2']
 ['3-3' '1-1']
 ['5-5' '1-2']]

          Col1    Col2
group 0:  1-2     1-2
group 1:  3-3     1-1
group 2:  5-5     1-2