What is the process for forming a series of arrays from one singular array?

Question

What is the process for forming a series of arrays from one singular array?

If I have a large array consisting of numbers from 1 to 18, is there a simple method to split it into pairs like [1,2], [3,4], [5,6], [7,8], [9,10], [11,12], [13,14] for n=2? The special case of n=2 is all I need.

javascript arrays underscore.js

Answer 1

Answer №1

Here is a solution that should do the trick:

for (let i = 0; i < arr.length; i += 2) {
  result.push([arr[i], arr[i + 1]]);
}

I came up with a more versatile approach that can handle any number of "pockets" or whatever you might call them. It takes care of handling undefined values for odd numbers of items:

Array.prototype.pockets = function(n) {

  let result = [],
      pocket = [],
      i, j;

  for (i = 0; i < this.length; i += n) {
    pocket.length = 0;
    for (j = 1; j < n; j++) {
      if (this[i + j] !== null) {
        pocket.push(this[i + j]);
      }
    }
    result.push([this[i]].concat(pocket));
  }

  if (arguments.length > 1) {
    return result.pockets.apply(result, [].slice.call(arguments, 1));
  }

  return result;
};

// Example of usage:
let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];

arr.pockets(2); //=> [[1, 2], [3, 4], [5, 6], [7, 8], [9, 10], [11]]
arr.pockets(3); //=> [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11]]

// Recursive:
arr.pockets(1, 3); //=> [ [[1], [2], [3]], [[4], [5], [6]], [[7], [8], [9]], [[10], [11]] ]

Answer 2

Here is a solution that should do the trick:

for (let i = 0; i < arr.length; i += 2) {
  result.push([arr[i], arr[i + 1]]);
}

I came up with a more versatile approach that can handle any number of "pockets" or whatever you might call them. It takes care of handling undefined values for odd numbers of items:

Array.prototype.pockets = function(n) {

  let result = [],
      pocket = [],
      i, j;

  for (i = 0; i < this.length; i += n) {
    pocket.length = 0;
    for (j = 1; j < n; j++) {
      if (this[i + j] !== null) {
        pocket.push(this[i + j]);
      }
    }
    result.push([this[i]].concat(pocket));
  }

  if (arguments.length > 1) {
    return result.pockets.apply(result, [].slice.call(arguments, 1));
  }

  return result;
};

// Example of usage:
let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];

arr.pockets(2); //=> [[1, 2], [3, 4], [5, 6], [7, 8], [9, 10], [11]]
arr.pockets(3); //=> [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11]]

// Recursive:
arr.pockets(1, 3); //=> [ [[1], [2], [3]], [[4], [5], [6]], [[7], [8], [9]], [[10], [11]] ]

Answer 3

Answer №2

A simpler way to achieve this task is by utilizing the Array.slice method:

function groupItems(arr, chunkSize) {
    var result = [], index = 0, length = arr.length;
    while(index < length) {
        result.push(arr.slice(index, index + chunkSize));
        index += chunkSize;
    }
    return result;
}

This method is not only simpler but also more efficient. You can check out the performance comparison here: http://jsperf.com/groupItems

Answer 4

A simpler way to achieve this task is by utilizing the Array.slice method:

function groupItems(arr, chunkSize) {
    var result = [], index = 0, length = arr.length;
    while(index < length) {
        result.push(arr.slice(index, index + chunkSize));
        index += chunkSize;
    }
    return result;
}

This method is not only simpler but also more efficient. You can check out the performance comparison here: http://jsperf.com/groupItems

Answer 5

Answer №3

To achieve the underscore variant, you can utilize _.groupBy() and group by the index of the item:

var doubles = _.groupBy(singles, function (num, i) {
    return Math.floor(i / 2);
});

However, since _.groupBy() returns an Object, converting it to an Array requires some additional steps:

_.mixin({
    segment: function (coll, per) {
        var result = [];
        _.chain(coll)
            .groupBy(function (item, i) { return Math.floor(i / per)})
            .each(function (group, key) { result[key] = group; })
        return result;
    }
});

var singles = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18];

var doubles = _.segment(singles, 2);
var triples = _.segment(singles, 3);

Answer 6

To achieve the underscore variant, you can utilize _.groupBy() and group by the index of the item:

var doubles = _.groupBy(singles, function (num, i) {
    return Math.floor(i / 2);
});

However, since _.groupBy() returns an Object, converting it to an Array requires some additional steps:

_.mixin({
    segment: function (coll, per) {
        var result = [];
        _.chain(coll)
            .groupBy(function (item, i) { return Math.floor(i / per)})
            .each(function (group, key) { result[key] = group; })
        return result;
    }
});

var singles = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18];

var doubles = _.segment(singles, 2);
var triples = _.segment(singles, 3);

Answer 7

Answer №4

To achieve this in Python, one can use the zip(*[iter(xs)]*n) method. As a fun exercise, below is an implementation in JavaScript:

Let's begin with a basic generator function (until ES6 becomes more widely used):

StopIteration = {"name": "StopIteration"}

function iter(xs) {
    if('next' in xs)
        return xs;
    var i = 0;
    return {
        next: function() {
            if(i >= xs.length)
                throw StopIteration;
            return xs[i++];
        }
    }
}

next = function(it) { return it.next() }

The implementation of zip() is quite straightforward:

zip = function() {
    var args = [].map.call(arguments, iter), chunks = [];
    while(1) {
        try {
            chunks.push(args.map(next));
        } catch(StopIteration) {
            return chunks;
        }
    }
}

To create pairs that are linked, simply pass the same iter function twice to zip:

xs = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

it = iter(xs)
a = zip(it, it)

console.log(a)
// [[1,2],[3,4],[5,6],[7,8],[9,10],[11,12]]

For creating N-pairs, an additional utility function is needed:

repeat = function(x, n) {
    for(var a = []; n; n--)
        a.push(x);
    return a;
}

a = zip.apply(this, repeat(iter(xs), 5))

console.log(a) 
// [[1,2,3,4,5],[6,7,8,9,10]]

It's important to note that like in Python, incomplete chunks are trimmed.

Answer 8

To achieve this in Python, one can use the zip(*[iter(xs)]*n) method. As a fun exercise, below is an implementation in JavaScript:

Let's begin with a basic generator function (until ES6 becomes more widely used):

StopIteration = {"name": "StopIteration"}

function iter(xs) {
    if('next' in xs)
        return xs;
    var i = 0;
    return {
        next: function() {
            if(i >= xs.length)
                throw StopIteration;
            return xs[i++];
        }
    }
}

next = function(it) { return it.next() }

The implementation of zip() is quite straightforward:

zip = function() {
    var args = [].map.call(arguments, iter), chunks = [];
    while(1) {
        try {
            chunks.push(args.map(next));
        } catch(StopIteration) {
            return chunks;
        }
    }
}

To create pairs that are linked, simply pass the same iter function twice to zip:

xs = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

it = iter(xs)
a = zip(it, it)

console.log(a)
// [[1,2],[3,4],[5,6],[7,8],[9,10],[11,12]]

For creating N-pairs, an additional utility function is needed:

repeat = function(x, n) {
    for(var a = []; n; n--)
        a.push(x);
    return a;
}

a = zip.apply(this, repeat(iter(xs), 5))

console.log(a) 
// [[1,2,3,4,5],[6,7,8,9,10]]

It's important to note that like in Python, incomplete chunks are trimmed.

Answer 9

Answer №5

If you want to split an array into groups using reduce and Modulus (%), here's how you can achieve it:

const arrayGrouper = (arr, size) => arr.reduce((acc, val, i) => {
        i % size === 0 ?
            acc.push([val])
            :
            acc[acc.length - 1].push(val);
        return acc;
    }, []);

let groupedArray = arrayGrouper([10, 20, 30, 40, 50, 60], 3);

console.log(groupedArray);
//[[10, 20, 30], [40, 50, 60]]

For more information on JavaScript reduce, you can visit this link.

Answer 10

If you want to split an array into groups using reduce and Modulus (%), here's how you can achieve it:

const arrayGrouper = (arr, size) => arr.reduce((acc, val, i) => {
        i % size === 0 ?
            acc.push([val])
            :
            acc[acc.length - 1].push(val);
        return acc;
    }, []);

let groupedArray = arrayGrouper([10, 20, 30, 40, 50, 60], 3);

console.log(groupedArray);
//[[10, 20, 30], [40, 50, 60]]

For more information on JavaScript reduce, you can visit this link.

What is the process for forming a series of arrays from one singular array?

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Similar questions

Limit the selection of 'pickable' attributes following selections in the picking function (TypeScript)

Unable to manipulate the marker with the leaflet library

What is the best way to export a React Material-UI component with two separate styles objects in Material-UI V1?

How can I implement a page switch in vuejs by clicking on a name in a table list?

Is Asp.net 4 resetting the array with every new run?

An issue has been detected in the constants.json file located in the constants-browserify

Guide on executing a Python script using Node.js

Tips on how to remove the first column from a jQuery hover effect

Having trouble seeing the filtered results in the React Redux search filter?

Enhancing my Javascript code by adjusting the styling of a link within the page navigation bar

Transform an array containing objects into a single object

What could be causing appendChild to malfunction?

Transform ISO-8859-1 encoding into UTF-8

How can PHP be used to iterate through numbers and retrieve letters?

Error: Unable to locate module '@emotion/react' in 'E:frontend ode_modules@muistyled-engine' directory

"Adding an Image within a Textbox Using Ext JS: A Step-by-Step

Adjusting Headers Using Buttons

"Using a PHP array with Highcharts for dynamic data visualization

Hear and register keypress in Angular service

Intersection observer automatically removes images from carousel (Siema) after they have been viewed