We Keep Coding

How to double the values in odd possitions on an array, and return the full array? [duplicate]

I have some array of numbers:
var arr = [1, 7, 1, 4];
I want to increase only every first value, such that the expected output would be: [2, 7, 2, 4]
I tried some combination of map and filter but I don't understand how it can work together...
var mapuj = arr.map(x => x *2);

You can use map() and use second argument which is idnex to determine if it's at event index or not
let arr = [1, 7, 1, 4];
let output = arr.map((n, index) => index % 2 === 0 ? n * 2 : n);
console.log(output);

Counting ocurrencies / frequency of given value in a array inside object in an array of objects [duplicate]

In Javascript, I'm trying to take an initial array of number values and count the elements inside it. Ideally, the result would be two new arrays, the first specifying each unique element, and the second containing the number of times each element occurs. However, I'm open to suggestions on the format of the output.
For example, if the initial array was:
5, 5, 5, 2, 2, 2, 2, 2, 9, 4
Then two new arrays would be created. The first would contain the name of each unique element:
5, 2, 9, 4
The second would contain the number of times that element occurred in the initial array:
3, 5, 1, 1
Because the number 5 occurs three times in the initial array, the number 2 occurs five times and 9 and 4 both appear once.
I've searched a lot for a solution, but nothing seems to work, and everything I've tried myself has wound up being ridiculously complex. Any help would be appreciated!
Thanks :)

You can use an object to hold the results:
const arr = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
const counts = {};
for (const num of arr) {
counts[num] = counts[num] ? counts[num] + 1 : 1;
}
console.log(counts);
console.log(counts[5], counts[2], counts[9], counts[4]);
So, now your counts object can tell you what the count is for a particular number:
console.log(counts[5]); // logs '3'
If you want to get an array of members, just use the keys() functions
keys(counts); // returns ["5", "2", "9", "4"]

const occurrences = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4].reduce(function (acc, curr) {
return acc[curr] ? ++acc[curr] : acc[curr] = 1, acc
}, {});
console.log(occurrences) // => {2: 5, 4: 1, 5: 3, 9: 1}

const arr = [2, 2, 5, 2, 2, 2, 4, 5, 5, 9];
function foo (array) {
let a = [],
b = [],
arr = [...array], // clone array so we don't change the original when using .sort()
prev;
arr.sort();
for (let element of arr) {
if (element !== prev) {
a.push(element);
b.push(1);
}
else ++b[b.length - 1];
prev = element;
}
return [a, b];
}
const result = foo(arr);
console.log('[' + result[0] + ']','[' + result[1] + ']')
console.log(arr)

One line ES6 solution. So many answers using object as a map but I can't see anyone using an actual Map
const map = arr.reduce((acc, e) => acc.set(e, (acc.get(e) || 0) + 1), new Map());
Use map.keys() to get unique elements
Use map.values() to get the occurrences
Use map.entries() to get the pairs [element, frequency]
var arr = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]
const map = arr.reduce((acc, e) => acc.set(e, (acc.get(e) || 0) + 1), new Map());
console.info([...map.keys()])
console.info([...map.values()])
console.info([...map.entries()])

If using underscore or lodash, this is the simplest thing to do:
_.countBy(array);
Such that:
_.countBy([5, 5, 5, 2, 2, 2, 2, 2, 9, 4])
=> Object {2: 5, 4: 1, 5: 3, 9: 1}
As pointed out by others, you can then execute the _.keys() and _.values() functions on the result to get just the unique numbers, and their occurrences, respectively. But in my experience, the original object is much easier to deal with.

Don't use two arrays for the result, use an object:
a = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
result = { };
for(var i = 0; i < a.length; ++i) {
if(!result[a[i]])
result[a[i]] = 0;
++result[a[i]];
}
Then result will look like:
{
2: 5,
4: 1,
5: 3,
9: 1
}

How about an ECMAScript2015 option.
const a = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
const aCount = new Map([...new Set(a)].map(
x => [x, a.filter(y => y === x).length]
));
aCount.get(5) // 3
aCount.get(2) // 5
aCount.get(9) // 1
aCount.get(4) // 1
This example passes the input array to the Set constructor creating a collection of unique values. The spread syntax then expands these values into a new array so we can call map and translate this into a two-dimensional array of [value, count] pairs - i.e. the following structure:
Array [
[5, 3],
[2, 5],
[9, 1],
[4, 1]
]
The new array is then passed to the Map constructor resulting in an iterable object:
Map {
5 => 3,
2 => 5,
9 => 1,
4 => 1
}
The great thing about a Map object is that it preserves data-types - that is to say aCount.get(5) will return 3 but aCount.get("5") will return undefined. It also allows for any value / type to act as a key meaning this solution will also work with an array of objects.
function frequencies(/* {Array} */ a){
return new Map([...new Set(a)].map(
x => [x, a.filter(y => y === x).length]
));
}
let foo = { value: 'foo' },
bar = { value: 'bar' },
baz = { value: 'baz' };
let aNumbers = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4],
aObjects = [foo, bar, foo, foo, baz, bar];
frequencies(aNumbers).forEach((val, key) => console.log(key + ': ' + val));
frequencies(aObjects).forEach((val, key) => console.log(key.value + ': ' + val));

I think this is the simplest way how to count occurrences with same value in array.
var a = [true, false, false, false];
a.filter(function(value){
return value === false;
}).length

const data = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]
function count(arr) {
return arr.reduce((prev, curr) => (prev[curr] = ++prev[curr] || 1, prev), {})
}
console.log(count(data))

2021's version
The more elegant way is using Logical nullish assignment (x ??= y) combined with Array#reduce() with O(n) time complexity.
The main idea is still using Array#reduce() to aggregate with output as object to get the highest performance (both time and space complexity) in terms of searching & construct bunches of intermediate arrays like other answers.
const arr = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];
const result = arr.reduce((acc, curr) => {
acc[curr] ??= {[curr]: 0};
acc[curr][curr]++;
return acc;
}, {});
console.log(Object.values(result));
Clean & Refactor code
Using Comma operator (,) syntax.
The comma operator (,) evaluates each of its operands (from left to
right) and returns the value of the last operand.
const arr = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];
const result = arr.reduce((acc, curr) => (acc[curr] = (acc[curr] || 0) + 1, acc), {});
console.log(result);
Output
{
"2": 5,
"4": 1,
"5": 3,
"9": 1
}

If you favour a single liner.
arr.reduce(function(countMap, word) {countMap[word] = ++countMap[word] || 1;return countMap}, {});
Edit (6/12/2015):
The Explanation from the inside out.
countMap is a map that maps a word with its frequency, which we can see the anonymous function. What reduce does is apply the function with arguments as all the array elements and countMap being passed as the return value of the last function call. The last parameter ({}) is the default value of countMap for the first function call.

ES6 version should be much simplifier (another one line solution)
let arr = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
let acc = arr.reduce((acc, val) => acc.set(val, 1 + (acc.get(val) || 0)), new Map());
console.log(acc);
// output: Map { 5 => 3, 2 => 5, 9 => 1, 4 => 1 }
A Map instead of plain Object helping us to distinguish different type of elements, or else all counting are base on strings

Edit 2020: this is a pretty old answer (nine years). Extending the native prototype will always generate discussion. Although I think the programmer is free to choose her own programming style, here's a (more modern) approach to the problem without extending Array.prototype:
{
// create array with some pseudo random values (1 - 5)
const arr = Array.from({length: 100})
.map( () => Math.floor(1 + Math.random() * 5) );
// frequencies using a reducer
const arrFrequencies = arr.reduce((acc, value) =>
({ ...acc, [value]: acc[value] + 1 || 1}), {} )
console.log(arrFrequencies);
console.log(`Value 4 occurs ${arrFrequencies[4]} times in arrFrequencies`);
// bonus: restore Array from frequencies
const arrRestored = Object.entries(arrFrequencies)
.reduce( (acc, [key, value]) => acc.concat(Array(value).fill(+key)), [] );
console.log(arrRestored.join());
}
.as-console-wrapper { top: 0; max-height: 100% !important; }
The old (2011) answer: you could extend Array.prototype, like this:
{
Array.prototype.frequencies = function() {
var l = this.length,
result = {
all: []
};
while (l--) {
result[this[l]] = result[this[l]] ? ++result[this[l]] : 1;
}
// all pairs (label, frequencies) to an array of arrays(2)
for (var l in result) {
if (result.hasOwnProperty(l) && l !== 'all') {
result.all.push([l, result[l]]);
}
}
return result;
};
var freqs = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4].frequencies();
console.log(`freqs[2]: ${freqs[2]}`); //=> 5
// or
var freqs = '1,1,2,one,one,2,2,22,three,four,five,three,three,five'
.split(',')
.frequencies();
console.log(`freqs.three: ${freqs.three}`); //=> 3
// Alternatively you can utilize Array.map:
Array.prototype.frequencies = function() {
var freqs = {
sum: 0
};
this.map(function(a) {
if (!(a in this)) {
this[a] = 1;
} else {
this[a] += 1;
}
this.sum += 1;
return a;
}, freqs);
return freqs;
}
}
.as-console-wrapper { top: 0; max-height: 100% !important; }

Based on answer of #adamse and #pmandell (which I upvote), in ES6 you can do it in one line:
2017 edit: I use || to reduce code size and make it more readable.
var a=[7,1,7,2,2,7,3,3,3,7,,7,7,7];
alert(JSON.stringify(
a.reduce((r,k)=>{r[k]=1+r[k]||1;return r},{})
));
It can be used to count characters:
var s="ABRACADABRA";
alert(JSON.stringify(
s.split('').reduce((a, c)=>{a[c]++?0:a[c]=1;return a},{})
));

A shorter version using reduce and tilde (~) operator.
const data = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];
function freq(nums) {
return nums.reduce((acc, curr) => {
acc[curr] = -~acc[curr];
return acc;
}, {});
}
console.log(freq(data));

If you are using underscore you can go the functional route
a = ['foo', 'foo', 'bar'];
var results = _.reduce(a,function(counts,key){ counts[key]++; return counts },
_.object( _.map( _.uniq(a), function(key) { return [key, 0] })))
so your first array is
_.keys(results)
and the second array is
_.values(results)
most of this will default to native javascript functions if they are available
demo : http://jsfiddle.net/dAaUU/

So here's how I'd do it with some of the newest javascript features:
First, reduce the array to a Map of the counts:
let countMap = array.reduce(
(map, value) => {map.set(value, (map.get(value) || 0) + 1); return map},
new Map()
)
By using a Map, your starting array can contain any type of object, and the counts will be correct. Without a Map, some types of objects will give you strange counts.
See the Map docs for more info on the differences.
This could also be done with an object if all your values are symbols, numbers, or strings:
let countObject = array.reduce(
(map, value) => { map[value] = (map[value] || 0) + 1; return map },
{}
)
Or slightly fancier in a functional way without mutation, using destructuring and object spread syntax:
let countObject = array.reduce(
(value, {[value]: count = 0, ...rest}) => ({ [value]: count + 1, ...rest }),
{}
)
At this point, you can use the Map or object for your counts (and the map is directly iterable, unlike an object), or convert it to two arrays.
For the Map:
countMap.forEach((count, value) => console.log(`value: ${value}, count: ${count}`)
let values = countMap.keys()
let counts = countMap.values()
Or for the object:
Object
.entries(countObject) // convert to array of [key, valueAtKey] pairs
.forEach(([value, count]) => console.log(`value: ${value}, count: ${count}`)
let values = Object.keys(countObject)
let counts = Object.values(countObject)

var array = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
function countDuplicates(obj, num){
obj[num] = (++obj[num] || 1);
return obj;
}
var answer = array.reduce(countDuplicates, {});
// answer => {2:5, 4:1, 5:3, 9:1};
If you still want two arrays, then you could use answer like this...
var uniqueNums = Object.keys(answer);
// uniqueNums => ["2", "4", "5", "9"];
var countOfNums = Object.keys(answer).map(key => answer[key]);
// countOfNums => [5, 1, 3, 1];
Or if you want uniqueNums to be numbers
var uniqueNums = Object.keys(answer).map(key => +key);
// uniqueNums => [2, 4, 5, 9];

Here's just something light and easy for the eyes...
function count(a,i){
var result = 0;
for(var o in a)
if(a[o] == i)
result++;
return result;
}
Edit: And since you want all the occurences...
function count(a){
var result = {};
for(var i in a){
if(result[a[i]] == undefined) result[a[i]] = 0;
result[a[i]]++;
}
return result;
}

Solution using a map with O(n) time complexity.
var arr = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];
const countOccurrences = (arr) => {
const map = {};
for ( var i = 0; i < arr.length; i++ ) {
map[arr[i]] = ~~map[arr[i]] + 1;
}
return map;
}
Demo: http://jsfiddle.net/simevidas/bnACW/

There is a much better and easy way that we can do this using ramda.js.
Code sample here
const ary = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
R.countBy(r=> r)(ary)
countBy documentation is at documentation

I know this question is old but I realized there are too few solutions where you get the count array as asked with a minimal code so here is mine
// The initial array we want to count occurences
var initial = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
// The count array asked for
var count = Array.from(new Set(initial)).map(val => initial.filter(v => v === val).length);
// Outputs [ 3, 5, 1, 1 ]
Beside you can get the set from that initial array with
var set = Array.from(new Set(initial));
//set = [5, 2, 9, 4]

My solution with ramda:
const testArray = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]
const counfFrequency = R.compose(
R.map(R.length),
R.groupBy(R.identity),
)
counfFrequency(testArray)
Link to REPL.

Using MAP you can have 2 arrays in the output: One containing the occurrences & the other one is containing the number of occurrences.
const dataset = [2,2,4,2,6,4,7,8,5,6,7,10,10,10,15];
let values = [];
let keys = [];
var mapWithOccurences = dataset.reduce((a,c) => {
if(a.has(c)) a.set(c,a.get(c)+1);
else a.set(c,1);
return a;
}, new Map())
.forEach((value, key, map) => {
keys.push(key);
values.push(value);
});
console.log(keys)
console.log(values)

This question is more than 8 years old and many, many answers do not really take ES6 and its numerous advantages into account.
Perhaps is even more important to think about the consequences of our code for garbage collection/memory management whenever we create additional arrays, make double or triple copies of arrays or even convert arrays into objects. These are trivial observations for small applications but if scale is a long term objective then think about these, thoroughly.
If you just need a "counter" for specific data types and the starting point is an array (I assume you want therefore an ordered list and take advantage of the many properties and methods arrays offer), you can just simply iterate through array1 and populate array2 with the values and number of occurrences for these values found in array1.
As simple as that.
Example of simple class SimpleCounter (ES6) for Object Oriented Programming and Object Oriented Design
class SimpleCounter {
constructor(rawList){ // input array type
this.rawList = rawList;
this.finalList = [];
}
mapValues(){ // returns a new array
this.rawList.forEach(value => {
this.finalList[value] ? this.finalList[value]++ : this.finalList[value] = 1;
});
this.rawList = null; // remove array1 for garbage collection
return this.finalList;
}
}
module.exports = SimpleCounter;

Using Lodash
const values = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
const frequency = _.map(_.groupBy(values), val => ({ value: val[0], frequency: val.length }));
console.log(frequency);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.min.js"></script>

To return an array which is then sortable:
let array = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]
let reducedArray = array.reduce( (acc, curr, _, arr) => {
if (acc.length == 0) acc.push({item: curr, count: 1})
else if (acc.findIndex(f => f.item === curr ) === -1) acc.push({item: curr, count: 1})
else ++acc[acc.findIndex(f => f.item === curr)].count
return acc
}, []);
console.log(reducedArray.sort((a,b) => b.count - a.count ))
/*
Output:
[
{
"item": 2,
"count": 5
},
{
"item": 5,
"count": 3
},
{
"item": 9,
"count": 1
},
{
"item": 4,
"count": 1
}
]
*/

Check out the code below.
<html>
<head>
<script>
// array with values
var ar = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
var Unique = []; // we'll store a list of unique values in here
var Counts = []; // we'll store the number of occurances in here
for(var i in ar)
{
var Index = ar[i];
Unique[Index] = ar[i];
if(typeof(Counts[Index])=='undefined')
Counts[Index]=1;
else
Counts[Index]++;
}
// remove empty items
Unique = Unique.filter(function(){ return true});
Counts = Counts.filter(function(){ return true});
alert(ar.join(','));
alert(Unique.join(','));
alert(Counts.join(','));
var a=[];
for(var i=0; i<Unique.length; i++)
{
a.push(Unique[i] + ':' + Counts[i] + 'x');
}
alert(a.join(', '));
</script>
</head>
<body>
</body>
</html>

function countOcurrences(arr){
return arr.reduce((aggregator, value, index, array) => {
if(!aggregator[value]){
return aggregator = {...aggregator, [value]: 1};
}else{
return aggregator = {...aggregator, [value]:++aggregator[value]};
}
}, {})
}

You can simplify this a bit by extending your arrays with a count function. It works similar to Ruby’s Array#count, if you’re familiar with it.
Array.prototype.count = function(obj){
var count = this.length;
if(typeof(obj) !== "undefined"){
var array = this.slice(0), count = 0; // clone array and reset count
for(i = 0; i < array.length; i++){
if(array[i] == obj){ count++ }
}
}
return count;
}
Usage:
let array = ['a', 'b', 'd', 'a', 'c'];
array.count('a'); // => 2
array.count('b'); // => 1
array.count('e'); // => 0
array.count(); // => 5
Gist
Edit
You can then get your first array, with each occurred item, using Array#filter:
let occurred = [];
array.filter(function(item) {
if (!occurred.includes(item)) {
occurred.push(item);
return true;
}
}); // => ["a", "b", "d", "c"]
And your second array, with the number of occurrences, using Array#count into Array#map:
occurred.map(array.count.bind(array)); // => [2, 1, 1, 1]
Alternatively, if order is irrelevant, you can just return it as a key-value pair:
let occurrences = {}
occurred.forEach(function(item) { occurrences[item] = array.count(item) });
occurences; // => {2: 5, 4: 1, 5: 3, 9: 1}

I am not getting how to do it for a case when we have to remove elements listed in another array

function removeListedValues(arr) {
var what, a = arguments, L = a.length, ax;
while (L > 1 && arr.length) {
what = a[--L];
while ((ax= arr.indexOf(what)) !== -1) {
arr.splice(ax, 1);
}
}
return arr;
}
arr: The given array
without: A list of elements which are to be removed from arr.
Return the array after removing the listed values.
Input:
arr: [1, 2, 2, 3, 1, 2]
without: [2, 3]
Output:
[1, 1]

To remove something in array, suggest using .filter
const input = [1, 2, 2, 3, 1, 2];
const without = [2, 3];
const result = input.filter(value => !without.includes(value))
console.log(result)

You can create a Set (for faster lookup) from the array of values to exclude and use Array#filter along with Set#has.
const arr = [1, 2, 2, 3, 1, 2], without = new Set([2, 3]);
const res = arr.filter(x => !without.has(x));
console.log(res);

How to reorder multiple elements in an array?

I have an array arr:
arr = [0,1,2,3,4,5]
How can I reorder/ shift multiple elements? For example,
I want to remove 0,2 and 3 and put them at index 4 so that the final array should be [1,0,2,3,4,5]
In the above example, index 4 is with the perspective of the original array, not the final array.
I tried using splice like this:
items = [0,2,3] // items to be reordered
indexes = [0,2,3] // get their indexes
arr = [...arr.filter((it)=> !indexes.includes(arr.indexOf(it)))]
arr.splice(4, 0, ...items)
// result [1, 4, 5, 0, 2, 3]
The above is not the intended result

This solution mutates the array.
You could store the value at the inser position and remove the items and splice the removed items after the index of the stored item.
position v
index 0 1 2 3 4
array [0, 1, 2, 3, 4, 5] store value at index
1 4 5 rest array after removing items
1 4 [0 2 3] 5 splice with removed items
var array = [0, 1, 2, 3, 4, 5],
remove = [0, 2, 3],
insertAtIndex = 4,
valueAtPosition = array[insertAtIndex];
remove.forEach(v => array.splice(array.indexOf(v), 1));
array.splice(array.indexOf(valueAtPosition) + 1, 0, ...remove);
console.log(...array);

You can first remove the given elements and then use splice() to add them at the required index.
function shiftElementsTo(arr, inds, final){
let set = new Set(inds);
let removed = inds.map(x => arr[x]);
arr = arr.filter((x, i) => !set.has(i));
arr.splice(final, 0, ...removed);
return arr;
}
console.log(shiftElementsTo([0, 1, 2, 3, 4, 5], [0, 2, 3], 2))

Try to use this approach:
let arr = [0,1,2,3,4,5];
let rmv = [0, 2, 3];
const remove = (src, rem, i ) => {
const arrWithIndexes = src.map((a, i) => { return {value: a, index: i}});
const filtered = arrWithIndexes.filter(f => !rem.some(s=> s === f.value));
const indexToInsert = filtered.findIndex(f=>f.index === i);
const result = filtered.map(f=> f.value);
result.splice(indexToInsert, 0, ...rem);
console.log(result);
}
console.log(remove(arr, rmv, 4));
Or if you know the desired index:
let arr = [0,1,2,3,4,5];
let rmv = [0, 2, 3];
const remove = (src, rem ) => {
const filtered = src.filter(f=> !rem.some(s=> s === f));
filtered.splice(2, 0, ...rmv)
console.log(filtered);
}
console.log(remove(arr, rmv));

const temp = [0, 2, 3];
const arr = [0, 1, 2, 3, 4, 5];
const index = arr[4];
// Accepts an array (temp) and returns a function to be used
// as the callback for `filter` which accepts an element
// and returns whether that element is in the temp array
const filterUsing = (arr) => (el) => !arr.includes(el);
// `filter` the elements from the main array
const filtered = arr.filter(filterUsing(temp));
// Find the new position of the element in `index`
const newIndex = filtered.findIndex(el => el === index);
// Splice in the temp array back into the filtered array
filtered.splice(newIndex, 0, ...temp);
console.log(filtered);

Sequentially Pairing Items in an Array

Given an array, [1, 2, 3, 4, 5], what is the most efficient method for pairing up each of the items sequentially, like so: [[1,2], [2,3], [3,4], [4,5]]?
I've been trying to use the reduce method but to no avail and want something elegant.

Use simple for loop
var data = [1, 2, 3, 4, 5];
var res = [];
for (var i = 0; i < data.length-1; i++) {
res.push(data.slice(i, i+2));
}
console.log(res);
With Array#reduce method
console.log(
[1, 2, 3, 4, 5].reduce(function(a, b, i) {
if (i == 1) return [[a, b]];
a.push([a[a.length - 1][1], b]);
return a
})
)
With Array#reduce method with initial value as empty array
console.log(
[1, 2, 3, 4, 5].reduce(function(a, b, i, arr) {
arr[i + 1] !== undefined && a.push([b, arr[i + 1]])
return a
}, [])
)

To answer the "elegant" bit... ;)
let pairwise = function *(it) {
var
a = it[Symbol.iterator](),
b = it[Symbol.iterator]();
b.next();
for (var x of b) {
yield [a.next().value, x]
}
};
console.log(Array.from(pairwise([1,2,3,4,5])))

Using lodash for given array:
var result = _.chunk( _.sortBy(array.concat(_.slice(array, 1, array.length - 1))), 2);
Check jsfiddle
So if array = [1,2,3,4,5] we have steps:
_.slice(array, 1, array.length - 1)
// = [2,3,4]
array.concat(_.slice(array, 1, array.length - 1)
// = [1,2,3,4,5].concat([2,3,4]) = [1,2,3,4,5,2,3,4]
_.sortBy(array.concat(_.slice(array, 1, array.length - 1))
// _sortBy([1,2,3,4,5,2,3,4]) = [1,2,2,3,3,4,4,5]
_.chunk( _.sortBy(array.concat(_.slice(array, 1, array.length - 1))), 2)
// _chunk([1,2,2,3,3,4,4,5],2) = [[1,2],[2,3],[3,4],[4,5]]

Another short solution using Array.forEach and Array.push functions:
var arr = [1, 2, 3, 4, 5], pairs = [];
arr.forEach((v, k, arr) => arr[k+1] && pairs.push([v, arr[k+1]]));
console.log(JSON.stringify(pairs)); // [[1,2],[2,3],[3,4],[4,5]]

Using reduce:
const res = [1, 2, 3, 4, 5].reduce(
([b, acc], a) => [a, acc.concat([[b, a]])]
, [null, []])[1].slice(1)
console.log(res)
The seed of reduce is a tuple of two items: [null, []]. null represents the current element in the array and [] is the result.
In the first iteration of reduce:
([b, acc], a) => ... b = null and acc = []
The function produces a new tuple, the first item in the tuple is the current element of the array and the second item is the result. In the second iteration:
([b, acc], a) => ..., b = 1 and acc = [[null, 1]]
The second iteration will add (concat) [1, 2] to the result (acc).
In the third iteration:
([b, acc], a) => ..., b = 2 and acc = [[null, 1], [1, 2]]
And so on so forth:
const trace = (x, y) => {
console.log(x);
return y;
}
const json = x => JSON.stringify(x)
const res = [1, 2, 3, 4, 5].reduce(
([b, acc], a) => trace(
`a = ${a}, b = ${b} acc = ${json(acc)} ++ ${json([[b, a]])}`
, [a, acc.concat([[b, a]])]
)
, [null, []]) // seed
// the result is a tuple `[currentElement, finalResult], we extract finalResult here
[1]
// -1 element of the original array was null (check the seed), let's remove it from the result
.slice(1)
console.log(res)
We can think about the problem another way: we are kind of joining the elements of the same array with each other into tuples. Using Ramda zip function is elegant but has a performance tradeoff because we go thru the list twice:
const arr = [1, 2, 3, 4, 5]
const res = R.zip(arr, R.drop(1, arr))
console.log(res)
<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.21.0/ramda.min.js"></script>

Reduce is most elegant way to do that.
[1,2,3,4,5].reduce((a,b,c) => {
a.push([c,b]);
return a;
}, [])