Say you have the following array:
const ab = ['a', 'a', 'b', 'b', 'b', 'a', 'b', 'b', 'a'];
How would you change this array so that all the "b" items get grouped together, until you hit another "a".
So the result of the above array would look like:
['a', 'a', 'bbb', 'a', 'bb', 'a'];
I'm trying to solve a problem with wrapping span tags around words that match a patter in a React app, but this is essentially my problem.
I've been working at it for ages but can't come up with anything I'm happy with.
Any ideas?
Cheers.
Count repeating occurences, then build the result based on that:
const result = [];
let curr = array[0], count = 1;
for(const el of array.slice(1).concat(undefined)) {
if(el !== curr || el !== "b") {
result.push(curr.repeat(count));
curr = el, count = 1;
} else count++;
}
Assuming the elements will always be single letters, you can merge the elements, then match on either bs or non-bs:
ab.join('').match(/(b+|.)/g)
const ab = ['a', 'a', 'b', 'b', 'b', 'a', 'b', 'b', 'a'];
let output = ab.join('').match(/(b+|.)/g);
console.log(output);
Using Array#reduce you could do something like this.
I'm assuming the first two characters in your solution were a typo.
const data = ['a', 'a', 'b', 'b', 'b', 'a', 'b', 'b', 'a'];
const res = data
.reduce((a,c)=>{
const lastIndex = a.length - 1;
if(a[lastIndex] && a[lastIndex].includes('b') && c === 'b') a[lastIndex] += c;
else a.push(c);
return a;
}, []);
console.log(res);
I don't know how to give an explanation for this but using reduce you can do it like this:
const ab = ['a', 'a', 'b', 'b', 'b', 'a', 'b', 'b', 'a'];
function merge(arr) {
return arr.reduce((acc,cur, index) => {
if(arr[index - 1] === 'b' && cur !== 'a') {
acc[acc.length - 1] = acc[acc.length - 1] + cur;
return acc;
}
acc.push(cur);
return acc;
},[]);
}
console.log(merge(ab))
Here's what you're after:
var ab = ['a', 'a', 'b', 'b', 'b', 'a', 'b', 'b', 'a'];
var index = 0;
var groupedArray = [];
var firstOccurence = true;
var groupedString = "";
var grouping = false;
for (var a = 0; a < ab.length; a++) {
if (ab[a] == "a") {
if (grouping) {
grouping = false;
firstOccurence = true;
groupedArray.push(groupedString);
}
groupedArray.push(ab[a]);
} else {
if (firstOccurence) {
groupedString = "";
firstOccurence = false;
}
groupedString += ab[a];
grouping = true;
}
}
console.log(groupedArray);
If you just want to merge b then you could use reduce like this:
If the current item and the previous item are b, then append it to the last accumulator item. Else, push it to the accumulator
const ab = ['a', 'a', 'b', 'b', 'b', 'a', 'b', 'b', 'a']
const output = ab.reduce((acc, c, i, arr) => {
arr[i-1] === "b" && c === "b"
? acc[acc.length - 1] += c
: acc.push(c)
return acc;
},[])
console.log(output)
You can just map through the ab array and if the current element is a, push it to a new array but if the current element is b, check if the previous element is b or not, and if it is, merge the current element to the previous element else just push the b to the new array.
const ab = ['a', 'a', 'b', 'b', 'b', 'a', 'b', 'b', 'a'];
let arr = [];
ab.map((e,i) => {
if(i > 0) { // check if element is the first one or not
if(e == "b") {
if(ab[i - 1].indexOf('b') > -1) { // check if prev element is "b" or not
arr[arr.length-1] += e; // merge if prev element is "b"
} else {
arr.push(e);
}
} else {
arr.push(e);
}
} else {
arr.push(e);
}
});
console.log(arr);
N.B. The reduce() method approach and the regex approach shown in the other answers are cleaner and more concise as compared to the map() method approach shown above though.
you can stringify your array then split it with a match of b or more or any other character
const ab = ['a', 'a', 'b', 'b', 'b', 'a', 'b', 'b', 'a'];
const res = ab.join("").match(/(b{1,}|.)/g);
console.log(res)
This question already has answers here:
How to count duplicate value in an array in javascript
(35 answers)
Closed 4 years ago.
let array = ['a', 'a', 'a', 'b', 'b', 'c'];
i want to group double values from array, to get result like this:
result:
['3a, 2b, c']
(or something similar)
Any idea?
You can use .reduce() and .map() methods:
let array = ['a', 'a', 'a', 'b', 'b', 'c'];
let result = Object.entries(
array.reduce((r, c) => (r[c] = (r[c] || 0) + 1, r) , {})
).map(([k, v]) => v == 1 ? k : v + k);
console.log(result);
You can also use Map if you need items in specific order:
let array = ['a', 'a', 'a', 'b', 'b', 'c'];
let result = (((arr, map) => {
arr.forEach(s => map.set(s, (map.get(s) || 0) + 1));
return [...map.entries()].map(([k, v]) => v == 1 ? k : v + k);
})(array, new Map()));
console.log(result);
I would recommend using a dictionary to track the duplicate values in the array.
var dictionary = {};
for(var i = 0;i < array.length;i++){
var value = array[i];
if(dictionary[value] === undefined){
dictionary[value] = 0;
}
dictionary[value] = dictionary[value] + 1;
}
console.log(dictionary)
function count() {
array_elements = ["a", "b", "c", "d", "e", "a", "b", "c", "f", "g", "h", "h", "h", "e", "a"];
array_elements.sort();
var current_elements = null;
var count= 0;
for (var i = 0; i < array_elements.length; i++) {
if (array_elements[i] != current_elements) {
if (cnt > 0) {
console.log(current_elements+count);
}
current_elements= array_elements[i];
count= 1;
} else {
count++;
}
}
if (cnt > 0) {
console.log(current_elements+count);
}
}
A single loop approach for simple run-length encoding (RLE) function.
let array = ['a', 'a', 'a', 'b', 'b', 'c'],
result = array.reduce(
(r, c, i, a) => r.concat(c === a[i - 1]
? ((+r.pop().slice(0, -1) || 1) + 1) + c
: c),
[]
);
console.log(result);
You can use Array.reduce() to produce an object with key-value (array value and count) pairs, then Array.map() to produce an array using Object.keys().
let array = ['a', 'a', 'a', 'b', 'b', 'c'];
let countObj = array.reduce((acc, val) => (acc[val] = acc[val] ? acc[val] + 1 : 1, acc), {});
console.log(countObj);
let countArr = Object.keys(countObj).map(key => '' + countObj[key] + key);
console.log(countArr);
For "something similar" why not just have an object that maps letter to occurrence. You can then simply use dot notation to get the property value assigned to each key.
Here I've used reduce.
const arr = ['a', 'a', 'a', 'b', 'b', 'c'];
// `reduce` over the arr
const obj = arr.reduce((acc, c) => {
// If the object (acc) we passed in doesn't have
// a key assigned to the letter in the current
// iteration (c) add it, set it to 0, then
// add one, otherwise, if there is a key, add one
acc[c] = (acc[c] || 0) + 1;
// Return the object for the next iteration
return acc;
// Pass in an initial object
}, {});
console.log(obj);
// Grab the value of `a` property
console.log(obj.a);
Is it possible to sort and rearrange an array that looks like this:
itemsArray = [
['Anne', 'a'],
['Bob', 'b'],
['Henry', 'b'],
['Andrew', 'd'],
['Jason', 'c'],
['Thomas', 'b']
]
to match the arrangement of this array:
sortingArr = [ 'b', 'c', 'b', 'b', 'a', 'd' ]
Unfortunately, I don’t have any IDs to keep track on. I would need to priority the items-array to match the sortingArr as close as possible.
Update:
Here is the output I’m looking for:
itemsArray = [
['Bob', 'b'],
['Jason', 'c'],
['Henry', 'b'],
['Thomas', 'b']
['Anne', 'a'],
['Andrew', 'd'],
]
Any idea how this can be done?
One-Line answer.
itemsArray.sort(function(a, b){
return sortingArr.indexOf(a) - sortingArr.indexOf(b);
});
Or even shorter:
itemsArray.sort((a, b) => sortingArr.indexOf(a) - sortingArr.indexOf(b));
Something like:
items = [
['Anne', 'a'],
['Bob', 'b'],
['Henry', 'b'],
['Andrew', 'd'],
['Jason', 'c'],
['Thomas', 'b']
]
sorting = [ 'b', 'c', 'b', 'b', 'c', 'd' ];
result = []
sorting.forEach(function(key) {
var found = false;
items = items.filter(function(item) {
if(!found && item[1] == key) {
result.push(item);
found = true;
return false;
} else
return true;
})
})
result.forEach(function(item) {
document.writeln(item[0]) /// Bob Jason Henry Thomas Andrew
})
Here's a shorter code, but it destroys the sorting array:
result = items.map(function(item) {
var n = sorting.indexOf(item[1]);
sorting[n] = '';
return [n, item]
}).sort().map(function(j) { return j[1] })
If you use the native array sort function, you can pass in a custom comparator to be used when sorting the array. The comparator should return a negative number if the first value is less than the second, zero if they're equal, and a positive number if the first value is greater.
So if I understand the example you're giving correctly, you could do something like:
function sortFunc(a, b) {
var sortingArr = [ 'b', 'c', 'b', 'b', 'c', 'd' ];
return sortingArr.indexOf(a[1]) - sortingArr.indexOf(b[1]);
}
itemsArray.sort(sortFunc);
Case 1: Original Question (No Libraries)
Plenty of other answers that work. :)
Case 2: Original Question (Lodash.js or Underscore.js)
var groups = _.groupBy(itemArray, 1);
var result = _.map(sortArray, function (i) { return groups[i].shift(); });
Case 3: Sort Array1 as if it were Array2
I'm guessing that most people came here looking for an equivalent to PHP's array_multisort (I did) so I thought I'd post that answer as well. There are a couple options:
1. There's an existing JS implementation of array_multisort(). Thanks to #Adnan for pointing it out in the comments. It is pretty large, though.
2. Write your own. (JSFiddle demo)
function refSort (targetData, refData) {
// Create an array of indices [0, 1, 2, ...N].
var indices = Object.keys(refData);
// Sort array of indices according to the reference data.
indices.sort(function(indexA, indexB) {
if (refData[indexA] < refData[indexB]) {
return -1;
} else if (refData[indexA] > refData[indexB]) {
return 1;
}
return 0;
});
// Map array of indices to corresponding values of the target array.
return indices.map(function(index) {
return targetData[index];
});
}
3. Lodash.js or Underscore.js (both popular, smaller libraries that focus on performance) offer helper functions that allow you to do this:
var result = _.chain(sortArray)
.pairs()
.sortBy(1)
.map(function (i) { return itemArray[i[0]]; })
.value();
...Which will (1) group the sortArray into [index, value] pairs, (2) sort them by the value (you can also provide a callback here), (3) replace each of the pairs with the item from the itemArray at the index the pair originated from.
this is probably too late but, you could also use some modified version of the code below in ES6 style. This code is for arrays like:
var arrayToBeSorted = [1,2,3,4,5];
var arrayWithReferenceOrder = [3,5,8,9];
The actual operation :
arrayToBeSorted = arrayWithReferenceOrder.filter(v => arrayToBeSorted.includes(v));
The actual operation in ES5 :
arrayToBeSorted = arrayWithReferenceOrder.filter(function(v) {
return arrayToBeSorted.includes(v);
});
Should result in arrayToBeSorted = [3,5]
Does not destroy the reference array.
function sortFunc(a, b) {
var sortingArr = ["A", "B", "C"];
return sortingArr.indexOf(a.type) - sortingArr.indexOf(b.type);
}
const itemsArray = [
{
type: "A",
},
{
type: "C",
},
{
type: "B",
},
];
console.log(itemsArray);
itemsArray.sort(sortFunc);
console.log(itemsArray);
Why not something like
//array1: array of elements to be sorted
//array2: array with the indexes
array1 = array2.map((object, i) => array1[object]);
The map function may not be available on all versions of Javascript
ES6
const arrayMap = itemsArray.reduce(
(accumulator, currentValue) => ({
...accumulator,
[currentValue[1]]: currentValue,
}),
{}
);
const result = sortingArr.map(key => arrayMap[key]);
More examples with different input arrays
I would use an intermediary object (itemsMap), thus avoiding quadratic complexity:
function createItemsMap(itemsArray) { // {"a": ["Anne"], "b": ["Bob", "Henry"], …}
var itemsMap = {};
for (var i = 0, item; (item = itemsArray[i]); ++i) {
(itemsMap[item[1]] || (itemsMap[item[1]] = [])).push(item[0]);
}
return itemsMap;
}
function sortByKeys(itemsArray, sortingArr) {
var itemsMap = createItemsMap(itemsArray), result = [];
for (var i = 0; i < sortingArr.length; ++i) {
var key = sortingArr[i];
result.push([itemsMap[key].shift(), key]);
}
return result;
}
See http://jsfiddle.net/eUskE/
var sortedArray = [];
for(var i=0; i < sortingArr.length; i++) {
var found = false;
for(var j=0; j < itemsArray.length && !found; j++) {
if(itemsArray[j][1] == sortingArr[i]) {
sortedArray.push(itemsArray[j]);
itemsArray.splice(j,1);
found = true;
}
}
}
http://jsfiddle.net/s7b2P/
Resulting order: Bob,Jason,Henry,Thomas,Anne,Andrew
In case you get here needing to do this with an array of objects, here is an adaptation of #Durgpal Singh's awesome answer:
const itemsArray = [
{ name: 'Anne', id: 'a' },
{ name: 'Bob', id: 'b' },
{ name: 'Henry', id: 'b' },
{ name: 'Andrew', id: 'd' },
{ name: 'Jason', id: 'c' },
{ name: 'Thomas', id: 'b' }
]
const sortingArr = [ 'b', 'c', 'b', 'b', 'a', 'd' ]
Object.keys(itemsArray).sort((a, b) => {
return sortingArr.indexOf(itemsArray[a].id) - sortingArr.indexOf(itemsArray[b].id);
})
let a = ['A', 'B', 'C' ]
let b = [3, 2, 1]
let c = [1.0, 5.0, 2.0]
// these array can be sorted by sorting order of b
const zip = rows => rows[0].map((_, c) => rows.map(row => row[c]))
const sortBy = (a, b, c) => {
const zippedArray = zip([a, b, c])
const sortedZipped = zippedArray.sort((x, y) => x[1] - y[1])
return zip(sortedZipped)
}
sortBy(a, b, c)
For getting a new ordered array, you could take a Map and collect all items with the wanted key in an array and map the wanted ordered keys by taking sifted element of the wanted group.
var itemsArray = [['Anne', 'a'], ['Bob', 'b'], ['Henry', 'b'], ['Andrew', 'd'], ['Jason', 'c'], ['Thomas', 'b']],
sortingArr = [ 'b', 'c', 'b', 'b', 'a', 'd' ],
map = itemsArray.reduce((m, a) => m.set(a[1], (m.get(a[1]) || []).concat([a])), new Map),
result = sortingArr.map(k => (map.get(k) || []).shift());
console.log(result);
I hope that I am helping someone, but if you are trying to sort an array of objects by another array on the first array's key, for example, you want to sort this array of objects:
const foo = [
{name: 'currency-question', key: 'value'},
{name: 'phone-question', key: 'value'},
{name: 'date-question', key: 'value'},
{name: 'text-question', key: 'value'}
];
by this array:
const bar = ['text-question', 'phone-question', 'currency-question', 'date-question'];
you can do so by:
foo.sort((a, b) => bar.indexOf(a.name) - bar.indexOf(b.name));
This is what I was looking for and I did for sorting an Array of Arrays based on another Array:
It's On^3 and might not be the best practice(ES6)
function sortArray(arr, arr1){
return arr.map(item => {
let a = [];
for(let i=0; i< arr1.length; i++){
for (const el of item) {
if(el == arr1[i]){
a.push(el);
}
}
}
return a;
});
}
const arr1 = ['fname', 'city', 'name'];
const arr = [['fname', 'city', 'name'],
['fname', 'city', 'name', 'name', 'city','fname']];
console.log(sortArray(arr,arr1));
It might help someone
I had to do this for a JSON payload I receive from an API, but it wasn't in the order I wanted it.
Array to be the reference array, the one you want the second array sorted by:
var columns = [
{last_name: "last_name"},
{first_name: "first_name"},
{book_description: "book_description"},
{book_id: "book_id"},
{book_number: "book_number"},
{due_date: "due_date"},
{loaned_out: "loaned_out"}
];
I did these as objects because these will have other properties eventually.
Created array:
var referenceArray= [];
for (var key in columns) {
for (var j in columns[key]){
referenceArray.push(j);
}
}
Used this with result set from database. I don't know how efficient it is but with the few number of columns I used, it worked fine.
result.forEach((element, index, array) => {
var tr = document.createElement('tr');
for (var i = 0; i < referenceArray.length - 1; i++) {
var td = document.createElement('td');
td.innerHTML = element[referenceArray[i]];
tr.appendChild(td);
}
tableBody.appendChild(tr);
});
let sortedOrder = [ 'b', 'c', 'b', 'b' ]
let itemsArray = [
['Anne', 'a'],
['Bob', 'b'],
['Henry', 'b'],
['Andrew', 'd'],
['Jason', 'c'],
['Thomas', 'b']
]
a.itemsArray(function (a, b) {
let A = a[1]
let B = b[1]
if(A != undefined)
A = A.toLowerCase()
if(B != undefined)
B = B.toLowerCase()
let indA = sortedOrder.indexOf(A)
let indB = sortedOrder.indexOf(B)
if (indA == -1 )
indA = sortedOrder.length-1
if( indB == -1)
indB = sortedOrder.length-1
if (indA < indB ) {
return -1;
} else if (indA > indB) {
return 1;
}
return 0;
})
This solution will append the objects at the end if the sorting key is not present in reference array
const result = sortingArr.map((i) => {
const pos = itemsArray.findIndex(j => j[1] === i);
const item = itemsArray[pos];
itemsArray.splice(pos, 1);
return item;
});
this should works:
var i,search, itemsArraySorted = [];
while(sortingArr.length) {
search = sortingArr.shift();
for(i = 0; i<itemsArray.length; i++) {
if(itemsArray[i][1] == search) {
itemsArraySorted.push(itemsArray[i]);
break;
}
}
}
itemsArray = itemsArraySorted;
You could try this method.
const sortListByRanking = (rankingList, listToSort) => {
let result = []
for (let id of rankingList) {
for (let item of listToSort) {
if (item && item[1] === id) {
result.push(item)
}
}
}
return result
}
with numerical sortingArr:
itemsArray.sort(function(a, b){
return sortingArr[itemsArray.indexOf(a)] - sortingArr[itemsArray.indexOf(b)];
});
This seems to work for me:
var outputArray=['10','6','8','10','4','6','2','10','4','0','2','10','0'];
var template=['0','2','4','6','8','10'];
var temp=[];
for(i=0;i<template.length;i++) {
for(x=0;x<outputArray.length;x++){
if(template[i] == outputArray[x]) temp.push(outputArray[x])
};
}
outputArray = temp;
alert(outputArray)
Use the $.inArray() method from jQuery. You then could do something like this
var sortingArr = [ 'b', 'c', 'b', 'b', 'c', 'd' ];
var newSortedArray = new Array();
for(var i=sortingArr.length; i--;) {
var foundIn = $.inArray(sortingArr[i], itemsArray);
newSortedArray.push(itemsArray[foundIn]);
}
Use intersection of two arrays.
Ex:
var sortArray = ['a', 'b', 'c', 'd', 'e'];
var arrayToBeSort = ['z', 's', 'b', 'e', 'a'];
_.intersection(sortArray, arrayToBeSort)
=> ['a', 'b', 'e']
if 'z and 's' are out of range of first array, append it at the end of result
this.arrToBeSorted = this.arrToBeSorted.sort(function(a, b){
return uppthis.sorrtingByArray.findIndex(x => x.Id == a.ByPramaeterSorted) - uppthis.sorrtingByArray.findIndex(x => x.Id == b.ByPramaeterSorted);
});
You can do something like this:
function getSorted(itemsArray , sortingArr ) {
var result = [];
for(var i=0; i<arr.length; i++) {
result[i] = arr[sortArr[i]];
}
return result;
}
You can test it out here.
Note: this assumes the arrays you pass in are equivalent in size, you'd need to add some additional checks if this may not be the case.
refer link
refer