The question doesn't make much sense but not sure how to word it without an example. If someone can word it better, feel free to edit it.
Let's say I have an array of arrays such as this:
[ ['a', 'a', 'b', 'c'], [], ['d', 'a'], ['b', 'b', 'b', 'e'] ]
I would like the output to be:
['a', 'a', 'b', 'b', 'b', 'c', 'd', 'e']
Not sure if there is an easy way to do this in javascript/jquery/underscore. One way I could think of is to look through each of these arrays and count up the number of times each element shows up and keep track of the maximum amount of times it shows up. Then I can recreate it. But that seems pretty slow considering that my arrays can be very large.
You need to:
Loop over each inner array and count the values
Store each value and its count (if higher than current count) in a counter variable
In the end, convert the value and counts into an array
Following code shows a rough outline of the process. Remember to replace .forEach and for..in with appropriate code:
var input = [['a', 'a', 'b', 'c'], [], ['d', 'a'], ['b', 'b', 'b', 'e']],
inputCount = {};
input.forEach(function(inner) {
var innerCount = {};
inner.forEach(function(value) {
innerCount[value] = innerCount[value] ? innerCount[value] + 1 : 1;
});
var value;
for (value in innerCount) {
inputCount[value] = inputCount[value] ? Math.max(inputCount[value], innerCount[value]) : innerCount[value];
}
});
console.log(inputCount);
// Object {a: 2, b: 3, c: 1, d: 1, e: 1}
After messing around, I found a solution but not sure if I like it enough to use. I would probably use it if I can't think of another one.
I would use underscorejs countBy to get the count of all the elements.
var array = [ ['a', 'a', 'b', 'c'], [], ['d', 'a'], ['b', 'b', 'b', 'e'] ];
var count = _.map(array, function(inner) {
return _.countBy(inner, function(element) {
return element;
});
});
var total = {};
_.each(_.uniq(_.flatten(array)), function(element) {
var max = _.max(count, function(countedElement) {
return countedElement[element];
});
total[element] = max[element];
});
console.log(total); // {a: 2, b: 3, c: 1, d: 1, e: 1}
Then I would recreate the array with that total.
Here is example of simple nested loop approach:
var input = [ ['a', 'a', 'b', 'c'], [], ['d', 'a'], ['b', 'b', 'b', 'e'] ];
var countMap = {};
// iterate outer array
for (i=0; i < input.length; i++) {
// iterate inner array
for (j=0; j < input[i].length; j++) {
// increment map counter
var value = input[i][j];
if (countMap[input[i][j]] === undefined) {
countMap[value] = 1;
} else {
countMap[value]++;
}
}
}
console.log(countMap); // output such as {'a':2, 'b':4, 'c':1, 'd':1, 'e':1}
Not the most efficient solution but it should describe you the process:
var big = [ ['a', 'a', 'b', 'c'], [], ['d', 'a'], ['b', 'b', 'b', 'e'] ];
function map(arr){
var map = {}
for (var i=arr.length-1; i>-1; i--){
if(arr[i] in map) map[arr[i]]++;
else map[arr[i]] = 1;
}
return map;
}
function reduce(matrix){
var arrMap = {};
for (var i=matrix.length-1; i>-1; i--){
var arrRes = map(matrix[i]);
for (var key in arrRes){
if( !arrMap[key] || arrMap[key] < arrRes[key])
arrMap[key] = arrRes[key];
}
}
return arrMap;
}
function calc(matrix){
var res = [],
arrMap = reduce(matrix);
for (var key in arrMap){
while(arrMap[key] > 0 ){
res.push(key);
arrMap[key]--;
}
}
return res;
}
console.log(calc(big));
// Array [ "e", "b", "b", "b", "a", "a", "d", "c" ]
Related
Sorry if this is a duplicate or a dumb question!
Basically, I need to get the count for duplicate values in an array until the next value changes. I can't use reduce() in my project, so any plain JS would be helpful.
let array = [a,a,a,b,b,b,b,b,c,c,c,a,d,d];
Results:
a:3,
b:5,
c:3,
a:1,
d:2
I would appreciate it very much.
You can use regex to get the desired result.
/([a-z])\1*/gi
let array = ["a", "a", "a", "b", "b", "b", "b", "b", "c", "c", "c", "a", "d", "d"];
const result = array
.join("")
.match(/([a-z])\1*/gi)
.map((s) => `${s[0]}${s.length}`);
console.log(result);
Simply loop over chars and check if char in dict then increment it else set it to 1;
let chars = ['a', 'a', 'a', 'b', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'a', 'd', 'd'];
const dic={}
for(char of chars){
if(char in dic){
dic[char]++;
}else{
dic[char]=1;
}
}
console.log(dic);//{a: 4, b: 5, c: 3, d: 2}
Run iteration over array elements. Find the next non-equal current character's position. then difference the two indexes you will find the current character's continuous last position. Increase the iteration so you need not worry about multiple counts. If no non-equal character found, then the size of the sub-array is the rest of the main array size.
let ara = ['a', 'a', 'a', 'b', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'a', 'd', 'd'];
let index = 0;
for (var i = 0; i < ara.length; i++) {
let char = ara[i];
let size = 0;
let nextIndex = ara.findIndex(
(a, index) => a !== char && index > i);
if (nextIndex < 0) {
size = ara.length - i;
i = ara.length - 1;
} else {
size = nextIndex - i;
i = nextIndex - 1;
}
console.log(char + ' ' + size);
Consider the following arrays:
['a', 'b', 'a'] //method should return true
['a', 'b', 'c'] //method should return true
['a', 'c', 'c'] //method should return false
I want to write a method that most efficiently checks to see if both 'a' and 'b' exist in the array. I know I can do this in a simple for loop
let a_counter = 0;
let b_counter = 0;
for (let i = 0; i < array.length; i++) {
if (array[i] === 'a') {
a_counter++;
}
if (array[i] === 'b') {
b_counter++;
}
}
return (a_counter > 0 && b_counter > 0);
But this isn't very short. I can do indexOf but that will loop through twice. I have also considered using a set as below:
const letter_set = new Set(array)
return (letter_set.has('a') && letter_set.has('b'))
But I am pretty unfamiliar with sets and don't know if this solution could potentially be more expensive than just looping. I know that has() operations should be faster than array iterations but constructing the set probably takes at least O(N) time (I'm assuming).
Is there a clean and efficient way to find multiple elements in an array? ES6 answers welcome
You can use every and includes to do this check.
So we are saying every item must be included in the array.
function contains(arr, ...items) {
return items.every(i => arr.includes(i))
}
console.log(contains(['a', 'b', 'a'], 'a', 'b'))
console.log(contains(['a', 'c', 'c'], 'a', 'b'))
console.log(contains(['a', 'b', 'c'], 'a', 'b', 'c'))
console.log(contains(['a', 'b', 'c', 'd'], 'a', 'b', 'c', 'd', 'e'))
You could use just the Set and check if the wanted items are in the items array.
const
check = (items, wanted) => wanted.every(Set.prototype.has, new Set(items));
console.log(check(['a', 'b', 'a'], ['a', 'b'])); // true
console.log(check(['a', 'b', 'c'], ['a', 'b'])); // true
console.log(check(['a', 'c', 'c'], ['a', 'b'])); // false
array.includes('a') && array.includes('b')
includes seems like a real handy way to check for specific elements, even if there is more than one.
Not as compact as the other examples, but it does do the job in single run.
const arr1 = ['a', 'b', 'a']; //method should return true
const arr2 = ['a', 'c', 'c']; //method should return false
const arr3 = ['a', 'b', 'c']; //method should return true
const reducer = ({ a, b }, char) => ({
a: a || char === 'a',
b: b || char === 'b'
});
const includesAnB = arr => {
const { a, b } = arr.reduce(reducer, {});
return a && b;
}
console.log(includesAnB(arr1));
console.log(includesAnB(arr2));
console.log(includesAnB(arr3));
I have arrays as following
var A = ['C', 'D', 'E', 'F', 'G'];
var B = [3, 0, 4, 1, 2];
I need to rearrange array A with the given index values in array B. My solution to the problem is following
function reArrange(A,B){
var num;
var letter;
for(var i = 0; i < A.length; i++){
num = B[i];
letter = A[i];
A[num] = letter;
}
return A;
}
reArrange(A, B);
I get an output of ['D', 'C', 'E', 'C', 'E'] when it should be ['D', 'F', 'G', 'C', 'E']
Hope this will help.
var A = ['C', 'D', 'E', 'F', 'G'];
var B = [3, 0, 4, 1, 2];
var C = []
function reArrange(A,B){
var num;
var letter;
for(var i = 0; i < A.length; i++){
num = B[i];
letter = A[i];
C[num] = letter;
}
return C;
}
reArrange(A, B);
console.log(C)
You could use the second array for the indices for assinging the actual value of the given array.
var array = ['C', 'D', 'E', 'F', 'G'],
order = [3, 0, 4, 1, 2],
result = order.reduce((r, a, i) => (r[a] = array[i], r), []);
console.log(result); // ['D', 'F', 'G', 'C', 'E']
Simple and short
var A = ['C', 'D', 'E', 'F', 'G'];
var B = [3, 0, 4, 1, 2];
function reorderAB(first, second){
var result= new Array();
for(var i=0;i<second.length;i++){
result[second[i]]=first[i];
}
console.log(result);
}
reorderAB(A,B);
The problem is that you replaced array A with new value.
Example with the first i in your loop.
i = 0, num = 3 and letter = C
The original A[3] = F
Then you assign A[num] = letter, mean that A[3] (F) has new value/replaced C and so on.
You can create a Map from B -> A and then order the keys accordingly to output your expected array, however I am not sure what use case you are using it for.
const A = ['C', 'D', 'E', 'F', 'G'];
const B = [3, 0, 4, 1, 2];
let bToA = new Map();
B.forEach((a, i) => bToA.set(a, A[i]))
const final = [...bToA.keys()].sort().map(x => bToA.get(x))
console.log(final);
I'm trying to count duplicates in an array of dates and add them to a new array.
But i'm only getting the duplicates and the amount of times they exist in the array.
What I want is:
[a, a, b, c, c] => [a: 2, b: 1, c: 2]
Code:
$scope.result = { };
for(var i = 0; i < $scope.loginArray.length; ++i) {
if(! $scope.result[$scope.loginArray[i]]){
$scope.result[$scope.loginArray[i]] = 0;
++ $scope.result[$scope.loginArray[i]];}
}
Any suggestions?
You might need an object for this, not an array. So what you are doing is already great, but the if condition is messing up:
$scope.result = {};
for (var i = 0; i < $scope.loginArray.length; ++i) {
if (!$scope.result[$scope.loginArray[i]])
$scope.result[$scope.loginArray[i]] = 0;
++$scope.result[$scope.loginArray[i]];
}
Snippet
var a = ['a', 'a', 'b', 'c', 'c'];
var r = {};
for (var i = 0; i < a.length; ++i) {
if (!r[a[i]])
r[a[i]] = 0;
++r[a[i]];
}
console.log(r);
Or in better way, you can use .reduce like how others have given. A simple reduce function will be:
var a = ['a', 'a', 'b', 'c', 'c'];
var r = a.reduce(function(c, e) {
c[e] = (c[e] || 0) + 1;
return c;
}, {});
console.log(r);
For that, you can use .reduce:
var arr = ['a','a', 'b', 'c', 'c'];
var result = arr.reduce(function(p,c){
if(p[c] === undefined)
p[c] = 0;
p[c]++;
return p;
},{});
console.log(result);
You can use reduce and return object
var ar = ['a', 'a', 'b', 'c', 'c'];
var result = ar.reduce(function(r, e) {
r[e] = (r[e] || 0) + 1;
return r;
}, {});
console.log(result)
You can also first create Object and then use forEach add properties and increment values
var ar = ['a', 'a', 'b', 'c', 'c'], result = {}
ar.forEach(e => result[e] = (result[e] || 0)+1);
console.log(result)
lodash's countBy function will do the trick:
_.countBy(['a', 'a', 'b', 'c', 'c']) will evaluate to: {a: 2, b: 1, c: 2}
It does involve adding lodash as a dependency though.
I would do like this
var arr = ["a", "a", "b", "c", "c", "a"],
red = arr.reduce((p,c) => (p[c]++ || (p[c]=1),p),{});
console.log(red);
var arr = ['a', 'b', 'c', 'd', 'e', 'f'];
var point = 'c';
How can I split the "arr" into two arrays based on the "point" variable, like:
['a', 'b']
and
['d', 'e', 'f']
var arr2 = ['a', 'b', 'c', 'd', 'e', 'f'];
arr = arr2.splice(0, arr2.indexOf('c'));
To remove 'c' from arr2:
arr2.splice(0,1);
arr contains the first two elements and arr2 contains the last three.
This makes some assumptions (like arr2 will always contain the 'point' at first assignment), so add some correctness checking for border cases as necessary.
Use indexOf and slice
var arr = ['a', 'b', 'c', 'd', 'e', 'f'];
var indexToSplit = arr.indexOf('c');
var first = arr.slice(0, indexToSplit);
var second = arr.slice(indexToSplit + 1);
console.log({first, second});
Sharing this convenience function that I ended up making after visiting this page.
function chunkArray(arr,n){
var chunkLength = Math.max(arr.length/n ,1);
var chunks = [];
for (var i = 0; i < n; i++) {
if(chunkLength*(i+1)<=arr.length)chunks.push(arr.slice(chunkLength*i, chunkLength*(i+1)));
}
return chunks;
}
Sample usage:
chunkArray([1,2,3,4,5,6],2);
//returns [[1,2,3],[4,5,6]]
chunkArray([1,2,3,4,5,6,7],2);
//returns [[1,2,3],[4,5,6,7]]
chunkArray([1,2,3,4,5,6],3);
//returns [[1,2],[3,4],[5,6]]
chunkArray([1,2,3,4,5,6,7,8],3);
//returns [[1,2],[3,4,5],[6,7,8]]
chunkArray([1,2,3,4,5,6,7,8],42);//over chunk
//returns [[1],[2],[3],[4],[5],[6],[7],[8]]
Try this one:
var arr = ['a', 'b', 'c', 'd', 'e', 'f'];
var point = 'c';
var idx = arr.indexOf(point);
arr.slice(0, idx) // ["a", "b"]
arr.slice(idx + 1) // ["d", "e", "f"]
var arr = ['a', 'b', 'c', 'd', 'e', 'f'];
var point = 'c';
Array.prototype.exists = function(search){
for (var i=0; i<this.length; i++) {
if (this[i] == search) return i;
}
return false;
}
if(i=arr.exists(point))
{
var neewArr=arr.splice(i);
neewArr.shift(0);
console.log(arr); // output: ["a", "b"]
console.log(neewArr); // output: ["d", "e", "f"]
}
Here is an example.
var arr = ['a', 'b', 'c', 'd', 'e', 'f'];
var point = 'c';
var i = arr.indexOf(point);
var firstHalf, secondHalf, end, start;
if (i>0) {
firstHalf = arr.slice(0, i);
secondHalf = arr.slice(i + 1, arr.length);
}
//this should get you started. Can you think of what edge cases you should test for to fix?
//what happens when point is at the start or the end of the array?
When splitting the array you are going to want to create two new arrays that will include what you are splitting, for example arr1 and arr2. To populate this arrays you are going to want to do something like this:
var arr1, arr2; // new arrays
int position = 0; // start position of second array
for(int i = 0; i <= arr.length(); i++){
if(arr[i] = point){ //when it finds the variable it stops adding to first array
//starts adding to second array
for(int j = i+1; j <= arr.length; j++){
arr2[position] = arr[j];
position++; //because we want to add from beginning of array i used this variable
}
break;
}
// add to first array
else{
arr1[i] = arr[i];
}
}
There are different ways to do this! good luck!
Yet another suggestion:
var segments = arr.join( '' ).split( point ).map(function( part ) {
return part.split( '' );
});
now segments contains an array of arrays:
[["a", "b"], ["d", "e", "f"]]
and can get accessed like
segments[ 0 ]; // ["a", "b"]
segments[ 1 ]; // ["d", "e", "f"]
if you want to split into equal half; why no simple while loop ?
var arr = ['a', 'b', 'c', 'd', 'e', 'f'];
var c=[];
while(arr.length > c.length){
c.push(arr.splice(arr.length-1)[0]);
}
Kaboom :).
Separate two arrays with given array elements as string array and number array;
let arr = [21,'hh',33,'kk',55,66,8898,'rtrt'];
arrStrNum = (arr) => {
let str = [],num = [];
for(let i = 0;i<arr.length;i++){
if(typeof arr[i] == "string"){
str.push(arr[i]);
}else if(typeof arr[i] == "number"){
num.push(arr[i]);
}
}
return [str, num]
}
let ans = arrStrNum(arr);
let str = ans[0];
let num = ans[1];
console.log(str);
console.log(num);