I need to construct a function intersection that compares input arrays and returns a new array with elements found in all of the inputs.
The following solution works if in each array the numbers only repeat once, otherwise it breaks. Also, I don't know how to simplify and not use messy for loops:
function intersection(arrayOfArrays) {
let joinedArray = [];
let reducedArray = [];
for (let iOuter in arrayOfArrays) {
for (let iInner in arrayOfArrays[iOuter]) {
joinedArray.push(arrayOfArrays[iOuter][iInner]);
}
return joinedArray;
}
for (let i in joinedArray.sort()) {
if (joinedArray[i] === joinedArray[ i - (arrayOfArrays.length - 1)]) {
reducedArray.push(joinedArray[i]);
}
}
return reducedArray;
}
Try thhis:-
function a1(ar,ar1){
x = new Set(ar)
y = new Set(ar1)
var result = []
for (let i of x){
if (y.has(i)){
result.push(i)
}
}
if (result){return result}
else{ return 0}
}
var a= [3,4,5,6]
var b = [8,5,6,1]
console.log(a1(a,b)) //output=> [5,6]
Hopefully this snippet will be useful
var a = [2, 3, 9];
var b = [2, 8, 9, 4, 1];
var c = [3, 4, 5, 1, 2, 1, 9];
var d = [1, 2]
function intersect() {
// create an empty array to store any input array,All the comparasion
// will be done against this one
var initialArray = [];
// Convert all the arguments object to array
// there can be n number of supplied input array
// sorting the array by it's length. the shortest array
//will have at least all the elements
var x = Array.prototype.slice.call(arguments).sort(function(a, b) {
return a.length - b.length
});
initialArray = x[0];
// loop over remaining array
for (var i = 1; i < x.length; i++) {
var tempArray = x[i];
// now check if every element of the initial array is present
// in rest of the arrays
initialArray.forEach(function(item, index) {
// if there is some element which is present in intial arrat but not in current array
// remove that eleemnt.
//because intersection requires element to present in all arrays
if (x[i].indexOf(item) === -1) {
initialArray.splice(index, 1)
}
})
}
return initialArray;
}
console.log(intersect(a, b, c, d))
There is a nice way of doing it using reduce to intersect through your array of arrays and then filter to make remaining values unique.
function intersection(arrayOfArrays) {
return arrayOfArrays
.reduce((acc,array,index) => { // Intersect arrays
if (index === 0)
return array;
return array.filter((value) => acc.includes(value));
}, [])
.filter((value, index, self) => self.indexOf(value) === index) // Make values unique
;
}
You can iterate through each array and count the frequency of occurrence of the number in an object where the key is the number in the array and its property being the array of occurrence in an array. Using the generated object find out the lowest frequency of each number and check if its value is more than zero and add that number to the result.
function intersection(arrayOfArrays) {
const frequency = arrayOfArrays.reduce((r, a, i) => {
a.forEach(v => {
if(!(v in r))
r[v] = Array.from({length:arrayOfArrays.length}).fill(0);
r[v][i] = r[v][i] + 1;
});
return r;
}, {});
return Object.keys(frequency).reduce((r,k) => {
const minCount = Math.min(...frequency[k]);
if(minCount) {
r = r.concat(Array.from({length: minCount}).fill(+k));
}
return r;
}, []);
}
console.log(intersection([[2,3, 45, 45, 5],[4,5,45, 45, 45, 6,7], [3, 7, 5,45, 45, 45, 45,7]]))
I have two arrays:
a = [2,2,3,0,6]
b = [6,3,2,2,0]
I am trying use for loop to match values and get the index of a in a new array c. How can we do this? Notice that there are multiple values which match and so I think the previous match must be skipped.
This is a proposal which respects the last index and looks further.
How it works:
It uses Array#map for iterating array b with a callback. map gets an own this space with an really empty object Object.create(null).
The callback has on parameter bb which is one element of `b.
Next is to find the element is in array a with a Array#indexOf and a fromIndex, based on the former searches. The former index is stored in the this object, as long as the result is not -1, because this would reset the fromIndex to zero.
If there is no this[bb] or a falsy value of this[bb] take zero as fromIndex.
Later, a found index is incremented and stored in this[bb].
At least, the index is returned.
var a = [2, 2, 3, 0, 6],
b = [6, 3, 2, 2, 0],
c = b.map(function (bb) {
var index = a.indexOf(bb, this[bb] || 0);
if (index !== -1) {
this[bb] = index + 1;
}
return index;
}, Object.create(null));
console.log(c);
Another solution could be first generate an object with all indices of a and use it in the iteration of b for returning the indices.
The example is a bit extended, to show what happen if there is no more than two indices (2) and one without being in a (7).
The content of aObj with all indices of a:
{
"0": [3],
"2": [0, 1],
"3": [2],
"6": [4]
}
var a = [2, 2, 3, 0, 6],
b = [6, 3, 2, 2, 0, 7, 2],
aObj = Object.create(null),
c;
a.forEach(function (aa, i) {
aObj[aa] = aObj[aa] || [];
aObj[aa].push(i);
});
c = b.map(function (bb) {
return aObj[bb] && aObj[bb].length ? aObj[bb].shift() : -1;
});
console.log(c);
As far I Understand, You can try this:
var a = [2,2,3,0,6];
var b = [6,3,2,2,0];
var c = new Array();
for(i = 0; i < b.length; i++)
{
for(j = 0; j < a.length; j++)
{
if(b[i] === a[j] && c.indexOf(j) < 0)
{
c.push(j);
break;
}
}
}
console.log(c); // [4, 2, 0, 1, 3]
FIDDLE DEMO HERE
If I understand correct.
let c = a.map(i => b.indexOf(i))
or
var c = a.map(function(i) { return b.indexOf(i); });
loop .map function and check same value by indexOf
indexOf will return a number,representing the position where the specified search value occurs for the first time, or -1 if it never occurs
var arr = [];
a.map(function(v){
if(b.indexOf(v) > -1){
arr.push(v);
}
});
console.log(arr);
try something like this
var a = [2,2,3,0,6];
var b = [6,3,2,2,0];
var arrayLength_a = a.length;
var arrayLength_b = b.length;
var new_array=[];
for (var i = 0; i < arrayLength_a; i++)
{
for (var j = 0; j < arrayLength_b; j++)
{
if (a[i] == b[j])
{
if(new_array.indexOf(a[i]) === -1)
{
new_array.push(a[i]);
}
}
}
}
I am saving some data in order using arrays, and I want to add a function that the user can reverse the list. I can't think of any possible method, so if anybody knows how, please help.
Javascript has a reverse() method that you can call in an array
var a = [3,5,7,8];
a.reverse(); // 8 7 5 3
Not sure if that's what you mean by 'libraries you can't use', I'm guessing something to do with practice. If that's the case, you can implement your own version of .reverse()
function reverseArr(input) {
var ret = new Array;
for(var i = input.length-1; i >= 0; i--) {
ret.push(input[i]);
}
return ret;
}
var a = [3,5,7,8]
var b = reverseArr(a);
Do note that the built-in .reverse() method operates on the original array, thus you don't need to reassign a.
Array.prototype.reverse() is all you need to do this work. See compatibility table.
var myArray = [20, 40, 80, 100];
var revMyArr = [].concat(myArray).reverse();
console.log(revMyArr);
// [100, 80, 40, 20]
Heres a functional way to do it.
const array = [1,2,3,4,5,6,"taco"];
function reverse(array){
return array.map((item,idx) => array[array.length-1-idx])
}
20 bytes
let reverse=a=>[...a].map(a.pop,a)
const original = [1, 2, 3, 4];
const reversed = [...original].reverse(); // 4 3 2 1
Concise and leaves the original unchanged.
reveresed = [...array].reverse()
The shortest reverse method I've seen is this one:
let reverse = a=>a.sort(a=>1)
**
Shortest reverse array method without using reverse method:
**
var a = [0, 1, 4, 1, 3, 9, 3, 7, 8544, 4, 2, 1, 2, 3];
a.map(a.pop,[...a]);
// returns [3, 2, 1, 2, 4, 8544, 7, 3, 9, 3, 1, 4, 1, 0]
a.pop method takes an last element off and puts upfront with spread operator ()
MDN links for reference:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/pop
two ways:
counter loop
function reverseArray(a) {
var rA = []
for (var i = a.length; i > 0; i--) {
rA.push(a[i - 1])
}
return rA;
}
Using .reverse()
function reverseArray(a) {
return a.reverse()
}
This is what you want:
array.reverse();
DEMO
Here is a version which does not require temp array.
function inplaceReverse(arr) {
var i = 0;
while (i < arr.length - 1) {
arr.splice(i, 0, arr.pop());
i++;
}
return arr;
}
// Useage:
var arr = [1, 2, 3];
console.log(inplaceReverse(arr)); // [3, 2, 1]
I've made some test of solutions that not only reverse array but also makes its copy. Here is test code. The reverse2 method is the fastest one in Chrome but in Firefox the reverse method is the fastest.
var array = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
var reverse1 = function() {
var reversed = array.slice().reverse();
};
var reverse2 = function() {
var reversed = [];
for (var i = array.length - 1; i >= 0; i--) {
reversed.push(array[i]);
}
};
var reverse3 = function() {
var reversed = [];
array.forEach(function(v) {
reversed.unshift(v);
});
};
console.time('reverse1');
for (var x = 0; x < 1000000; x++) {
reverse1();
}
console.timeEnd('reverse1'); // Around 184ms on my computer in Chrome
console.time('reverse2');
for (var x = 0; x < 1000000; x++) {
reverse2();
}
console.timeEnd('reverse2'); // Around 78ms on my computer in Chrome
console.time('reverse3');
for (var x = 0; x < 1000000; x++) {
reverse3();
}
console.timeEnd('reverse3'); // Around 1114ms on my computer in Chrome
53 bytes
function reverse(a){
for(i=0,j=a.length-1;i<j;)a[i]=a[j]+(a[j--]=a[i++],0)
}
Just for fun, here's an alternative implementation that is faster than the native .reverse method.
You can do
var yourArray = ["first", "second", "third", "...", "etc"]
var reverseArray = yourArray.slice().reverse()
console.log(reverseArray)
You will get
["etc", "...", "third", "second", "first"]
> var arr = [1,2,3,4,5,6];
> arr.reverse();
[6, 5, 4, 3, 2, 1]
array.reverse()
Above will reverse your array but modifying the original.
If you don't want to modify the original array then you can do this:
var arrayOne = [1,2,3,4,5];
var reverse = function(array){
var arrayOne = array
var array2 = [];
for (var i = arrayOne.length-1; i >= 0; i--){
array2.push(arrayOne[i])
}
return array2
}
reverse(arrayOne)
function reverseArray(arr) {
let reversed = [];
for (i = 0; i < arr.length; i++) {
reversed.push((arr[arr.length-1-i]))
}
return reversed;
}
Using .pop() method and while loop.
var original = [1,2,3,4];
var reverse = [];
while(original.length){
reverse.push(original.pop());
}
Output: [4,3,2,1]
I'm not sure what is meant by libraries, but here are the best ways I can think of:
// return a new array with .map()
const ReverseArray1 = (array) => {
let len = array.length - 1;
return array.map(() => array[len--]);
}
console.log(ReverseArray1([1,2,3,4,5])) //[5,4,3,2,1]
// initialize and return a new array
const ReverseArray2 = (array) => {
const newArray = [];
let len = array.length;
while (len--) {
newArray.push(array[len]);
}
return newArray;
}
console.log(ReverseArray2([1,2,3,4,5]))//[5,4,3,2,1]
// use swapping and return original array
const ReverseArray3 = (array) => {
let i = 0;
let j = array.length - 1;
while (i < j) {
const swap = array[i];
array[i++] = array[j];
array[j--] = swap;
}
return array;
}
console.log(ReverseArray3([1,2,3,4,5]))//[5,4,3,2,1]
// use .pop() and .length
const ReverseArray4 = (array) => {
const newArray = [];
while (array.length) {
newArray.push(array.pop());
}
return newArray;
}
console.log(ReverseArray4([1,2,3,4,5]))//[5,4,3,2,1]
As others mentioned, you can use .reverse() on the array object.
However if you care about preserving the original object, you may use reduce instead:
const original = ['a', 'b', 'c'];
const reversed = original.reduce( (a, b) => [b].concat(a) );
// ^
// |
// +-- prepend b to previous accumulation
// original: ['a', 'b', 'c'];
// reversed: ['c', 'b', 'a'];
Pure functions to reverse an array using functional programming:
var a = [3,5,7,8];
// ES2015
function immutableReverse(arr) {
return [ ...a ].reverse();
}
// ES5
function immutableReverse(arr) {
return a.concat().reverse()
}
It can also be achieved using map method.
[1, 2, 3].map((value, index, arr) => arr[arr.length - index - 1])); // [3, 2, 1]
Or using reduce (little longer approach)
[1, 2, 3].reduce((acc, curr, index, arr) => {
acc[arr.length - index - 1] = curr;
return acc;
}, []);
reverse in place with variable swapping (mutative)
const myArr = ["a", "b", "c", "d"];
for (let i = 0; i < (myArr.length - 1) / 2; i++) {
const lastIndex = myArr.length - 1 - i;
[myArr[i], myArr[lastIndex]] = [myArr[lastIndex], myArr[i]]
}
Reverse by using the sort method
This is a much more succinct method.
const resultN = document.querySelector('.resultN');
const resultL = document.querySelector('.resultL');
const dataNum = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
const dataLetters = ['a', 'b', 'c', 'd', 'e'];
const revBySort = (array) => array.sort((a, b) => a < b);
resultN.innerHTML = revBySort(dataNum);
resultL.innerHTML = revBySort(dataLetters);
<div class="resultN"></div>
<div class="resultL"></div>
Using ES6 rest operator and arrow function.
const reverse = ([x, ...s]) => x ? [...reverse(s), x] : [];
reverse([1,2,3,4,5]) //[5, 4, 3, 2, 1]
Use swapping and return the original array.
const reverseString = (s) => {
let start = 0, end = s.length - 1;
while (start < end) {
[s[start], s[end]] = [s[end], s[start]]; // swap
start++, end--;
}
return s;
};
console.log(reverseString(["s", "t", "r", "e", "s", "s", "e", "d"]));
Infact the reverse() may not work in some cases, so you have to make an affectation first as the following
let a = [1, 2, 3, 4];
console.log(a); // [1,2,3,4]
a = a.reverse();
console.log(a); // [4,3,2,1]
or use concat
let a = [1, 2, 3, 4];
console.log(a, a.concat([]).reverse()); // [1,2,3,4], [4,3,2,1]
What about without using push() !
Solution using XOR !
var myARray = [1,2,3,4,5,6,7,8];
function rver(x){
var l = x.length;
for(var i=0; i<Math.floor(l/2); i++){
var a = x[i];
var b = x[l-1-i];
a = a^b;
b = b^a;
a = a^b;
x[i] = a;
x[l-1-i] = b;
}
return x;
}
console.log(rver(myARray));
JavaScript already has reverse() method on Array, so you don't need to do that much!
Imagine you have the array below:
var arr = [1, 2, 3, 4, 5];
Now simply just do this:
arr.reverse();
and you get this as the result:
[5, 4, 3, 2, 1];
But this basically change the original array, you can write a function and use it to return a new array instead, something like this:
function reverse(arr) {
var i = arr.length, reversed = [];
while(i) {
i--;
reversed.push(arr[i]);
}
return reversed;
}
Or simply chaning JavaScript built-in methods for Array like this:
function reverse(arr) {
return arr.slice().reverse();
}
and you can call it like this:
reverse(arr); //return [5, 4, 3, 2, 1];
Just as mentioned, the main difference is in the second way, you don't touch the original array...
How about this?:
function reverse(arr) {
function doReverse(a, left, right) {
if (left >= right) {
return a;
}
const temp = a[left];
a[left] = a[right];
a[right] = temp;
left++;
right--;
return doReverse(a, left, right);
}
return doReverse(arr, 0, arr.length - 1);
}
console.log(reverse([1,2,3,4]));
https://jsfiddle.net/ygpnt593/8/
Assume we have the following arrays:
a = [1, 2, 3, 4, 5]
and
b = [2, 3]
How can I subtract b from a? So that we have c = a - b which should be equal to [1, 4, 5]. jQuery solution would also be fine.
Assuming you're on a browser that has Array.prototype.filter and Array.prototype.indexOf, you could use this:
var c = a.filter(function(item) {
return b.indexOf(item) === -1;
});
If the browser in question does not have those methods, you may be able to shim them.
This is a modified version of the answer posted by #icktoofay.
In ES6 we can make use of:
Array.prototype.contains()
Array.prototype.filter()
Arrow functions
This will simplify our code to:
var c = a.filter(x => !b.includes(x));
Demo:
var a = [1, 2, 3, 4, 5];
var b = [2, 3];
var c = a.filter(x => !b.includes(x));
console.log(c);
For code that would work in all browsers, you would have to manually find each element from b in a and remove it.
var a = [1, 2, 3, 4, 5];
var b = [2, 3];
var result = [], found;
for (var i = 0; i < a.length; i++) {
found = false;
// find a[i] in b
for (var j = 0; j < b.length; j++) {
if (a[i] == b[j]) {
found = true;
break;
}
}
if (!found) {
result.push(a[i]);
}
}
// The array result now contains just the items from a that are not in b
Working example here: http://jsfiddle.net/jfriend00/xkBzR/
And, here's a version that could be faster for large arrays because it puts everything into an object for hashed lookups rather than brute force array searching:
var a = [1, 2, 3, 4, 5];
var b = [2, 3];
function filterArray(src, filt) {
var temp = {}, i, result = [];
// load contents of filt into object keys for faster lookup
for (i = 0; i < filt.length; i++) {
temp[filt[i]] = true;
}
// go through src
for (i = 0; i < src.length; i++) {
if (!(src[i] in temp)) {
result.push(src[i]);
}
}
return(result);
}
var filtered = filterArray(a, b);
Working example here: http://jsfiddle.net/jfriend00/LUcx6/
For the ones struggling with Objects, like Date, you'll find out that two different objects are never equal to each other, even if they have the same values, so the answers above wouldn't work.
Here is an answer to this problem in ES6.
const c = a.filter(aObject => b.findIndex(bObject => aObject.valueOf() === bObject.valueOf()) === -1)
Here an implementation for try works in all browsers:
if('filter' in Array == false) {
Array.prototype.filter =
function(callback) {
if(null == this || void 0 == this) {
return;
}
var filtered = [];
for(i = 0, len = this.length; i < len; i++) {
var tmp = this[i];
if(callback(tmp)) {
filtered.push(tmp);
}
}
return filtered;
}
}
a = [1, 2, 3, 4, 5];
b = [2, 3];
var c = a.filter(function(item) { /*implementation of icktoofay */
return b.indexOf(item) === -1;
});
Might be an outdated query but i thought this might be useful to someone.
let first = [1,2,3,4,5,6,7,9];
let second = [2,4,6,8];
const difference = first.filter(item=>!second.includes(item));
console.log(difference);//[ 1, 3, 6,7]
/*
the above will not work for objects with properties
This might do the trick
*/
const firstObj = [{a:1,b:2},{a:3,b:4},{a:5,b:6},{a:7,b:8}]//not ideal. I know
const secondObj = [{a:3,b:4},{a:7,b:8}]
const objDiff = firstObj.filter(obj=>
!secondObj.find(sec=>//take note of the "!"
sec.a===obj.a
&&//or use || if you want to check for either or
sec.b===obj.b
)//this is formatted so that it is easily readable
);
console.log(objDiff)/*
[
{
"a": 1,
"b": 2
},
{
"a": 5,
"b": 6
}
]
*/