I'm trying to animate the sorting of a few elements with jQuery and my sorting map is defined like this:
var mapping = [
[0, 5],
[1, 4],
[2, 3],
[3, 2],
[4, 1],
[5, 0]
];
mapping[0][0] is the element's index and mapping[0][1] is the target index.
My solution was basically this:
var elements = [1, 2, 3, 4, 5, 6];
for (var i = 0; i < elements.length; i++) {
var clone1 = elements[mapping[i][0]];
var clone2 = elements[mapping[i][1]];
elements[mapping[i][0]] = clone2;
elements[mapping[i][1]] = clone1;
}
console.log(elements);
The problem is, I am working on the object I'm modifying and the operations undo themselves; instead of reversing the array, I get the same array back.
How would I swap elements like this in JavaScript? Here's a JSFiddle example of the code.
// [source, target]
var mapping = [
[0, 5],
[1, 4],
[2, 3],
[3, 2],
[4, 1],
[5, 0]
];
var elements = [1, 2, 3, 4, 5, 6];
function swapElements(elements, mapping) {
var tmp = new Array(elements.length);
for(var i = 0, l = mapping.length; i < l; i++) {
tmp[mapping[i][1]] = elements[mapping[i][0]];
}
for(var i = 0, l = elements.length; i < l; i++) {
elements[i] = tmp[i];
}
}
swapElements(elements, mapping);
console.log(elements);
Some functional fun with accessor function
var at = function(arr) { return function(i) { return arr[i] } };
[5,4,3,2,0,1].map( at(['a','b','c','d','e','f']) )
this will returns ['f','e','d','c','a','b']
You swap elements twice from 0 to 5 and from 5 to 0 again :) Try
for (var i = 0; i < elements.length / 2; i++)
and it will be ok.
Look here
You're changing the array during a loop of it. And it's swapped twice for each pair, so the result is the same. You may need another arr to store mapped value:
var result = [];
for(var i = 0; i<mapping.length; i++){
result[mapping[i][0]] = elements[mapping[i][1]];
}
console.log(result);
Just define a temporary array:
var elements = [1, 2, 3, 4, 5, 6];
var temp = [];
for (var i = 0; i < elements.length; i++) {
temp.push(elements[mapping[i][1]]);
}
elements = temp;
Just_Mad's answer is correct if you are only trying to reverse the list. But if you want to be able to handle any mapping configuration you will need a different approach.
If you don't care about memory space you could just iterate through the mapping list and copy the element at index "mapping[i][0]" to a new array at index "mapping [i][1]". Then just return the new array. This should handle whatever mapping you put into the system (you just have to decide if and how you want to handle mappings that leave empty spaces and/or override spaces)
Related
I am trying to write a function that will take an array and n as parameters,
it will return all subsets of that array with n elements, have tried a couple things, couldn't yet succeed.
thanks to whoever put it here, this functions is way too complicated and doesn't do the job, basically what I tried to do here is to pick out one element from a 4 element array to create its 3 element subsets. It doesn't even take N as parameter. it returns all 3 element subsets but also identical ones, so I have to filter them out as well, in any case I will keep trying.
function findSubsets(array) {
var answers = [];
var firstArray = array;
for (i = 0; i < array.length; i++) {
array = firstArray;
for (var k = 0; k < array.length; k++) {
if (k != i) {
var subset = array.splice(k, 1);
answers.push(array); array.splice(k, 0, subset[0]);
}
}
}
}
That not as complicated as it seems. This one is optimized because it doesn't creates useless temporary arrays during the process.
function findSubsets(array, n) {
var answers = [];
for(var i = 0 ; i < array.length ; i += n) {
answers.push(array.slice(i, i + n));
}
return answers;
}
findSubsets([1, 2, 3, 4, 5, 6, 7, 8, 9], 2) // --> [[1, 2], [3, 4], [5, 6], [7, 8], [9]]
findSubsets([1, 2, 3, 4, 5, 6, 7, 8, 9], 3) // --> [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
You can try this solution
var subsetArray = (function() {
return {
getResult: getResult
}
function getResult(array, n) {
function isBigEnough(value) {
return value.length === n;
}
var ps = [
[]
];
for (var i = 0; i < array.length; i++) {
for (var j = 0, len = ps.length; j < len; j++) {
ps.push(ps[j].concat(array[i]));
}
}
return ps.filter(isBigEnough);
}
})();
var arr = [1, 2, 3, 4,5,6,7,8,9];
console.log(subsetArray.getResult(arr,2));
the title explains it all
somehow I'd like to use the method "Combination" that math has,
this is the Wikipedia page to be clear: https://en.wikipedia.org/wiki/Combination
I have already found the solution with two loops, I want to do it in one loop
example:
const arr = [1, 2, 3, 4]
function getPairs(arr) {
/*
desired return:
[
[1, 2], [1, 3], [1, 4],
[2, 3], [2, 4],
[3, 4]
]
*/
}
You can use Array.flatMap() to iterate the array, and Array.map() to iterate all items after the current (by slicing from index + 1), and return the pair.
const getPairs = arr => arr.flatMap((a, i) => arr.slice(i + 1).map(b => [a, b]))
const arr = [1, 2, 3, 4]
const result = getPairs(arr)
console.log(result)
You can use simple for-loops.
function getPairs(arr = []) {
if (!arr.length) return []
const pairs = []
for (let i = 0; i < arr.length; i++) {
for (let j = i + 1; j < arr.length; j++) {
pairs.push([arr[i], arr[j]])
}
}
return pairs
}
console.log(getPairs([1, 2, 3, 4]))
For a simple combination algorithm, you need to loop twice through the array.
Looking for the math expression, we see that we need to get all elements from the array (let's call this element i) and match all subsequent elements from i with it (let's call those j)
Following it, what we have is something like:
let array = [1, 2, 3, 4];
let results = [];
// Loop through the items on your array (get `i`)
for (let i = 0; i < array.length - 1; i++) {
// Loop after the current 1st loop item (get `j`) and push them to a new array
for (let j = i + 1; j < array.length; j++) {
results.push([array[i], array[j]]);
}
}
console.log(results);
Using slice and map
const arr = [1, 2, 3, 4]
console.log(arr.slice(0, -1).map((x, i) => arr.slice(i+1).map(y => [x, y])))
So I want to create a function in js which takes two arrays and compare them and give a score depending on how many spots in the two arrays match. Is it right or it's bd written. I am new on coding.
Another problem I have is when I try to execute it on chrome's console. It says that compare is not defined
let score = 0;
function compare(arr1, arr2) {
for (let i = 0; i < arr2.length; i++) {
for (let j = 0; j < arr1.length; j++) {
if(arr1[j] === arr2[i]){
score++;
}
}
}
You can use .reduce() to find count of matched elements:
let arr1 = [1, 2, 3, 4, 5],
arr2 = [3, 4, 5, 6, 7],
compare = (a1, a2) => arr1.reduce((a, c) => a + arr2.includes(c), 0);
console.log(compare(arr1, arr2));
Alternatively, you can use .filter() to find array of matched elements and use its length to determine the count.
let arr1 = [1, 2, 3, 4, 5],
arr2 = [3, 4, 5, 6, 7],
compare = (a1, a2) => arr1.filter(v => arr2.includes(v)).length;
console.log(compare(arr1, arr2));
Docs:
Array.prototype.reduce()
Array.prototype.includes()
Array.prototype.filter()
Arrow Functions
This should work for you:
Updated the code as required by you :
let arr1 = [1, 2, 3, 4];
let arr2 = [3,2,3,4,5];
function compare(arr1,arr2){
let count=0;
const max=arr1.length>arr2.length ? arr2.length : arr1.length;
for(var i=0;i<max;i++){
if(arr1[i]==arr2[i]){
count++;
}
}
return count;
}
console.log(compare(arr1,arr2));
What is a fast way to find combinations that aren't present in an array yet?
E.g, I have list of points: [1, 2, 4, 9]
And I have a list of connections [[1,2], [1,4], [1,9], [2,4], [4,9]]
So the missing connection in this list is [2,9]. As there is one requirement: every integer must be connected to a bigger integer.
var points = [1, 2, 4, 9];
var connections = [[1,2], [1,4], [1,9], [2,4], [4,9]];
var missing = [];
for(i = 0; i < points.length; i++){
for(j = i + 1; j < points.length; j++){
var found = false;
for(var a = 0; a < connections.length; a++){
if(connections[a][0] == points[i] && connections[a][1] == points[j]){
found = true;
break;
}
}
if(!found) missing.push([points[i], points[j]]);
}
}
console.log(missing);
The above code works, but the amount of for loops makes me think it is reasonably slow. Is there any faster way to do this? View jsfiddle
By sorting the array, you can do it with 2 nests. Sorting takes O(n log n), and the loops are basically O(n ^ 2).
var points = [1, 2, 4, 9];
var connections = [
[1, 2],
[1, 4],
[1, 9],
[2, 4],
[4, 9]
];
connections.sort();
var missing = [];
var currentIndex = 0;
for (var i = 0; i < points.length; i++) {
for (var j = i + 1; j < points.length; j++) {
if (connections[currentIndex][0] == points[i] && connections[currentIndex][1] == points[j]) {
currentIndex++;
} else {
missing.push([points[i], points[j]]);
}
}
}
console.log(missing);
You can use .reduce method in order to generate all the combination of two elements.Then the only thing that will remain is to get the difference from two arrays.
For this, you can use filter method which accepts a callback method.
var points = [1, 2, 4, 9];
points=points.sort();
var connections = [[1,2], [1,4], [1,9], [2,4], [4,9]];
var combinations = points.reduce(function(arr,elem,i){
for(j=i+1;j<points.length;j++)
arr.push([elem,points[j]]);
return arr;
},[]);
var diff=combinations.filter(function(elem,i){
return connections.find(a=>a[0]==elem[0] && a[1]==elem[1])==undefined;
});
console.log(diff);
You could iterate only the outer loop until length - 2 and use a hash table for inserted connections. The sort order of connections does not matter.
var points = [1, 2, 4, 9],
connections = [[1, 2], [1, 4], [1, 9], [2, 4], [4, 9]],
missing = [],
i, j,
pair,
connected = Object.create(null);
connections.forEach(function (a) {
connected[a.join()] = true;
});
for (i = 0; i < points.length - 1; i++) {
for (j = i + 1; j < points.length; j++) {
pair = [points[i], points[j]];
connected[pair.join()] || missing.push(pair);
}
}
console.log(missing);
.as-console-wrapper { max-height: 100% !important; top: 0; }
I'm building an app to test different icons. Admins upload a number of icons and input how many icons must be shown at the same time. The app then displays all possible sets of icons in sequence until all combination of icons have been shown.
Now, I need a function to generate all unique icons combinations based on two number:
the number of total icons (i)
the number of icons in each set (s)
If i = 6 and s = 3, I want the output to look as follows:
[
[1, 2, 3],
[1, 2, 4],
[1, 2, 5],
[1, 2, 6],
[1, 3, 4],
[1, 3, 5],
[1, 3, 6],
[1, 4, 5],
[1, 4, 6],
[1, 5, 6],
[2, 3, 4],
[2, 3, 5],
[2, 3, 6],
[2, 4, 5],
[2, 4, 6],
[2, 5, 6],
[3, 4, 5],
[3, 4, 6],
[3, 5, 6],
[4, 5, 6],
]
Requirements:
All sets have to be unique
A number can only occur one time in a set
I have been trying to code a recursive function, but I havent anything really anything to show. I can't get my head around it :(
Based off the idea given as an answer to this question:
Computing all n-sized permutations without repetitions and without "classic" ordering
Then use C++ std::next_permutation like algorithms which work as
follows:
Go from left and find rightmost one preceeded by zero. Put one in
place of zero and sort the rest of array.
Disclaimer: My javascript is very, very rusty so I'm sure there is a much more elegant way of implementing this.
function combine(n, k) {
var result = [];
// initialize array of values
var values = [];
for (var i = 1; i <= n; i++) {
values[i - 1] = i;
}
// initialize permutations
var perm = [];
for (var i = 0; i < n; i++) {
if (i < k) {
perm[i] = 1;
} else {
perm[i] = 0;
}
}
perm.sort();
whileloop:
while (true) {
// save subresult
var subresult = [];
for (var i = 0; i < n; i++) {
if (perm[i] == 1) {
subresult.push(values[i]);
}
}
result.push(subresult);
// get next permutation
for (var i = n - 1; i > 0; i--) {
if (perm[i - 1] == 1) {
continue;
}
if (perm[i] == 1) {
perm[i - 1] = 1;
perm[i] = 0;
perm = perm.slice(0, i).concat(perm.slice(i).sort())
continue whileloop;
}
}
// no additional permutations exist
break whileloop;
}
return result;
}
Combining n elements in to number of sets with k elements each without repetitions, how to do it.
The algorithm is relatively simple, main idea: we consequently feed first set of k elements and then try to increment each element from the end of set to populate another k-set and so on.
When we can't do that we leave the process (all possible sets are ready).
function combine(n,k) {
var result = Array();
var a = Array();
// make initial (first) k-set
for (var i=1; i<=k; i++) {
a[i-1] = i;
}
j = k-1;
while (j >= 1) {
// submit current results
result.push(a.slice());
if (a[k-1] == n) {
j = j - 1;
} else {
j = k-1;
}
if (j >= 1) {
// make next k-set based on previous one
for (var i=k; i>=j; i--) {
a[i-1] = a[j-1] + i - j + 1;
}
}
}
return result;
}
Note: JavaScript arrays have start index 0 so in code we have -1 correction for array indices (cause set of possible values from 1 to n)