Develop Reference

How do I check if 2 numbers are the same from Math.random [duplicate]

Can't seem to find an answer to this, say I have this:
setInterval(function() {
m = Math.floor(Math.random()*7);
$('.foo:nth-of-type('+m+')').fadeIn(300);
}, 300);
How do I make it so that random number doesn't repeat itself. For example if the random number is 2, I don't want 2 to come out again.

There are a number of ways you could achieve this.
Solution A:
If the range of numbers isn't large (let's say less than 10), you could just keep track of the numbers you've already generated. Then if you generate a duplicate, discard it and generate another number.
Solution B:
Pre-generate the random numbers, store them into an array and then go through the array. You could accomplish this by taking the numbers 1,2,...,n and then shuffle them.
shuffle = function(o) {
for(var j, x, i = o.length; i; j = parseInt(Math.random() * i), x = o[--i], o[i] = o[j], o[j] = x);
return o;
};
var randorder = shuffle([0,1,2,3,4,5,6]);
var index = 0;
setInterval(function() {
$('.foo:nth-of-type('+(randorder[index++])+')').fadeIn(300);
}, 300);
Solution C:
Keep track of the numbers available in an array. Randomly pick a number. Remove number from said array.
var randnums = [0,1,2,3,4,5,6];
setInterval(function() {
var m = Math.floor(Math.random()*randnums.length);
$('.foo:nth-of-type('+(randnums[m])+')').fadeIn(300);
randnums = randnums.splice(m,1);
}, 300);

You seem to want a non-repeating random number from 0 to 6, so similar to tskuzzy's answer:
var getRand = (function() {
var nums = [0,1,2,3,4,5,6];
var current = [];
function rand(n) {
return (Math.random() * n)|0;
}
return function() {
if (!current.length) current = nums.slice();
return current.splice(rand(current.length), 1);
}
}());
It will return the numbers 0 to 6 in random order. When each has been drawn once, it will start again.

could you try that,
setInterval(function() {
m = Math.floor(Math.random()*7);
$('.foo:nth-of-type(' + m + ')').fadeIn(300);
}, 300);

I like Neal's answer although this is begging for some recursion. Here it is in java, you'll still get the general idea. Note that you'll hit an infinite loop if you pull out more numbers than MAX, I could have fixed that but left it as is for clarity.
edit: saw neal added a while loop so that works great.
public class RandCheck {
private List<Integer> numbers;
private Random rand;
private int MAX = 100;
public RandCheck(){
numbers = new ArrayList<Integer>();
rand = new Random();
}
public int getRandomNum(){
return getRandomNumRecursive(getRand());
}
private int getRandomNumRecursive(int num){
if(numbers.contains(num)){
return getRandomNumRecursive(getRand());
} else {
return num;
}
}
private int getRand(){
return rand.nextInt(MAX);
}
public static void main(String[] args){
RandCheck randCheck = new RandCheck();
for(int i = 0; i < 100; i++){
System.out.println(randCheck.getRandomNum());
}
}
}

Generally my approach is to make an array containing all of the possible values and to:
Pick a random number <= the size of the array
Remove the chosen element from the array
Repeat steps 1-2 until the array is empty
The resulting set of numbers will contain all of your indices without repetition.
Even better, maybe something like this:
var numArray = [0,1,2,3,4,5,6];
numArray.shuffle();
Then just go through the items because shuffle will have randomized them and pop them off one at a time.

Here's a simple fix, if a little rudimentary:
if(nextNum == lastNum){
if (nextNum == 0){nextNum = 7;}
else {nextNum = nextNum-1;}
}
If the next number is the same as the last simply minus 1 unless the number is 0 (zero) and set it to any other number within your set (I chose 7, the highest index).
I used this method within the cycle function because the only stipulation on selecting a number was that is musn't be the same as the last one.
Not the most elegant or technically gifted solution, but it works :)

Use sets. They were introduced to the specification in ES6. A set is a data structure that represents a collection of unique values, so it cannot include any duplicate values. I needed 6 random, non-repeatable numbers ranging from 1-49. I started with creating a longer set with around 30 digits (if the values repeat the set will have less elements), converted the set to array and then sliced it's first 6 elements. Easy peasy. Set.length is by default undefined and it's useless that's why it's easier to convert it to an array if you need specific length.
let randomSet = new Set();
for (let index = 0; index < 30; index++) {
randomSet.add(Math.floor(Math.random() * 49) + 1)
};
let randomSetToArray = Array.from(randomSet).slice(0,6);
console.log(randomSet);
console.log(randomSetToArray);

An easy way to generate a list of different numbers, no matter the size or number:
function randomNumber(max) {
return Math.floor(Math.random() * max + 1);
}
const list = []
while(list.length < 10 ){
let nbr = randomNumber(500)
if(!list.find(el => el === nbr)) list.push(nbr)
}
console.log("list",list)

I would like to add--
var RecordKeeper = {};
SRandom = function () {
currTimeStamp = new Date().getTime();
if (RecordKeeper.hasOwnProperty(currTimeStamp)) {
RecordKeeper[currTimeStamp] = RecordKeeper[currTimeStamp] + 1;
return currTimeStamp.toString() + RecordKeeper[currTimeStamp];
}
else {
RecordKeeper[currTimeStamp] = 1;
return currTimeStamp.toString() + RecordKeeper[currTimeStamp];
}
}
This uses timestamp (every millisecond) to always generate a unique number.

you can do this. Have a public array of keys that you have used and check against them with this function:
function in_array(needle, haystack)
{
for(var key in haystack)
{
if(needle === haystack[key])
{
return true;
}
}
return false;
}
(function from: javascript function inArray)
So what you can do is:
var done = [];
setInterval(function() {
var m = null;
while(m == null || in_array(m, done)){
m = Math.floor(Math.random()*7);
}
done.push(m);
$('.foo:nth-of-type('+m+')').fadeIn(300);
}, 300);
This code will get stuck after getting all seven numbers so you need to make sure it exists after it fins them all.

Get the intersection of n arrays

Using ES6's Set, given two arrays we can get the intersection like so:
let a = new Set([1,2,3])
let b = new Set([1,2,4])
let intersect = new Set([...a].filter(i => b.has(i)));
How can we get the intersection of n arrays?
Update:
I'm trying to wrap my head around this for the following use case. I have a two dimensional array with at least one element.
parts.forEach(part => {
intersection = new Set()
})
How would you get the intersection of each element (array) in parts?

Assuming you have some function function intersect(set1, set2) {...} that can intersect two sets, you can get the intersection of an array of sets using reduce:
function intersect(a, b) {
return new Set(a.filter(i => b.has(i)));
}
var sets = [new Set([1,2,3]), ...];
var intersection = sets.reduce(intersect);

You can create an intersect helper function using a combination of Array methods like .filter(), .map(), and .every().
This answer is inspired by the comment above from Xufox, who mentioned using Array#every in a filter predicate.
function intersect (first = [], ...rest) {
rest = rest.map(array => new Set(array))
return first.filter(e => rest.every(set => set.has(e)))
}
let parts = [
[1, 2, 3],
[1, 2, 4],
[1, 5, 2]
]
console.log(
intersect(...parts)
)

ES6 still has a while
This is the type of function that can easily cause long lags due to excessive amounts of processing. This is more true with the unquestioning and even preferential use of ES6 and array methods like reduce, filter etc, over simple old fashioned loops like while and for.
When calculating the intersection of many sets the amount of work done per iteration should go down if an item has been found not to be part of the intersection. Because forEach can not break you are forced to still iterate all elements. Adding some code to avoid doing the search if the current item has been found to not belong can improve the performance, but it is a real kludge.
The is also the tendency to just create whole new datasets just to remove a single item from an array, set, or map. This is a very bad habit that i see more and more of as people adopt the ES5 way.
Get the intersection of n sets.
So to the problem at hand. Find the intersection of many sets.
Solution B
A typical ES6 solution
function intersectB(firstSet, ...sets) {
// function to intercept two sets
var intersect = (a,b) => {
return new Set([...a].filter(item => b.has(item)))
};
// iterate all sets comparing the first set to each.
sets.forEach(sItem => firstSet = intersect(firstSet, sItem));
// return the result.
return firstSet;
}
var sets = [new Set([1,2,3,4]), new Set([1,2,4,6,8]), new Set([1,3,4,6,8])];
var inter = intersectB(...sets);
console.log([...inter]);
Works well and for the simple test case execution time is under a millisecond. But in my book it is a memory hogging knot of inefficiency, creating arrays, and sets at every line almost and iterating whole sets when the outcome is already known.
Let's give it some more work. 100 sets, with up to 10000 items over 10 tests each with differing amount of matching items. Most of the intercepts will return empty sets.
Warning will cause page to hang up to one whole second... :(
// Create a set of numbers from 0 and < count
// With a odds for any number occurring to be odds
// return as a new set;
function createLargeSet(count,odds){
var numbers = new Set();
while(count-- > 0){
if(Math.random() < odds){
numbers.add(count);
}
}
return numbers;
}
// create a array of large sets
function bigArrayOfSets(setCount,setMaxSize,odds){
var bigSets = [];
for(var i = 0; i < setCount; i ++){
bigSets.push(createLargeSet(setMaxSize,odds));
}
return bigSets;
}
function intersectB(firstSet, ...sets) {
var intersect = (a,b) => {
return new Set([...a].filter(item => b.has(item)))
};
sets.forEach(sItem => firstSet = intersect(firstSet, sItem));
return firstSet;
}
var testSets = [];
for(var i = 0.1; i <= 1; i += 0.1){
testSets.push(bigArrayOfSets(100,10000,i));
}
var now = performance.now();
testSets.forEach(testDat => intersectB(...testDat));
var time = performance.now() - now;
console.log("Execution time : " + time);
Solution A
A better way, not as fancy but much more efficient.
function intersectA(firstSet,...sets) {
var count = sets.length;
var result = new Set(firstSet); // Only create one copy of the set
firstSet.forEach(item => {
var i = count;
var allHave = true;
while(i--){
allHave = sets[i].has(item)
if(!allHave) { break } // loop only until item fails test
}
if(!allHave){
result.delete(item); // remove item from set rather than
// create a whole new set
}
})
return result;
}
Compare
So now let's compare both, if you are feeling lucky try and guess the performance difference, it's a good way to gage your understanding of Javascript execution.
// Create a set of numbers from 0 and < count
// With a odds for any number occurring to be odds
// return as a new set;
function createLargeSet(count,odds){
var numbers = new Set();
while(count-- > 0){
if(Math.random() < odds){
numbers.add(count);
}
}
return numbers;
}
// create a array of large sets
function bigArrayOfSets(setCount,setMaxSize,odds){
var bigSets = [];
for(var i = 0; i < setCount; i ++){
bigSets.push(createLargeSet(setMaxSize,odds));
}
return bigSets;
}
function intersectA(firstSet,...sets) {
var count = sets.length;
var result = new Set(firstSet); // Only create one copy of the set
firstSet.forEach(item => {
var i = count;
var allHave = true;
while(i--){
allHave = sets[i].has(item)
if(!allHave) { break } // loop only until item fails test
}
if(!allHave){
result.delete(item); // remove item from set rather than
// create a whole new set
}
})
return result;
}
function intersectB(firstSet, ...sets) {
var intersect = (a,b) => {
return new Set([...a].filter(item => b.has(item)))
};
sets.forEach(sItem => firstSet = intersect(firstSet, sItem));
return firstSet;
}
var testSets = [];
for(var i = 0.1; i <= 1; i += 0.1){
testSets.push(bigArrayOfSets(100,10000,i));
}
var now = performance.now();
testSets.forEach(testDat => intersectB(...testDat));
var time = performance.now() - now;
console.log("Execution time 'intersectB' : " + time);
var now = performance.now();
testSets.forEach(testDat => intersectA(...testDat));
var time = performance.now() - now;
console.log("Execution time 'intersectA' : " + time);
As you can see using a simple while loop may not be a cool as using filter but the performance benefit is huge, and something to keep in mind next time you are writing that perfect 3 line ES6 array manipulation function. Dont forget about for and while.

The most efficient algorithm for intersecting n arrays is the one implemented in fast_array_intersect. It runs in O(n), where n is the total number of elements in all the arrays.
The base principle is simple: iterate over all the arrays, storing the number of times you see each element in a map. Then filter the smallest array, to return only the elements that have been seen in all the arrays. (source code).
You can use the library with a simple :
import intersect from 'fast_array_intersect'
intersect([[1,2,3], [1,2,6]]) // --> [1,2]

OK i guess the most efficient way of performing the Array intersection is by utilizing a Map or Hash object. Here I test 1000 arrays each with ~1000 random integer items among 1..175 for an intersection. The result is obtained in less than 100msec.
function setIntersection(a){
var m = new Map(),
r = new Set(),
l = a.length;
a.forEach(sa => new Set(sa).forEach(n => m.has(n) ? m.set(n,m.get(n)+1)
: m.set(n,1)));
m.forEach((v,k) => v === l && r.add(k));
return r;
}
var testSets = Array(1000).fill().map(_ => Array(1000).fill().map(_ => ~~(Math.random()*175+1)));
console.time("int");
result = setIntersection(testSets);
console.timeEnd("int");
console.log(JSON.stringify([...result]));

Javascript challenge - which basket contains the last apple?

I'm presented with the following challenge question:
There are a circle of 100 baskets in a room; the baskets are numbered
in sequence from 1 to 100 and each basket contains one apple.
Eventually, the apple in basket 1 will be removed but the apple in
basket 2 will be skipped. Then the apple in basket 3 will be removed.
This will continue (moving around the circle, removing an apple from a
basket, skipping the next) until only one apple in a basket remains.
Write some code to determine in which basket the remaining apple is
in.
I concluded that basket 100 will contain the last apple and here's my code:
var allApples = [];
var apples = [];
var j = 0;
var max = 100;
var o ='';
while (j < max) {
o += ++j;
allApples.push(j);
}
var apples = allApples.filter(function(val) {
return 0 == val % 2;
});
while (apples.length > 1) {
for (i = 0; i < apples.length; i += 2) {
apples.splice(i, 1);
}
}
console.log(apples);
My question is: did I do this correctly? What concerns me is the description of "a circle" of baskets. I'm not sure this is relevant at all to how I code my solution. And would the basket in which the remaining apple reside be one that would otherwise be skipped?
I hope someone can let me know if I answered this correctly, answered it partially correct or my answer is entirely wrong. Thanks for the help.

So, ... I got WAY too into this question :)
I broke out the input/output of my last answer and that revealed a pretty simple pattern.
Basically, if the total number of items is a power of 2, then it will be the last item. An additional item after that will make the second item the last item. Each additional item after that will increase the last item by 2, until you reach another item count that is again divisible by a power of 2. Rinse and repeat.
Still not a one-liner, but will be much faster than my previous answer. This will not work for 1 item.
var items = 100;
function greatestPowDivisor(n, p) {
var i = 1;
while(n - Math.pow(p, i) > 0) {
i++;
}
return Math.pow(p, (i - 1));
}
var d = greatestPowDivisor(items, 2)
var last_item = (items - d) * 2;

I believe Colin DeClue is right that there is a single statement that will solve this pattern. I would be really interested to know that answer.
Here is my brute force solution. Instead of moving items ("apples") from their original container ("basket") into a discard pile, I am simply changing the container values from true or false to indicate that an item is no longer present.
var items = 100;
var containers = [];
// Just building the array of containers
for(i=0; i<items; i++) {
containers.push(true);
}
// count all containers with value of true
function countItemsLeft(containers) {
total = 0;
for(i=0; i<containers.length; i++) {
if(containers[i]) {
total++;
}
}
return total;
}
// what is the index of the first container
// with a value of true - hopefully there's only one
function getLastItem(containers) {
for(i=0; i<containers.length; i++) {
if(containers[i]) {
return(i);
}
}
// shouldn't get here if the while loop did it's job
return false;
}
var skip = false;
// loop through the items,
// setting every other to false,
// until there is only 1 left
while(countItemsLeft(containers) > 1) {
for(i=0; i<containers.length; i++) {
if(containers[i]) {
if(skip) {
skip = false;
} else {
containers[i] = false;
skip = true;
}
}
}
}
// what's the last item? add one to account for 0 index
// to get a human readable answer
var last_item = getLastItem(containers) + 1;
Needs error checking, etc... but it should get the job done assuming items is an integer.

javascript autoupdate array

I'd like to do the following;
create a 2 -column array of arbitrary finite length (say, 10 rows)
populate it sequentially from a constant-rate datastream
once its populated, update it from the same datastream ( ie replace element 0, move 1-9 down, discard old 9)
(optimally) output an average for each column
I can probably do 4 myself, but have no idea how to do 1-3.
If it helps, I'm trying to translate this;
http://kindohm.com/2011/03/01/KinectCursorControl.html (see under the dreaded shaking cursor).

This should work ok (nice question, by the way - fun little challenge, since there are tons of ways to do it)
var pointSmoother = (function(){
var pointCount = 10, // number of points to keep
componentCount = 2, // number of components per point (i.e. 2 for x and y)
points = [];
function clear() {
var i, l;
for( i = 0, l = componentCount ; i < l ; i++ ) {
if( typeof points[i] === 'undefined' ) {
points.push([]);
} else {
points[i].splice(0, pointCount);
}
}
}
clear();
function pushPoint( /* point components */ ) {
var i, l;
for( i = 0 ; i < componentCount ; i++ ) {
points[i].unshift(arguments[i]);
points[i] = points[i].slice(0, pointCount);
}
}
function average() {
var i, j, l, sum, averages = [];
l = points[0].length;
for( i = 0 ; i < componentCount ; i++ ) {
sum = 0;
for( j = 0 ; j < l ; j++ ) {
sum += points[i][j];
}
averages.push(sum/l);
}
return averages;
}
function getPoints() {
return points;
}
return {
getPoints: getPoints,
pushPoint: pushPoint,
clear: clear,
average: average
};
}());
What it basically does is it creates an object and assigns it to pointSmoother. The object has 4 methods: pushPoint(), clear(), getPoints() and average(). At the top of the thing you can set how many coordinates a point has, and how many points (maximum) to keep. I used your example of 2 coordinates per point, and 10 points at a time.
Now, I've made the assumption that you get your values in sets of 2 at a time. Let's call those 2 values x and y. When you receive them, add them to the thing by using pointSmoother.pushPoint(x, y);. You can then get the "average point", by calling pointSmoother.average() which will return an array with (in this case) 2 values: average x and average y.
If you want to look at the array yourself, you can call pointSmoother.getPoints() which will return the points array. And lastly, pointSmoother.clear() will empty the array of previous values.
Here's a demo, of sorts: http://jsfiddle.net/tapqs/1/

var nums = [[1,2],[1,3],[2,1]];
alert(nums[0][0]); //1
alert(nums[0][1]); //2
That's a two dimensional array. Loading data works just like with any other array.
To update it sequentially you are looking at queue behavior in javascript. Use unshift() and pop().
Javascript Arrays: http://www.w3schools.com/jsref/jsref_obj_array.asp
Finally for averaging the columns assuming there are 10 positions:
var num1 = 0;
var num2 = 0;
for(int i = 0;i<10;i++)
{
num1 +=array[i][0];
num2 +=array[i][1];
}
num1 = num1/10;
num2 = num2/10;
//num1 is now the average of the first column
//num2 is now the average of the second column

Pick random property from a Javascript object

Suppose you have a Javascript object like:
{cat: 'meow', dog: 'woof', snake: 'hiss'}
Is there a more concise way to pick a random property from the object than this long winded way I came up with:
function pickRandomProperty(obj) {
var prop, len = 0, randomPos, pos = 0;
for (prop in obj) {
if (obj.hasOwnProperty(prop)) {
len += 1;
}
}
randomPos = Math.floor(Math.random() * len);
for (prop in obj) {
if (obj.hasOwnProperty(prop)) {
if (pos === randomPos) {
return prop;
}
pos += 1;
}
}
}

The chosen answer will work well. However, this answer will run faster:
var randomProperty = function (obj) {
var keys = Object.keys(obj);
return obj[keys[ keys.length * Math.random() << 0]];
};

Picking a random element from a stream
function pickRandomProperty(obj) {
var result;
var count = 0;
for (var prop in obj)
if (Math.random() < 1/++count)
result = prop;
return result;
}

I didn't think any of the examples were confusing enough, so here's a really hard to read example doing the same thing.
Edit: You probably shouldn't do this unless you want your coworkers to hate you.
var animals = {
'cat': 'meow',
'dog': 'woof',
'cow': 'moo',
'sheep': 'baaah',
'bird': 'tweet'
};
// Random Key
console.log(Object.keys(animals)[Math.floor(Math.random()*Object.keys(animals).length)]);
// Random Value
console.log(animals[Object.keys(animals)[Math.floor(Math.random()*Object.keys(animals).length)]]);
Explanation:
// gets an array of keys in the animals object.
Object.keys(animals)
// This is a number between 0 and the length of the number of keys in the animals object
Math.floor(Math.random()*Object.keys(animals).length)
// Thus this will return a random key
// Object.keys(animals)[0], Object.keys(animals)[1], etc
Object.keys(animals)[Math.floor(Math.random()*Object.keys(animals).length)]
// Then of course you can use the random key to get a random value
// animals['cat'], animals['dog'], animals['cow'], etc
animals[Object.keys(animals)[Math.floor(Math.random()*Object.keys(animals).length)]]
Long hand, less confusing:
var animalArray = Object.keys(animals);
var randomNumber = Math.random();
var animalIndex = Math.floor(randomNumber * animalArray.length);
var randomKey = animalArray[animalIndex];
// This will course this will return the value of the randomKey
// instead of a fresh random value
var randomValue = animals[randomKey];

If you are capable of using libraries, you may find that Lo-Dash JS library has lots of very useful methods for such cases. In this case, go ahead and check _.sample().
(Note Lo-Dash convention is naming the library object _.
Don't forget to check installation in the same page to set it up for your project.)
_.sample([1, 2, 3, 4]);
// → 2
In your case, go ahead and use:
_.sample({
cat: 'meow',
dog: 'woof',
mouse: 'squeak'
});
// → "woof"

You can just build an array of keys while walking through the object.
var keys = [];
for (var prop in obj) {
if (obj.hasOwnProperty(prop)) {
keys.push(prop);
}
}
Then, randomly pick an element from the keys:
return keys[keys.length * Math.random() << 0];

If you're using underscore.js you can do:
_.sample(Object.keys(animals));
Extra:
If you need multiple random properties add a number:
_.sample(Object.keys(animals), 3);
If you need a new object with only those random properties:
const props = _.sample(Object.keys(animals), 3);
const newObject = _.pick(animals, (val, key) => props.indexOf(key) > -1);

You can use the following code to pick a random property from a JavaScript object:
function randomobj(obj) {
var objkeys = Object.keys(obj)
return objkeys[Math.floor(Math.random() * objkeys.length)]
}
var example = {foo:"bar",hi:"hello"}
var randomval = example[randomobj(example)] // will return to value
// do something

Another simple way to do this would be defining a function that applies Math.random() function.
This function returns a random integer that ranges from the 'min'
function getRandomArbitrary(min, max) {
return Math.floor(Math.random() * (max - min) + min);
}
Then, extract either a 'key' or a 'value' or 'both' from your Javascript object each time you supply the above function as a parameter.
var randNum = getRandomArbitrary(0, 7);
var index = randNum;
return Object.key(index); // Returns a random key
return Object.values(index); //Returns the corresponding value.

A lot of great answers here, so let me just try to spread the awareness of the bitwise NOT (~) operator in its double-trouble variant (which I'm pretty sure I learned about on StackOverflow, anways).
Typically, you'd pick a random number from one to ten like this:
Math.floor(Math.random()*10) + 1
But bitwise operation means rounding gets done faster, so the following implementation has the potential to be noticeably more performant, assuming you're doing enough truckloads of these operations:
~~(Math.random()*10) + 1