How can you, in using a random number generator, stop a number from appearing if it has already appeared once?
Here is the current code:
var random = Math.ceil(Math.random() * 24);
But the numbers appear more than once.
You can use an array of possible values ( I think in your case it will be 24 ) :
var values = [];
for (var i = 1; i <= 24; ++i){
values.push(i);
}
When you want to pick a random number you just do:
var random = values.splice(Math.random()*values.length,1)[0];
If you know how many numbers you want then it's easy, first create an array:
var arr = [];
for (var i = 0; i <= 24; i++) arr.push(i);
Then you can shuffle it with this little function:
function shuffle(arr) {
return arr.map(function(val, i) {
return [Math.random(), i];
}).sort().map(function(val) {
return val[1];
});
}
And use it like so:
console.log(shuffle(arr)); //=> [2,10,15..] random array from 0 to 24
You can always use an hashtable and before using the new number, check if it is in there or not. That would work for bigger numbers. Now for 24, you can always shuffle an array.
You could put the numbers you generate in an array and then check against that. If the value is found, try again:
var RandomNumber = (function()
{
// Store used numbers here.
var _used = [];
return {
get: function()
{
var random = Math.ceil(Math.random() * 24);
for(var i = 0; i < _used.length; i++)
{
if(_used[i] === random)
{
// Do something here to prevent stack overflow occuring.
// For example, you could just reset the _used list when you
// hit a length of 24, or return something representing that.
return this.get();
}
}
_used.push(random);
return random;
}
}
})();
You can test being able to get all unique values like so:
for(var i = 0; i < 24; i++)
{
console.log( RandomNumber.get() );
}
The only issue currently is that you will get a stack overflow error if you try and get a random number more times than the amount of possible numbers you can get (in this case 24).
Related
I'm building a dc.js / d3.js dashboard in which I often have to make crossfilter groups containing a quantity for each key and a percentage of the total value for each key. That is why I want to make generic reduceAdd, reduceRemove and reduceInitial functions. I managed doing the first 2, but I don't understand reduceInitial behaviour :
function reduceAdd(dim,grouping,col) {
var keys = getKeys(dim); // get the keys name in an array of string
return function(p,v) {
p.total += parseInt(v[col]); // get the running total
for (var i = 0; i < keys.length; i++) {
if(v[grouping] == keys[i]) {p[keys[i]] += +v[col];}
p[keys[i]+"perc"] = p[keys[i]]/p.total; // calculate a percentage for ech key
}
return p;
}
}
function reduceRemove(dim,grouping,col) {
var keys = getKeys(dim);
return function(p,v) {
p.total -= parseInt(v[col]);
for (var i = 0; i < keys.length; i++) {
if(v[grouping] == keys[i]) {p[keys[i]] -= +v[col];}
p[keys[i]+"perc"] = p[keys[i]]/p.total;
}
return p;
}
}
This is the working non generic function reduceInitFC() {
return {total:0, LILOU:0, MARIUS:0,ORIANE:0,LILOUperc:0,MARIUSperc:0,ORIANEperc:0};
}
This is what I tried :
function reduceInit(dim) {
var keys = getKeys(dim);
var initArray= {};
initArray["total"] = 0;
for (var i = 0; i < keys.length; i++) {
initArray[keys[i]] = 0;
initArray[keys[i]+"perc"] = 0;
}
console.log(initArray); // (1)
console.log({total:0, LILOU:0, MARIUS:0,ORIANE:0,LILOUperc:0,MARIUSperc:0,ORIANEperc:0});
return function() {
return initArray;
}
}
The result is :
(1) The output gives 0 for all the keys every two iterations and some non zero values for the other iterations
When I use this function the resulting values in the group are constant respect with the keys what is not the case in reality and not the case when I hand write the zero values.
If anyone can help, it would be super kind and useful.
Best,
Theo
I think your reduceInit looks fine, but I don't think this is possible through the reduce. The reduce calculates totals incrementally -- and don't make sense until the group runs through all the rows in the dimension. So the percentages for each key can't be calculated until reduce is finished. (Likewise, while the reduce can be used to calculate averages across a single key's values, they can't be used for averages across all keys because that would be dependent on the reduced total.) (UPDATED, also see this answer's comments)
But since you already have the reduced total and the total for the key / category, you can calculate the percentage in your chart's valueAccessor.
function valueAccessor(d){
return d.value[keyName] / d.value.total
}
I would like to ask about one tiny riddle. I want to draw lots within in this example 15 figures that should always be different like in the lottery. I have already checked first condition. If for example 1 is equal 1 - please draw lots again, but there is possibility that next sorting will show again number 1. I would like to avoid this some way, to exclude it. There cannot be drawn lots figures the same. Some ideas? :-)
var figureShowed = document.querySelector("div");
var button = document.getElementById("random-figure");
var allFigures = [];
button.addEventListener("click", drawAmount);
function drawAmount() {
for (i = 0; i <= 5; i++) {
var random = Math.floor(Math.random() * 15 + 1);
if (allFigures[0] === allFigures[1, 2, 3, 4, 5]) {
allFigures[0] = random;
} else {
allFigures[i] = random;
}
}
figureShowed.textContent = allFigures.join(" | ");
}
You can use includes method to check if random number is already in allFigures array. If it is, then you draw lots again. Check the code:
var figureShowed = document.querySelector("div");
var button = document.getElementById("random-figure");
var allFigures = [];
button.addEventListener("click", drawAmount);
function drawAmount() {
allFigures = [];
for (i = 0; i <= 5; i++) {
let random;
do {
random = Math.floor(Math.random() * 15 + 1);
} while (allFigures.includes(random));
allFigures.push(random);
}
figureShowed.textContent = allFigures.join(" | ");
}
<button id="random-figure">random figure</button>
<div></div>
Add each random number to an array, alreadyDrawn, and compare subsequent randoms to the numbers in that array. If you have a match, discard and re-random until you don't have a match.
This is probably a naive or non-optimal solution to your problem, but it should work.
I have seen this question answered here, but I have an additional question that is related.
Im trying to achieve:
the same thing, but, with the output being a selection of more than 1 number, the above works fine if you only want a single value returned.
How can I return (x) amount of outputs #7 in this case into a new var or array ...? Some guidance on best practice will also be appreciated ;)
Thanks a bunch....
Just for fun,
Objective:
Create a teeny weenie web App that returns 7 variable numbers in a range [ 1 - 49 ] into an array.
`
Think return a list of Lotto Numbers
Create new array from selection using _underscore.js [Sample function]
**** I know this is easier, but im trying to get an understanding
of using Vanilla JS to accomplish this
_.sample([1, 2, 3, 4, 5, 6], 3); => [1, 6, 2]
var getLoot = Array.from(Array(50).keys()) // Create array of 49 numbers.
console.info(getLoot);
var pick = getLoot[Math.floor(Math.random() * getLoot.length)];
pick;
// pick returns a single which is fine if you want a single but, ..
// I want something like this :
var pick = function() {
// code that will return 7 numbers from the array into a new Array
// will randomize every time pick is called...
}
If you want to return more than just 1 value you can store your results into a data structure like an array. Here is a solution to the problem
assuming you can pass in your array of 50 numbers into the pick() funciton.:
var getRandomArbitrary = function(min, max) {
return Math.floor(Math.random() * (max - min) + min);
}
var pick = function(fiftyNumberArray, numberOfValuesWanted) {
var randomNums = [];
for(var i = 0; i < numberOfValuesWanted; i++) {
randomNums.push(
fiftyNumberArray[getRandomArbitrary(0, 50)]
);
}
return randomNums;
};
var fiftyNumbers = [] // <- set your array of fifty numbers
pick(fiftyNumbers, 7);
Javascript's Math.random() will return a value in between 0 and 1 (exclusive). So to get an index scaled up to the correct value to look into your array, you would want to multiply that by the formula (max - min) + min
You can use Array.prototype.splice(), Math.floor(), Math.random(), for loop to remove elements from input array, return an array containing pseudo randomly picked index from input array without duplicate indexes being selected.
function rand(n) {
var res = []; var a = Array.from(Array(n).keys());
for (;a.length;res.push(a.splice(Math.floor(Math.random()*a.length),1)[0]));
return res
}
console.log(rand(50));
One good way of doing this job is shuffling the array and picking the first n values. Such as;
function shuffle(a){
var i = a.length,
j,
tmp;
while (i > 1) {
j = Math.floor(Math.random()*i--);
tmp = a[i];
a[i] = a[j];
a[j] = tmp;
}
return a;
};
var arr = Array(50).fill().map((_,i) => i+1); //[1,2,3...49,50]
randoms = shuffle(arr).slice(0,7) // to get 7 random numbers from arrary arr
console.log(randoms)
This is probably what you want.
$(function()
{
var lol = [1,4,5,6,7,8,9];
function getPart(array,number)
{
var part = [],
index;
while(true)
{
if(part.length == number)
{
break;
}
index = $.random(0,part.length);
part.push(lol.splice(index,1));
}
return part;
}
});
$.random = function(min,max,filter)
{
var i,
n = Math.floor(Math.random()*(max-min+1)+min);
if(filter != undefined && filter.constructor == Array)
{
for(i=filter.length-1;i>-1;i--)
{
if(n == filter[i])
{
n = Math.floor(Math.random()*(max-min+1)+min)
i = filter.length;
}
}
}
return n;
}
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.
I have:
function getRandomInt(min, max){
return Math.floor(Math.random() * (max - min + 1)) + min;
}
But the problem is I want randomise the population of something with elements in an array (so they do not appear in the same order every time in the thing I am populating) so I need to ensure the number returned is unique compared to the other numbers so far.
So instead of:
for(var i = 0; i < myArray.length; i++) {
}
I have:
var i;
var count = 0;
while(count < myArray.length){
count++;
i = getRandomInt(0, myArray.length); // TODO ensure value is unique
// do stuff with myArray[i];
}
It looks like rather than independent uniform random numbers you rather want a random permutation of the set {1, 2, 3, ..., N}. I think there's a shuffle method for arrays that will do that for you.
As requested, here's the code example:
function shuffle(array) {
var top = array.length;
while (top--) {
var current = Math.floor(Math.random() * top);
var tmp = array[current];
array[current] = array[top - 1];
array[top - 1] = tmp;
}
return array;
}
Sometimes the best way to randomize something (say a card deck) is to not shuffle it before pulling it out, but to shuffle it as you pull it out.
Say you have:
var i,
endNum = 51,
array = new Array(52);
for(i = 0; i <= endNum; i++) {
array[i] = i;
}
Then you can write a function like this:
function drawNumber() {
// set index to draw from
var swap,
drawIndex = Math.floor(Math.random() * (endNum+ 1));
// swap the values at the drawn index and at the "end" of the deck
swap = array[drawIndex];
array[drawIndex] = array[endNum];
array[endNum] = swap;
endNum--;
}
Since I decrement the end counter the drawn items will be "discarded" at the end of the stack and the randomize function will only treat the items from 0 to end as viable.
This is a common pattern I've used, I may have adopted it into js incorrectly since the last time I used it was for writing a simple card game in c#. In fact I just looked at it and I had int ____ instead of var ____ lol
If i understand well, you want an array of integers but sorted randomly.
A way to do it is described here
First create a rand function :
function randOrd(){
return (Math.round(Math.random())-0.5); }
Then, randomize your array. The following example shows how:
anyArray = new Array('1','2','3','4','5');
anyArray.sort( randOrd );
document.write('Random : ' + anyArray + '<br />';);
Hope that will help,
Regards,
Max
You can pass in a function to the Array.Sort method. If this function returns a value that is randomly above or below zero then your array will be randomly sorted.
myarray.sort(function() {return 0.5 - Math.random()})
should do the trick for you without you having to worry about whether or not every random number is unique.
No loops and very simple.