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I have an array like this:
var arr = [0,1,2,3,4,5]
What I need is to write a function, which will generate a new array of random length (but less or equal than arr.length) and won't generate repeated numbers. Something like:
var arrNew = [2,4,0]
but not
var arrNew = [2,4,2,2,0]
I came up with this, but it generates repeated elements:
var list = [0,1,2,3,4,5];
var getRandomNumberRange = function (min, max) {
return Math.floor(Math.random() * (max - min) + min);
};
var getRandomArr = function (list) {
var arr = new Array(getRandomNumberRange(1, 6));
for (var i = 0; i < arr.length; i++) {
arr[i] = list[getRandomNumber(list.length)];
}
return arr;
};
Appreciate your help!
NOTE: there's nothing stopping this code from giving you an empty array as your random array.
var arr = [0,1,2,3,4,5, 36, 15, 25]
function generateRandomArrayFromSeed(arr) {
// Create a new set to keep track of what values we've used
const used = new Set();
// generate a new array.
const newArr = [];
// what random length will we be using today?
const newLength = getRandomNumber(arr.length);
// while our new array is still less than our newly randomized length, we run this block of code
while(newArr.length < newLength) {
// first, we generate a random index
const randomNumber = getRandomNumber(arr.length);
// then we check if the value at this index is in our set; if it is, we continue to the next iteration of the loop, and none of the remaining code is ran.
if(used.has(arr[randomNumber])) continue;
// otherwise, push this new value to our array
newArr.push(arr[randomNumber]);
// and update our used set by adding our random number
used.add(arr[randomNumber]);
}
// finally return this new random array.
return newArr;
}
function getRandomNumber(limit = 10) {
return Math.floor(Math.random() * limit);
}
console.log(generateRandomArrayFromSeed(arr));
Moving the data vrom an array datastructure to a Set could solve the problem.
Sets dont accept duplicate values.
For more on Set in javascript, see MDN here
To convert the Set back to an array, use Array.from()
To avoid duplicates in you final array just use Set and convert it back to an array using Array.from
Save you result inside set as below:
arr = Array.from(new Set(<result array>));
Please, how do you populate an array say ‘num’ with numbers not in a second array say ‘fig’? I’m trying to use a loop to have the values of the already populated array ‘fig’ compared to ‘num’ which is to be populated with integers not found in ‘fig’. I’m a bit confused.
If you need to do an array with n numbers you can use this two ways.
const arrayLength = 100;
const numberArray = [...new Array(arrayLength).keys()]
const anotherWay = new Array(arrayLength).fill().map((_, idx) => idx + 1);
console.log(numberArray, anotherWay)
so to do this we have to do a few things:
1) define an existing array with numbers to avoid
2) define length on new array
3) generate a random number and make it an integer
4) check to see if we need to avoid
5) if it's a new value add it to the second array
var first=[55,45,35,1,2,3,4,5];
var second = [];
var i = 7;
var x;
while (i != 0){
x = ~~(Math.random()*100);
var check = false;
for(let j=0; j<first.length;j++){
if(x == first[j]){
check = true;
}
}
if(!check){
second.push(x);
i--;
}
}
console.log(second);
const fig = [-21, 0, 3, 6, 7, 42]
const min = Math.min(...fig) // or fig[0] if the the array is already sorted
const max = Math.max(...fig) // or fig[fig.length - 1]
const num = Array.from({ length: max - min }, (_, i) => i + min)
.filter(el => !fig.includes(el))
or, saving one loop
const num = Array.from({ length: max - min }).reduce((acc, _, i) => {
const curr = i + min
if (!fig.includes(curr)) {
return acc.concat(curr)
}
return acc
}, [])
This is assuming your range is from the smallest number in fig to the largest in fig.
I have 2 array:
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
Want to get 1 merged array with the sum of corresponding keys;
var array1 = [[1,10],[2,10],[3,10],[4,10],[5,50],[6,50],[7,10],[8,10],[9,10]];
Both arrays have unique keys, but the corresponding keys needs to be summed.
I tried loops, concat, etc but can't get the result i need.
anybody done this before?
You can use .reduce() to pass along an object that tracks the found sets, and does the addition.
DEMO: http://jsfiddle.net/aUXLV/
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
var result =
array1.concat(array2)
.reduce(function(ob, ar) {
if (!(ar[0] in ob.nums)) {
ob.nums[ar[0]] = ar
ob.result.push(ar)
} else
ob.nums[ar[0]][1] += ar[1]
return ob
}, {nums:{}, result:[]}).result
If you need the result to be sorted, then add this to the end:
.sort(function(a,b) {
return a[0] - b[0];
})
This is one way to do it:
var sums = {}; // will keep a map of number => sum
// for each input array (insert as many as you like)
[array1, array2].forEach(function(array) {
//for each pair in that array
array.forEach(function(pair) {
// increase the appropriate sum
sums[pair[0]] = pair[1] + (sums[pair[0]] || 0);
});
});
// now transform the object sums back into an array of pairs
var results = [];
for(var key in sums) {
results.push([key, sums[key]]);
}
See it in action.
a short routine can be coded using [].map()
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
array1=array2.concat(array1).map(function(a){
var v=this[a[0]]=this[a[0]]||[a[0]];
v[1]=(v[1]||0)+a[1];
return this;
},[])[0].slice(1);
alert(JSON.stringify(array1));
//shows: [[1,10],[2,10],[3,10],[4,10],[5,50],[6,50],[7,10],[8,10],[9,10]]
i like how it's just 3 line of code, doesn't need any internal function calls like push() or sort() or even an if() statement.
Try this:
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
var res = [];
someReasonableName(array1, res);
someReasonableName(array2, res);
function someReasonableName(arr, res) {
var arrLen = arr.length
, i = 0
;
for(i; i < arrLen; i++) {
var ar = arr[i]
, index = ar[0]
, value = ar[1]
;
if(!res[index]) {
res[index] = [index, 0];
}
res[index][1] += value;
}
}
console.log(JSON.stringify(res, null, 2));
So, the result may have holes. Just like 0th index. Use the below function if you want to ensure there are no holes.
function compact(arr) {
var i = 0
, arrLen = arr.length
, res = []
;
for(i; i < arrLen; i++) {
var v = arr[i]
;
if(v) {
res[res.length] = v;
}
}
return res;
}
So, you can do:
var holesRemoved = compact(res);
And finally if you don't want the 0th elem of res. Do res.shift();
Disclaimer: I am not good with giving reasonable names.
The simple solution is like this.
function sumArrays(...arrays) {
const n = arrays.reduce((max, xs) => Math.max(max, xs.length), 0);
const result = Array.from({ length: n });
return result.map((_, i) => arrays.map(xs => xs[i] || 0).reduce((sum, x) => sum + x, 0));
}
console.log(...sumArrays([0, 1, 2], [1, 2, 3, 4], [1, 2])); // 2 5 5 4
What is a clean way of taking a random sample, without replacement from an array in javascript? So suppose there is an array
x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
and I want to randomly sample 5 unique values; i.e. generate a random subset of length 5. To generate one random sample one could do something like:
x[Math.floor(Math.random()*x.length)];
But if this is done multiple times, there is a risk of a grabbing the same entry multiple times.
I suggest shuffling a copy of the array using the Fisher-Yates shuffle and taking a slice:
function getRandomSubarray(arr, size) {
var shuffled = arr.slice(0), i = arr.length, temp, index;
while (i--) {
index = Math.floor((i + 1) * Math.random());
temp = shuffled[index];
shuffled[index] = shuffled[i];
shuffled[i] = temp;
}
return shuffled.slice(0, size);
}
var x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];
var fiveRandomMembers = getRandomSubarray(x, 5);
Note that this will not be the most efficient method for getting a small random subset of a large array because it shuffles the whole array unnecessarily. For better performance you could do a partial shuffle instead:
function getRandomSubarray(arr, size) {
var shuffled = arr.slice(0), i = arr.length, min = i - size, temp, index;
while (i-- > min) {
index = Math.floor((i + 1) * Math.random());
temp = shuffled[index];
shuffled[index] = shuffled[i];
shuffled[i] = temp;
}
return shuffled.slice(min);
}
A little late to the party but this could be solved with underscore's new sample method (underscore 1.5.2 - Sept 2013):
var x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];
var randomFiveNumbers = _.sample(x, 5);
In my opinion, I do not think shuffling the entire deck necessary. You just need to make sure your sample is random not your deck. What you can do, is select the size amount from the front then swap each one in the sampling array with another position in it. So, if you allow replacement you get more and more shuffled.
function getRandom(length) { return Math.floor(Math.random()*(length)); }
function getRandomSample(array, size) {
var length = array.length;
for(var i = size; i--;) {
var index = getRandom(length);
var temp = array[index];
array[index] = array[i];
array[i] = temp;
}
return array.slice(0, size);
}
This algorithm is only 2*size steps, if you include the slice method, to select the random sample.
More Random
To make the sample more random, we can randomly select the starting point of the sample. But it is a little more expensive to get the sample.
function getRandomSample(array, size) {
var length = array.length, start = getRandom(length);
for(var i = size; i--;) {
var index = (start + i)%length, rindex = getRandom(length);
var temp = array[rindex];
array[rindex] = array[index];
array[index] = temp;
}
var end = start + size, sample = array.slice(start, end);
if(end > length)
sample = sample.concat(array.slice(0, end - length));
return sample;
}
What makes this more random is the fact that when you always just shuffling the front items you tend to not get them very often in the sample if the sampling array is large and the sample is small. This would not be a problem if the array was not supposed to always be the same. So, what this method does is change up this position where the shuffled region starts.
No Replacement
To not have to copy the sampling array and not worry about replacement, you can do the following but it does give you 3*size vs the 2*size.
function getRandomSample(array, size) {
var length = array.length, swaps = [], i = size, temp;
while(i--) {
var rindex = getRandom(length);
temp = array[rindex];
array[rindex] = array[i];
array[i] = temp;
swaps.push({ from: i, to: rindex });
}
var sample = array.slice(0, size);
// Put everything back.
i = size;
while(i--) {
var pop = swaps.pop();
temp = array[pop.from];
array[pop.from] = array[pop.to];
array[pop.to] = temp;
}
return sample;
}
No Replacement and More Random
To apply the algorithm that gave a little bit more random samples to the no replacement function:
function getRandomSample(array, size) {
var length = array.length, start = getRandom(length),
swaps = [], i = size, temp;
while(i--) {
var index = (start + i)%length, rindex = getRandom(length);
temp = array[rindex];
array[rindex] = array[index];
array[index] = temp;
swaps.push({ from: index, to: rindex });
}
var end = start + size, sample = array.slice(start, end);
if(end > length)
sample = sample.concat(array.slice(0, end - length));
// Put everything back.
i = size;
while(i--) {
var pop = swaps.pop();
temp = array[pop.from];
array[pop.from] = array[pop.to];
array[pop.to] = temp;
}
return sample;
}
Faster...
Like all of these post, this uses the Fisher-Yates Shuffle. But, I removed the over head of copying the array.
function getRandomSample(array, size) {
var r, i = array.length, end = i - size, temp, swaps = getRandomSample.swaps;
while (i-- > end) {
r = getRandom(i + 1);
temp = array[r];
array[r] = array[i];
array[i] = temp;
swaps.push(i);
swaps.push(r);
}
var sample = array.slice(end);
while(size--) {
i = swaps.pop();
r = swaps.pop();
temp = array[i];
array[i] = array[r];
array[r] = temp;
}
return sample;
}
getRandomSample.swaps = [];
Or... if you use underscore.js...
_und = require('underscore');
...
function sample(a, n) {
return _und.take(_und.shuffle(a), n);
}
Simple enough.
You can get a 5 elements sample by this way:
var sample = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
.map(a => [a,Math.random()])
.sort((a,b) => {return a[1] < b[1] ? -1 : 1;})
.slice(0,5)
.map(a => a[0]);
You can define it as a function to use in your code:
var randomSample = function(arr,num){ return arr.map(a => [a,Math.random()]).sort((a,b) => {return a[1] < b[1] ? -1 : 1;}).slice(0,num).map(a => a[0]); }
Or add it to the Array object itself:
Array.prototype.sample = function(num){ return this.map(a => [a,Math.random()]).sort((a,b) => {return a[1] < b[1] ? -1 : 1;}).slice(0,num).map(a => a[0]); };
if you want, you can separate the code for to have 2 functionalities (Shuffle and Sample):
Array.prototype.shuffle = function(){ return this.map(a => [a,Math.random()]).sort((a,b) => {return a[1] < b[1] ? -1 : 1;}).map(a => a[0]); };
Array.prototype.sample = function(num){ return this.shuffle().slice(0,num); };
While I strongly support using the Fisher-Yates Shuffle, as suggested by Tim Down, here's a very short method for achieving a random subset as requested, mathematically correct, including the empty set, and the given set itself.
Note solution depends on lodash / underscore:
Lodash v4
const _ = require('loadsh')
function subset(arr) {
return _.sampleSize(arr, _.random(arr.length))
}
Lodash v3
const _ = require('loadsh')
function subset(arr) {
return _.sample(arr, _.random(arr.length));
}
If you're using lodash the API changed in 4.x:
const oneItem = _.sample(arr);
const nItems = _.sampleSize(arr, n);
https://lodash.com/docs#sampleSize
A lot of these answers talk about cloning, shuffling, slicing the original array. I was curious why this helps from a entropy/distribution perspective.
I'm no expert but I did write a sample function using the indexes to avoid any array mutations — it does add to a Set though. I also don't know how the random distribution on this but the code was simple enough to I think warrant an answer here.
function sample(array, size = 1) {
const { floor, random } = Math;
let sampleSet = new Set();
for (let i = 0; i < size; i++) {
let index;
do { index = floor(random() * array.length); }
while (sampleSet.has(index));
sampleSet.add(index);
}
return [...sampleSet].map(i => array[i]);
}
const words = [
'confused', 'astonishing', 'mint', 'engine', 'team', 'cowardly', 'cooperative',
'repair', 'unwritten', 'detailed', 'fortunate', 'value', 'dogs', 'air', 'found',
'crooked', 'useless', 'treatment', 'surprise', 'hill', 'finger', 'pet',
'adjustment', 'alleged', 'income'
];
console.log(sample(words, 4));
Perhaps I am missing something, but it seems there is a solution that does not require the complexity or potential overhead of a shuffle:
function sample(array,size) {
const results = [],
sampled = {};
while(results.length<size && results.length<array.length) {
const index = Math.trunc(Math.random() * array.length);
if(!sampled[index]) {
results.push(array[index]);
sampled[index] = true;
}
}
return results;
}
Here is another implementation based on Fisher-Yates Shuffle. But this one is optimized for the case where the sample size is significantly smaller than the array length. This implementation doesn't scan the entire array nor allocates arrays as large as the original array. It uses sparse arrays to reduce memory allocation.
function getRandomSample(array, count) {
var indices = [];
var result = new Array(count);
for (let i = 0; i < count; i++ ) {
let j = Math.floor(Math.random() * (array.length - i) + i);
result[i] = array[indices[j] === undefined ? j : indices[j]];
indices[j] = indices[i] === undefined ? i : indices[i];
}
return result;
}
You can remove the elements from a copy of the array as you select them. Performance is probably not ideal, but it might be OK for what you need:
function getRandom(arr, size) {
var copy = arr.slice(0), rand = [];
for (var i = 0; i < size && i < copy.length; i++) {
var index = Math.floor(Math.random() * copy.length);
rand.push(copy.splice(index, 1)[0]);
}
return rand;
}
For very large arrays, it's more efficient to work with indexes rather than the members of the array.
This is what I ended up with after not finding anything I liked on this page.
/**
* Get a random subset of an array
* #param {Array} arr - Array to take a smaple of.
* #param {Number} sample_size - Size of sample to pull.
* #param {Boolean} return_indexes - If true, return indexes rather than members
* #returns {Array|Boolean} - An array containing random a subset of the members or indexes.
*/
function getArraySample(arr, sample_size, return_indexes = false) {
if(sample_size > arr.length) return false;
const sample_idxs = [];
const randomIndex = () => Math.floor(Math.random() * arr.length);
while(sample_size > sample_idxs.length){
let idx = randomIndex();
while(sample_idxs.includes(idx)) idx = randomIndex();
sample_idxs.push(idx);
}
sample_idxs.sort((a, b) => a > b ? 1 : -1);
if(return_indexes) return sample_idxs;
return sample_idxs.map(i => arr[i]);
}
My approach on this is to create a getRandomIndexes method that you can use to create an array of the indexes that you will pull from the main array. In this case, I added a simple logic to avoid the same index in the sample. this is how it works
const getRandomIndexes = (length, size) => {
const indexes = [];
const created = {};
while (indexes.length < size) {
const random = Math.floor(Math.random() * length);
if (!created[random]) {
indexes.push(random);
created[random] = true;
}
}
return indexes;
};
This function independently of whatever you have is going to give you an array of indexes that you can use to pull the values from your array of length length, so could be sampled by
const myArray = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']
getRandomIndexes(myArray.length, 3).map(i => myArray[i])
Every time you call the method you are going to get a different sample of myArray. at this point, this solution is cool but could be even better to sample different sizes. if you want to do that you can use
getRandomIndexes(myArray.length, Math.ceil(Math.random() * 6)).map(i => myArray[i])
will give you a different sample size from 1-6 every time you call it.
I hope this has helped :D
Underscore.js is about 70kb. if you don't need all the extra crap, rando.js is only about 2kb (97% smaller), and it works like this:
console.log(randoSequence([8, 6, 7, 5, 3, 0, 9]).slice(-5));
<script src="https://randojs.com/2.0.0.js"></script>
You can see that it keeps track of the original indices by default in case two values are the same but you still care about which one was picked. If you don't need those, you can just add a map, like this:
console.log(randoSequence([8, 6, 7, 5, 3, 0, 9]).slice(-5).map((i) => i.value));
<script src="https://randojs.com/2.0.0.js"></script>
D3-array's shuffle uses the Fisher-Yeates shuffle algorithm to randomly re-order arrays. It is a mutating function - meaning that the original array is re-ordered in place, which is good for performance.
D3 is for the browser - it is more complicated to use with node.
https://github.com/d3/d3-array#shuffle
npm install d3-array
//import {shuffle} from "d3-array"
let x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];
d3.shuffle(x)
console.log(x) // it is shuffled
<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/5.0.0/d3.min.js"></script>
If you don't want to mutate the original array
let x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];
let shuffled_x = d3.shuffle(x.slice()) //calling slice with no parameters returns a copy of the original array
console.log(x) // not shuffled
console.log(shuffled_x)
<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/5.0.0/d3.min.js"></script>
This question already has answers here:
Getting a random value from a JavaScript array
(28 answers)
Closed 7 years ago.
var items = Array(523, 3452, 334, 31, ..., 5346);
How do I get random item from items?
var item = items[Math.floor(Math.random()*items.length)];
1. solution: define Array prototype
Array.prototype.random = function () {
return this[Math.floor((Math.random()*this.length))];
}
that will work on inline arrays
[2,3,5].random()
and of course predefined arrays
var list = [2,3,5]
list.random()
2. solution: define custom function that accepts list and returns element
function get_random (list) {
return list[Math.floor((Math.random()*list.length))];
}
get_random([2,3,5])
Use underscore (or loDash :)):
var randomArray = [
'#cc0000','#00cc00', '#0000cc'
];
// use _.sample
var randomElement = _.sample(randomArray);
// manually use _.random
var randomElement = randomArray[_.random(randomArray.length-1)];
Or to shuffle an entire array:
// use underscore's shuffle function
var firstRandomElement = _.shuffle(randomArray)[0];
If you really must use jQuery to solve this problem (NB: you shouldn't):
(function($) {
$.rand = function(arg) {
if ($.isArray(arg)) {
return arg[$.rand(arg.length)];
} else if (typeof arg === "number") {
return Math.floor(Math.random() * arg);
} else {
return 4; // chosen by fair dice roll
}
};
})(jQuery);
var items = [523, 3452, 334, 31, ..., 5346];
var item = jQuery.rand(items);
This plugin will return a random element if given an array, or a value from [0 .. n) given a number, or given anything else, a guaranteed random value!
For extra fun, the array return is generated by calling the function recursively based on the array's length :)
Working demo at http://jsfiddle.net/2eyQX/
Here's yet another way:
function rand(items) {
// "~~" for a closest "int"
return items[~~(items.length * Math.random())];
}
Or as recommended below by #1248177:
function rand(items) {
// "|" for a kinda "int div"
return items[items.length * Math.random() | 0];
}
var random = items[Math.floor(Math.random()*items.length)]
jQuery is JavaScript! It's just a JavaScript framework. So to find a random item, just use plain old JavaScript, for example,
var randomItem = items[Math.floor(Math.random()*items.length)]
// 1. Random shuffle items
items.sort(function() {return 0.5 - Math.random()})
// 2. Get first item
var item = items[0]
Shorter:
var item = items.sort(function() {return 0.5 - Math.random()})[0];
Even shoter (by José dB.):
let item = items.sort(() => 0.5 - Math.random())[0];
var rndval=items[Math.floor(Math.random()*items.length)];
var items = Array(523,3452,334,31,...5346);
function rand(min, max) {
var offset = min;
var range = (max - min) + 1;
var randomNumber = Math.floor( Math.random() * range) + offset;
return randomNumber;
}
randomNumber = rand(0, items.length - 1);
randomItem = items[randomNumber];
credit:
Javascript Function: Random Number Generator
If you are using node.js, you can use unique-random-array. It simply picks something random from an array.
An alternate way would be to add a method to the Array prototype:
Array.prototype.random = function (length) {
return this[Math.floor((Math.random()*length))];
}
var teams = ['patriots', 'colts', 'jets', 'texans', 'ravens', 'broncos']
var chosen_team = teams.random(teams.length)
alert(chosen_team)
const ArrayRandomModule = {
// get random item from array
random: function (array) {
return array[Math.random() * array.length | 0];
},
// [mutate]: extract from given array a random item
pick: function (array, i) {
return array.splice(i >= 0 ? i : Math.random() * array.length | 0, 1)[0];
},
// [mutate]: shuffle the given array
shuffle: function (array) {
for (var i = array.length; i > 0; --i)
array.push(array.splice(Math.random() * i | 0, 1)[0]);
return array;
}
}