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
Related
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;
}
}
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;
}
}
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.
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]));
I am currently using FreeCodeCamp to try to learn basic JavaScript scripting. The problem that I am currently working on is:
http://www.freecodecamp.com/challenges/bonfire-map-the-debris.
The problem involves using OOP to solve a specific task (calculating orbital periods from the given altitude).
My code is as follows:
function orbitalPeriod(arr) {
var GM = 398600.4418;
var earthRadius = 6367.4447;
this.arr = arr;
for(var i = 0; i < arr.length; i++){
var altitude = this.arr[i]["avgAlt"] + earthRadius;
var calc = Math.round((2*Math.PI) * Math.sqrt(Math.pow(altitude,3) / GM),1);
this.arr[i]["avgAlt"] = calc;
}
return this.arr;
}
orbitalPeriod([{name : "sputkin", avgAlt : 35873.5553}]);
The issue is not with my calculations. Rather, when I submit my code, I get: "expected [ { name: 'sputkin', avgAlt: 86400 } ] to deeply equal [ Array (1) ]". Does anyone know why it is telling me that I should return an Array (1)?
The test suite is expecting the return array to contain an object with the properties name and orbitalPeriod - yours is returning an array containing an object with the properties name and avgAlt.
Side note, don't use the this keyword unless you're sure as to what it does - and I promise you it does not do what you think it does here.
Here's the solution, compare it with yours. Your calculations were correct, so good job on that part.
function orbitalPeriod(arr) {
var GM = 398600.4418,
earthRadius = 6367.4447,
output = [], altitude, calc;
for (var i = 0; i < arr.length; i++){
altitude = arr[i].avgAlt + earthRadius;
calc = Math.round((2*Math.PI) * Math.sqrt(Math.pow(altitude,3) / GM));
output.push({
name: arr[i].name,
orbitalPeriod: calc
});
}
return output;
}
orbitalPeriod([{name : "sputkin", avgAlt : 35873.5553}]);
Bonus note: Math.round() only takes one parameter.
Bonus answer:
Array.prototype.map() makes this super clean, if we're not tuning for performance.
function orbitalPeriod(arr) {
var GM = 398600.4418,
earthRadius = 6367.4447;
return arr.map(function (o) {
return {
name: o.name,
orbitalPeriod: Math.round((2 * Math.PI) * Math.sqrt(Math.pow(o.avgAlt + earthRadius, 3) / GM))
};
});
}
orbitalPeriod([{name : "sputkin", avgAlt : 35873.5553}]);