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I have a nested array like this:
[ [5, 10, 5, 15, 5, 10], [10, 15, 50, 200] ]
So basically I have index 0 that also has an array with the values [5,10,5,15,5,10] and index 1 with an array with values [10,15,50,350].
let array = [[],[]];
let x = 0;
let y = 0;
for (let i = 0; i < 10; i++) {
x = ... generate random number ...
y = ... generate random number ...
while (array[0].includes(x,y) {
x = ... generate new random number ...
y = ... generate new random number ...
}
array[0].push(x,y);
}
Is there a way for me to find if for example index 0 in the array already contains the two generated values in the same order they are being pushed?
For example, I'm adding value 5,15 to the array. And the array already contains index 0 with 5 and index 1 with, 15 in that order. So I want to generate a new number if the numbers already exist in that order.
I've tried looking for solutions, but haven't found anything that helps me with what I wanna do.
I recommend making a function that checks for a specific sequence of numbers, this way if you need it to find a X number sequence you're ready.
const max = 100;
let array = [[], []];
let x = 0,
y = 0;
for (let i = 0; i < 1000; i++) {
x = Math.floor(Math.random() * max);
y = Math.floor(Math.random() * max);
while (containsSequence(array[0], [x, y])) {
console.log(`Sequence ${x},${y} already exists.`);
x = Math.floor(Math.random() * max);
y = Math.floor(Math.random() * max);
}
array[0].push(x, y);
}
/**
* #param {Array<Number>} arr
* #param {Array<Number>} sequence */
function containsSequence(arr, sequence) {
if(arr.length === 0) { return false; }
const first = sequence[0];
let index = arr.indexOf(first, 0);
while(index > -1) {
if(sequence.every((v, i) => arr[index + i] === v)) {
return true;
}
index = arr.indexOf(first, index);
if(index == -1) { return false; }
index++;
}
return false;
}
You are putting x and y one after each other in the list, if you connect them to an object or an array, you can iterate over the outer array more easily:
for (let i = 0; i < 10; i++) {
x = ... generate random number ...
y = ... generate random number ...
while (array[0].filter(i=>i.x == x && i.y == y).length) {
x = ... generate new random number ...
y = ... generate new random number ...
}
array[0].push({x,y});
}
it will also make it more simle. Of course,only if you can make this change to the array structure
Assuming that this is just a general structural question as mentioned in the comments, you can adapt your code as below.
for (let i = 0; i < 10; i++) {
int index = 1;
while (index < array[0].length){
if ((array[0][i][index-1].includes(x) && array[0][i][index].includes(y)) {
x = ... generate new random number ...
y = ... generate new random number ...
index = 0;
}
else{
index++;
}
}
array[0].push(x,y);
}
}
I need help with writing some code that will create a random number from an array of 12 numbers and print it 9 times without dupes. This has been tough for me to accomplish. Any ideas?
var nums = [1,2,3,4,5,6,7,8,9,10,11,12];
var gen_nums = [];
function in_array(array, el) {
for(var i = 0 ; i < array.length; i++)
if(array[i] == el) return true;
return false;
}
function get_rand(array) {
var rand = array[Math.floor(Math.random()*array.length)];
if(!in_array(gen_nums, rand)) {
gen_nums.push(rand);
return rand;
}
return get_rand(array);
}
for(var i = 0; i < 9; i++) {
console.log(get_rand(nums));
}
The most effective and efficient way to do this is to shuffle your numbers then print the first nine of them. Use a good shuffle algorithm.What Thilo suggested will give you poor results. See here.
Edit
Here's a brief Knuth Shuffle algorithm example:
void shuffle(vector<int> nums)
{
for (int i = nums.size()-1; i >= 0; i--)
{
// this line is really shorthand, but gets the point across, I hope.
swap(nums[i],nums[rand()%i]);
}
}
Try this once:
//Here o is the array;
var testArr = [6, 7, 12, 15, 17, 20, 21];
shuffle = function(o){ //v1.0
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;
};
shuffle(testArr);
This is relatively simple to do, the theory behind it is creating another array which keeps track of which elements of the array you have used.
var tempArray = new Array(12),i,r;
for (i=0;i<9;i++)
{
r = Math.floor(Math.random()*12); // Get a random index
if (tempArray[r] === undefined) // If the index hasn't been used yet
{
document.write(numberArray[r]); // Display it
tempArray[r] = true; // Flag it as have been used
}
else // Otherwise
{
i--; // Try again
}
}
Other methods include shuffling the array, removing used elements from the array, or moving used elements to the end of the array.
If I understand you correctly, you want to shuffle your array.
Loop a couple of times (length of array should do), and in every iteration, get two random array indexes and swap the two elements there. (Update: if you are really serious about this, this may not be the best algorithm).
You can then print the first nine array elements, which will be in random order and not repeat.
Here is a generic way of getting random numbers between min and max without duplicates:
function inArray(arr, el) {
for(var i = 0 ; i < arr.length; i++)
if(arr[i] == el) return true;
return false;
}
function getRandomIntNoDuplicates(min, max, DuplicateArr) {
var RandomInt = Math.floor(Math.random() * (max - min + 1)) + min;
if (DuplicateArr.length > (max-min) ) return false; // break endless recursion
if(!inArray(DuplicateArr, RandomInt)) {
DuplicateArr.push(RandomInt);
return RandomInt;
}
return getRandomIntNoDuplicates(min, max, DuplicateArr); //recurse
}
call with:
var duplicates =[];
for (var i = 1; i <= 6 ; i++) {
console.log(getRandomIntNoDuplicates(1,10,duplicates));
}
const nums = [1,2,3,4,5,6,7,8,9,10,11,12];
for(var i = 1 ; i < 10; i++){
result = nums[Math.floor(Math.random()*nums.length)];
const index = nums.indexOf(result);
nums.splice(index, 1);
console.log(i+' - '+result);
}
My goal is to make a randomly generated 2D Array in Javascript, that has an X amount of the same one character value while the rest of the values are equal to another character.
In this example, there are 10 rows and 10 columns for the 2D Array. 20 out of the possible 100 values of the Array should be equal to 'Y' (for yes) and the 80 others should be 'N' (for no). I want the 'Y's to be randomly placed all over the Array, and I absolute need exactly 20 of them to be 'Y's and the rest 'N's.
I had a less efficient way before, and I thought to try this approach, where after I define the Array, I make the first X amount of values a 'Y' and then the rest all 'N's. Then I shuffle the array, (using the shuffle from the underscore library) so that the 'Y's are all spread out randomly everywhere.
Is this an efficient way of getting what I need done? Are there any better solutions? I tried making a JSFiddle with my example, but the site appears to be down at the moment.
(I was unable to test my code yet to see if the shuffle worked correctly on my 2D array)
var rows = 10;
var cols = 10;
var elements = 20;
//Define Empty Array
var test = new Array(rows);
for (var k = 0; k < rows; k++)
{
test[k] = Array(cols);
}
var i = 1;
for (var x = 0; x < rows; x++)
{
for (var y = 0; y < cols; y++)
{
if (i <= elements)
{
test[x][y] = "Y";
}
else
{
test[x][y] = "N";
}
}
}
//Shuffle all those values so they're no longer in order
var shuffledTest = _.shuffle(test);
//Print in rows
for (var x = 0; x < rows; x++)
{
console.log(shuffledTest[x]);
}
A very simple solution is to first create an array, fill it with a number of "N"s, insert the "Y"s at random indexes, and then finally splitting it into the 2-dimensional array that you want:
var tmpArr = [], // Temporary 1-dimensional array to hold all values
arr = [], // The final 2-dimensional array
rows = 10,
cols = 10,
elements = 20; // Number of "Y"s
// 1. Fill temporary array with "N"s
for (var i = 0; i < rows * cols - elements; i += 1) {
tmpArr.push("N");
}
// 2. Insert "Y"s at random indexes in the temporary array
for (var i = 0; i < elements; i += 1) {
var index = Math.round(Math.random() * (tmpArr.length + 1));
tmpArr.splice(index, 0, "Y");
}
// 3. Split temporary array into 10 seperate arrays
// and insert them into the final array
for (var i = 0; i < rows; i += 1) {
var row = tmpArr.slice(i * cols, (i + 1) * cols);
arr.push(row);
}
JSBin to illustrate: http://jsbin.com/luyacora/1/edit
You can try this solution, it uses underscores range to create a pair of arrays to use as iterators, though their values don't matter.
Play around with the randomizer function to get an even distribution of 'y's
JSBIN: http://jsbin.com/yaletape/1/
var rows = _.range(0, 10, 0);
var columns = _.range(0, 10, 0);
function randomizer(mult){
return Math.floor((Math.random()*mult)+1);
}
var y_count = 0;
var matrix = _.map(rows, function(){
return _.map(columns, function(v, i){
var value;
var y_allowed = randomizer(3);
var current_y_count = 0;
if(y_count < 20 && current_y_count < y_allowed){
var rand = randomizer(5);
if(rand > 4){
value = 'y';
current_y_count++;
y_count++;
}
}
if(!value){
value = 'n';
}
return value;
});
});
//The above could be simplified to
var matrix = _.range(0,10,0).map(function(){
return _.range(0,10,0).map(function(){
//put the logic code from above here
});
});
Maybe shuflle a 2D array is not the best way. As #Zeb mentioned, here is some code that fill random positions with the 'Y' value. After that, the other positions are filled with 'N'.
http://plnkr.co/edit/avyKfgsgOSdAkRa1WOsk
var arr = [];
var cols = 10;
var rows = 10;
var positions = rows*cols; // 100
var YQty = 10; // only 10 'Y' are needed
// 'Y' values.
for(i = 0; i < YQty; i++)
{
do
{
x = parseInt(Math.random() * cols);
y = parseInt(Math.random() * rows);
filled = false;
if (typeof(arr[x]) == "undefined")
{
arr[x] = [];
}
if (typeof(arr[x][y]) == "undefined")
{
arr[x][y] = 'Y';
filled = true;
}
}
while (!filled);
}
// 'N' values.
for (x = 0; x < cols; x++)
{
if (typeof(arr[x]) == "undefined")
{
arr[x] = [];
}
for (y = 0; y < rows; y++)
{
if (arr[x][y] != 'Y')
{
arr[x][y] = 'N';
}
}
}
Shuffling the multidimensional array is not the best approach. Seeing as any sort is worse than linear time complexity. The easiest solution would be to create your multidimensional array and then set each index value to the char you want the 'rest' of the values to be. Then for 1 -> the number of other char value choose a random index and set that to the char.
Note: If the randomly picked spot has already been changed you need to choose a new one to make sure you have the right amount at the end.
I've got an array that is constantly updating with analogue readings from an Arduino pin.
I'd like to create a function that takes the latest 100 values in the array, and returns an average of them (the array has a max length of 100,000 at which it starts 'shifting' and 'pushing' to make space for new values).
I created this function, but it returns 'NaN' every time:
function returnAverage(){
var averageArray = [];
var sum = 0;
var sampleEnd = values.length
for (var x = sampleEnd - 100; x < sampleEnd; x++) {
averageArray[x] = values[x]
}
for(var i = 0; i < averageArray.length; i++){
sum += parseInt(averageArray[i]);
}
var avg = sum/averageArray.length;
console.log(avg)
}
Any ideas?
If values is a array of numbers, last maximum 100 items average:
function returnAverage(values) {
var arr = values.slice(-100);
return arr.reduce(function(a, b){ return a + b; }, 0) / (arr.length || 1);
}
Issue number one is that the final value of sum and averageArray.lnegth is 0.
It seems this would happen because the "value" array is empty.
See example below:
var values = [0,1,2,3,4,5,6];
var averageArray = [];
var sum = 0;
var sampleEnd = values.length
for (var x = sampleEnd - 7; x < sampleEnd; x++) {
averageArray[x] = values[x]
}
for(var i = 0; i < averageArray.length; i++){
sum += parseInt(averageArray[i]);
}
var avg = sum/averageArray.length;
console.log(avg)
Edit: NaN is a result of division by zero. So you might want to check for that before calculating:
if(sum == 0 || averageArray.length == 0)
{
return 0;
}
I need help with writing some code that will create a random number from an array of 12 numbers and print it 9 times without dupes. This has been tough for me to accomplish. Any ideas?
var nums = [1,2,3,4,5,6,7,8,9,10,11,12];
var gen_nums = [];
function in_array(array, el) {
for(var i = 0 ; i < array.length; i++)
if(array[i] == el) return true;
return false;
}
function get_rand(array) {
var rand = array[Math.floor(Math.random()*array.length)];
if(!in_array(gen_nums, rand)) {
gen_nums.push(rand);
return rand;
}
return get_rand(array);
}
for(var i = 0; i < 9; i++) {
console.log(get_rand(nums));
}
The most effective and efficient way to do this is to shuffle your numbers then print the first nine of them. Use a good shuffle algorithm.What Thilo suggested will give you poor results. See here.
Edit
Here's a brief Knuth Shuffle algorithm example:
void shuffle(vector<int> nums)
{
for (int i = nums.size()-1; i >= 0; i--)
{
// this line is really shorthand, but gets the point across, I hope.
swap(nums[i],nums[rand()%i]);
}
}
Try this once:
//Here o is the array;
var testArr = [6, 7, 12, 15, 17, 20, 21];
shuffle = function(o){ //v1.0
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;
};
shuffle(testArr);
This is relatively simple to do, the theory behind it is creating another array which keeps track of which elements of the array you have used.
var tempArray = new Array(12),i,r;
for (i=0;i<9;i++)
{
r = Math.floor(Math.random()*12); // Get a random index
if (tempArray[r] === undefined) // If the index hasn't been used yet
{
document.write(numberArray[r]); // Display it
tempArray[r] = true; // Flag it as have been used
}
else // Otherwise
{
i--; // Try again
}
}
Other methods include shuffling the array, removing used elements from the array, or moving used elements to the end of the array.
If I understand you correctly, you want to shuffle your array.
Loop a couple of times (length of array should do), and in every iteration, get two random array indexes and swap the two elements there. (Update: if you are really serious about this, this may not be the best algorithm).
You can then print the first nine array elements, which will be in random order and not repeat.
Here is a generic way of getting random numbers between min and max without duplicates:
function inArray(arr, el) {
for(var i = 0 ; i < arr.length; i++)
if(arr[i] == el) return true;
return false;
}
function getRandomIntNoDuplicates(min, max, DuplicateArr) {
var RandomInt = Math.floor(Math.random() * (max - min + 1)) + min;
if (DuplicateArr.length > (max-min) ) return false; // break endless recursion
if(!inArray(DuplicateArr, RandomInt)) {
DuplicateArr.push(RandomInt);
return RandomInt;
}
return getRandomIntNoDuplicates(min, max, DuplicateArr); //recurse
}
call with:
var duplicates =[];
for (var i = 1; i <= 6 ; i++) {
console.log(getRandomIntNoDuplicates(1,10,duplicates));
}
const nums = [1,2,3,4,5,6,7,8,9,10,11,12];
for(var i = 1 ; i < 10; i++){
result = nums[Math.floor(Math.random()*nums.length)];
const index = nums.indexOf(result);
nums.splice(index, 1);
console.log(i+' - '+result);
}