I want to parse my array to 4 equal intervals. For example, array [12, 48] parse to 4 intervals [12,21], [21, 30], [30, 39], [39, 48]. I could only parse my Array to pairs (code here). But I don't know how to do what I want.
var arrayTest = [];
for (var i = 0; i < 1000; i++) arrayTest[i] = i;
var ab = [2, 6];
start = ab[0];
finish = ab[1];
var ab_new = [];
for (var i = start; i <= finish; i++) ab_new[i]= arrayTest[i];
var output = [];
for (var i = start; i < ab_new.length - 1; ++i) {
output[i] = [];
output[i].push(ab_new[i]);
output[i].push(ab_new[i + 1]);
}
console.log(output);
My output is:
[ , , [ 2, 3 ], [ 3, 4 ], [ 4, 5 ], [ 5, 6 ] ]
You can calculate the interval and use it to map in an array of length 4 over the index.
const ab = [12, 48]
const start = ab[0];
const end = ab[1];
const interval = (end - start) / 4;
const out = Array(4)
.fill()
.map((_, i) => [start + i*interval, start + (i + 1) * interval])
You could take a length for each interval and add the lenght to the start value for each part.
function getIntervals(range, parts) {
var result = [],
length = (range[1] - range[0]) / parts,
i = 0,
t;
while (i < parts) {
t = range[0] + i * length;
result.push([t, t + length]);
i++;
}
return result;
}
console.log(getIntervals([12, 48], 4));
Another approach by using length as increment value.
function getIntervals(range, parts) {
var result = [],
length = (range[1] - range[0]) / parts,
i = range[0];
while (i < range[1]) {
result.push([i, i += length]);
}
return result;
}
console.log(getIntervals([12, 48], 4));
ES6
function getIntervals(range, parts) {
var l = (range[1] - range[0]) / parts,
i = range[0];
return Array.from({ length: parts}, _ => [i, i += l]);
}
console.log(getIntervals([12, 48], 4));
let a=[12,48]; // your array
var interval = 4;
var diff =(a[1]-a[0])/interval; // calculating the diff
var start = a[0]; //start value
var b =[]; // output array
for(var i=0; i<interval; i++){
b.push( [start, start + diff]); // for each iteration push start and start+diff
start+=diff;
}
console.log(b)
function sep4(inp){
var diff = inp[1] - inp[0]
var add = diff/4
var ar = []
for(i = 1; i < 5; i++){
ar.push([inp[0], inp[0] + (add * i)])
}
return ar
}
// just call the function as sep4([12, 48])
Related
Input Arr=[1,2,3,4,5,6,7,8,9,10]
Expected output:-
Arr1 = [1,2,3,4,5,6,7] = 28
Arr2 = [8,9,10] = 27
The sum of arrays should be almost the same..
It can also be 3 or more parts
How to achieve this via custom function?
let Arr = [1,2,3,4,5,6,7,8,9,10]
const numberOfParts = 2
function SplitArr(Array, Parts){
/* ... */
}
let result = SplitArr(Arr,numberOfParts)
/* result should be [[1,2,3,4,5,6,7],[8,9,10]] */
/* output can be in any format as long as it can get the parts */
I think you can't do that directly by JS functions.
You have to create a custom function to achieve this.
I have considered dividing the array into 2 equal parts.
You can't always split the array equally. Here in this array, you can't partition array into more than 2 subparts, otherwise it will give more than 3 parts as some of the elements are present there having sum more than the partitioned Sum.
Note: I treated the array to be sorted, otherwise it depends on the usecase.
Note: I have updated the old implementation based on the updated question requirement
let arr=[1,2,3,4,5,6,7,8,9,10]
function splitArrayEqually(arr, parts=2){
//get the total sum of the array
let sum = arr.reduce((currentSum, value) => currentSum+value ,0);
//get the half sum of the array
let partitionedSum = Math.ceil(sum/parts);
let start=0, end=0, currentSum=0;
let splittedArray=[];
//get the index till which the sum is less then equal partitioned sum
while(end < arr.length){
if(currentSum+arr[end] > partitionedSum){
splittedArray.push(arr.slice(start,end));
start = end; //start new window from current index
currentSum = 0; //make sum =0
}
//add current end index to sum
currentSum += arr[end];
end++;
}
splittedArray.push(arr.slice(start));
return splittedArray;
}
splitted = splitArrayEqually(arr,3);
console.log(splitted)
let Arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
const numberOfParts = 3
function sumOfArray(arr) {
if (arr) {
if (arr.length > 0) {
let sum = 0
for (let i = 0; i < arr.length; i++) sum += arr[i]
return sum
} else {
return 0
}
} else {
return 0
}
}
function SplitArr(Array, Parts) {
let lastIndex = 0
let result = []
function getReamingSum(arr) {
let psum = sumOfArray(Array.slice(lastIndex)) / Parts
console.log('psum ' + psum)
return psum + Parts
}
let psum = getReamingSum(Array)
for (let j = 0; j < Parts; j++) {
let total = 0
for (let i = 0; i < Array.length; i++) {
if (i >= lastIndex) {
total += Array[i]
if (total < psum || j === Parts - 1) {
if (result[j]?.length > 0) {
result[j].push(Array[i])
} else {
let arr = []
arr.push(Array[i])
result[j] = arr
}
lastIndex = i + 1
}
}
}
}
return result
}
let result = SplitArr(Arr, numberOfParts)
console.log(result)
Assuming the array isn't sorted,using a 2D array, with each sub array with sum almost equal to (sum of array / n).
let arr = [9,2,10,4,5,6,7,8,1,3]
arr.sort(function(a, b) { return a - b; });
const sum = arr.reduce((a, b) => a + b, 0);
const n = 2;
const result = [];
let s = 0;
let j = 0;
result[j] = [];
for(let i=0; i<arr.length; i++){
if(s <= Math.floor(sum/n)){
result[j].push(arr[i]);
s +=arr[i];
}
else{
s = 0;
j = j + 1;
result[j] = [];
result[j].push(arr[i]);
}
}
console.log(result)
O/P:
[ [1, 2, 3, 4,5, 6, 7], [ 8, 9, 10 ] ]
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
const splitArray = (arr,parts) => {
const totalSum = arr.reduce((acc, item) => {
acc += item;
return acc;
}, 0)
const splitSum = Math.floor(totalSum / parts);
const arrObj = arr.reduce((acc, item,index) => {
acc.sum = acc.sum || 0;
acc.split = acc.split || {};
const pointer = Math.floor(acc.sum / splitSum);
//console.log(item,acc.sum, splitSum, pointer);
acc.split[pointer] = acc.split[pointer] || [];
acc.split[pointer].push(item);
acc.splitSum = splitSum;
acc.sum += item;
return acc;
}, {})
return arrObj;
}
console.log(splitArray(arr,2).split)
You're better off making a custom function:
let currentTotal = 0
let tempList = []
Arr.forEach(val => {
if (val >= 27) {
// push tempList to a new array
tempList = [];
currentTotal = val;
} else {
tempList.push(val);
currentTotal += val;
}
})
Here is the idea:
var a = [4, 5, 6];
for (var m = 0; m < a[0]; m++)
for (var n = 0; n < a[1]; n++)
for (var p = 0; p < a[2]; p++)
console.log(`${m} + ${n} + ${p} = ${m+n+p}`);
Live Copy:
// This just tells the Stack Snippets in-snippet console not
// to throw away entries once it reaches a max (the default max
// is just the last 50 logs).
console.config({maxEntries: Infinity});
var a = [4, 5, 6];
for (var m = 0; m < a[0]; m++)
for (var n = 0; n < a[1]; n++)
for (var p = 0; p < a[2]; p++)
console.log(`${m} + ${n} + ${p} = ${m+n+p}`);
/* This just makes the console take up the full output area */
.as-console-wrapper {
max-height: 100% !important;
}
The code would get longer if the array a has more indexes. Could the code be shorten using Array.map or filter or a function?
We can do this without taking up massive amounts of memory, and fairly simply, by using recursion:
const process = (array, n, numbers) => {
if (n < array.length) {
// Not done yet, recurse once for each number at this level
const max = array[n];
for (let i = 0; i < max; ++i) {
process(array, n + 1, [...numbers, i]);
}
} else {
// Done with this level, process the numbers we got
console.log(`${numbers.join(" + ")} = ${numbers.reduce((s, e) => s + e)}`);
}
}
process([4, 5, 6], 0, []);
Live Copy, with cross-checking against your results to ensure the above does the same thing:
// This just tells the Stack Snippets in-snippet console not
// to throw away entries once it reaches a max (the default max
// is just the last 50 logs).
console.config({maxEntries: Infinity});
function thisSolution() {
const results = [];
const process = (array, n, numbers) => {
if (n < array.length) {
// Not done yet, recurse once for each number at this level
const max = array[n];
for (let i = 0; i < max; ++i) {
process(array, n + 1, [...numbers, i]);
}
} else {
// Done with this level, process the numbers we got
const result = numbers.reduce((s, e) => s + e);
results.push(result);
console.log(`${numbers.join(" + ")} = ${result}`);
}
}
process([4, 5, 6], 0, []);
return results;
}
function yourSolution() {
const results = [];
var a = [4, 5, 6];
for (var m = 0; m < a[0]; m++)
for (var n = 0; n < a[1]; n++)
for (var p = 0; p < a[2]; p++)
results.push(m + n + p);
return results;
}
const thisResult = thisSolution();
const yourResult = yourSolution();
if (thisResult.some((entry, index) => entry !== yourResult[index])) {
console.log("WRONG");
} else {
console.log("RIGHT");
}
/* This just makes the console take up the full output area */
.as-console-wrapper {
max-height: 100% !important;
}
This never goes deep into the stack (a.length + 1 stack frames, to be precise, so four in the example case). It builds up a number of temporary arrays (145 in the example case) that max out at a.length entries, releasing them as soon as they aren't needed anymore (a max of four are retained at any given time). Here's the quick and dirty metrics on that:
let maxStack = 0;
let stack = 0;
let totalArrays = 0;
let maxArrays = 0;
let arrays = 0;
// A wrapper for counting stack frames
const process = (...args) => {
if (++stack > maxStack) {
maxStack = stack;
}
const result = process2(...args);
--stack;
return result;
};
const process2 = (array, n, numbers) => {
if (n < array.length) {
// Not done yet, recurse once for each number at this level
const max = array[n];
for (let i = 0; i < max; ++i) {
++totalArrays;
if (++arrays > maxArrays) {
maxArrays = arrays;
}
process(array, n + 1, [...numbers, i]);
--arrays;
}
} else {
// Done with this level, process the numbers we got
//console.log(`${numbers.join(" + ")} = ${numbers.reduce((s, e) => s + e)}`);
}
}
process([4, 5, 6], 0, []);
++maxArrays; // To account for the one in the last argument above
++totalArrays; // "
console.log(`Max stack: ${maxStack}, max arrays: ${maxArrays}, total arrays: ${totalArrays}`);
It's easier if you break it down. First, you need to create a series per every element of your array.
let series = num => Array.from({ length: num + 1 }, (n, i) => i); //creates an array with nums from 0 to num.
That's the first part of your question. Then you need to do a cross product of your series.
Basically for two series [1, 2, 3] and [1, 2, 3, 4] you'll end up with a set of 12 elements:
[2, 3, 4, 5, 3, 4, 5, 6, 4, 5, 6, 7]
And for that you could do:
let crossProduct = (a1, a2) => Array.prototype.concat.call(...a1.map(n1 => a2.map(n2 => n1 + n2)));
Now all you need to do is have a crossProduct for every series.
let final = numbers.map(series).reduce(crossProduct);
And there you have it:
let numbers = [4, 5, 6];
let series = num => Array.from({ length: num + 1 }, (n, i) => i);
let crossProduct = (a1, a2) => Array.prototype.concat.call(...a1.map(n1 => a2.map(n2 => n1 + n2)));
let final = numbers.map(series).reduce(crossProduct);
console.log(final);
Edit: If it's from 0 to the number before (e.g. 4 is [0, 1, 2, 3]) then just take the + 1 in the series function.
2nd Edit: Less objects created for your crossProduct:
let crossProduct = (a1, a2) => {
let resultingSet = [];
for(let i = 0; i < a1.length; i++)
for(let j = 0; j < a2.length; j++)
resultingSet.push(a1[i] + a2[j]);
return resultingSet;
} //only one array is created
And if you want to avoid having the series on memory all the time:
let numbers = [4, 5, 6];
let series = function* (num){
for(let i = 0; i < num; i++){
yield i;
}
}
let crossProduct = (set, num) => {
let resultingSet = [];
for(let i = 0; i < set.length; i++){
for(let j of series(num)){
resultingSet.push(set[i] + j);
}
}
return resultingSet;
}
let final = numbers.reduce(crossProduct, [0]);
console.log(final);
Another solution that doesn't consume alot of memory and fairly efficient is by using an array that represnt the value of the indexes and update it each iteration.
first you create an array that represent in each element the amount of iterations you need to run in order to update the indexes respectively for example for this array [1, 2, 3 ,4 ,5] you will get:
[280, 140, 20, 5, 1] this means that index[0] will be updated each 280 iterations, index[1] will be updated each 140 iterations and so on..
totally you will run arr[n] * arr[n-1] * arr[n-2] * .... * arr[0] iterations as you did with ordinary nested for loop.
var arr = [1, 2, 7, 4, 5];
var indexes = Array.from({length: arr.length}, () => 0);
iterationsPerElement = arr.map((_, i) => arr.slice(i+1).reduce((acc, elem) => acc * elem, 1));
var totalIterations = iterationsPerElement[0] * arr[0];
for(var iteration = 1; iteration <= totalIterations; iteration++) {
// sum those indexes
console.log(`sum = ${indexes.reduce((acc, index) => acc + index, 0)}`);
// update indexes
for(i = 0; i < indexes.length; i++) {
if(iteration % iterationsPerElement[i] == 0) {
indexes[i]++;
// empty the indexes on the right
for(var j=i+1; j <indexes.length; j++) {
indexes[j] = 0;
}
}
}
}
I have an Array which contains numbers. I want to find the pair of the number which sums equal to given value with o(n) complexity.
let data = [5,8,9,6];
var x = {};
function findSum(arr, sum){
data.forEach(function(item){
if(item > sum){
return null
}
var diff = sum - item;
x[item] = diff
})
console.log(x);
}
findSum(data, 7);
Working with O(n2) complexity.
let data = [2, 4, 11, 3, 5, 8, 9, 1, 6, 5]
function findSum(arr, sum) {
let sortArray = arr.sort(function(a, b) {
return a - b
});
let findIndex = arr.indexOf(arr.find(function(item) {
return item >= sum
}))
let iterateValues = sortArray.slice(0, findIndex);
var pairs = [];
console.log(iterateValues)
iterateValues.forEach(function(value, index) {
let getDiff = sum - value;
let findDiff = iterateValues.find(function(diff, index) {
return diff === getDiff
});
if (findDiff) {
let firstPair = value.toString()
let secondPair = findDiff.toString();
let merge = firstPair + ',' + secondPair;
pairs.push(merge)
}
})
console.log(pairs)
}
findSum(data, 7);
Try the following, it takes O(n) time to find the pair whose sum is equal to given value:
let data = [5,8,9,6];
var sum = 17;
var map = {};
var found = false;
for(var i = 0; i < data.length; i++){
map[data[i]] = i;
}
for(var i = 0; i < data.length; i++){
if(map[sum - data[i]] && map[sum - data[i]] != i){
found = true;
console.log(data[map[sum - data[i]]] + " "+data[i]);
break;
}
}
if(!found)
console.log("No pair found");
Using a Set and for of loop
function findSum(arr, sum) {
let set = new Set(),
res = null;
for (let n of arr) {
const diff = sum - n;
if (diff && set.has(diff)) {
res = [diff, n]
break;
}
set.add(n);
}
return res ? res.join(' + ') + ' = ' + sum : 'No Matches for ' + sum
}
let data = [2, 4, 11, 3, 5, 8, 9, 1, 6, 5];
console.log(findSum(data, 6))
console.log(findSum(data, 7))
console.log(findSum(data, 8))
console.log(findSum(data, 22))
I want to count the average value of the odd numbers from a list of numbers. I have a starting code to count the average, but I don't know how can I choose only the odd numbers from the list?
Here is my code:
var numberArray = [1,2,3,4,5,6], thisTotal=0,thisAverage=0;
for (var i=0; i<numberArray.length; i++) {
thisTotal += numberArray[i];
}
thisAverage = (thisTotal/numberArray.length);
alert(thisAverage)
You can use a filter function to return only the odd numbers:
var oddArray = numberArray.filter(function(val) {
return val % 2 !== 0;
});
Full example:
var numberArray = [1, 2, 3, 4, 5, 6];
var thisTotal = 0;
var thisAverage = 0;
var oddArray = numberArray.filter(function(val) {
return val % 2 !== 0;
});
console.log(oddArray); // [1, 3, 5]
var thisTotal = oddArray.reduce(function(accumulator, currentValue) { return accumulator + currentValue;
});
console.log(thisTotal); // 1 + 3 + 5 => 9
var thisAverage = thisTotal / oddArray.length;
console.log(thisAverage); // 9 / 3 => 3
var numberArray=[1,2,3,4,5,6], thisAverage=0,oddlength=0;
for(var i=0;i<numberArray.length;i++)
{
if(numberArray[i]%2!==0){
thisAverage+=numberArray[i];
oddlength++;
}
}
thisAverage=(thisAverage/oddlength);
alert(thisAverage)
Well, you can get what you want by this.
var numberArray=[1,2,3,4,5,6], thisTotal=0,thisAverage=0;
for(var i=0; i < 3; i++) {
thisTotal += numberArray[i * 2];
thisAverage= (thisTotal/numberArray.length);
}
console.log(thisAverage);
or if you want general solution, use this.
var numberArray=[1,2,3,4,5,6,7,8,...on and on], thisTotal=0,thisAverage=0;
for(var i=0; i < Math.ceil(numberArray.length() / 2); i++) {
thisTotal += numberArray[i * 2];
thisAverage= (thisTotal/numberArray.length);
}
console.log(thisAverage);
hope my code be helpful :)
You can use the function reduce to add and count.
var numberArray = [1, 2, 3, 4, 5, 6],
result = numberArray.reduce((a, n) => {
if (n % 2 !== 0) {
a.sum += n;
a.count++;
}
return a;
}, {sum: 0, count: 0}),
average = result.sum / result.count;
console.log(average);
Assuming the numbers are in an array, you can do this:
var numbers = [1, 2, 3, 4, 5, 6];
var info = numbers.filter(function(n) { return n % 2 !== 0})
.reduce(function(acc, item) {
return {sum: acc.sum + item, count: acc.count + 1}
}, {sum: 0, count: 0});
var avg = info.sum / info.count;
This example uses filter and reduce methods, which are declarative and more clear.
filter returns a new array with the items for which the function returns true, and then reduce, for each item, updates an 'accumulator'. The accumulator can be anything, and in this case is an object with the sum of the numbers and their count. For each item, we add add the current number to the sum property and add 1 to count. Finally, we just devide sum by count and done.
var acc = 0, oddCount = 0;
for(var i = 0; i < numberArray.length; i++) {
if(numberArray[i] % 2 !== 0) {
acc += numberArray[i];
oddCount++;
}
}
return acc / oddCount;
You can create a new array and store odd values in that array and after that you can apply your logic to that array.
var a=[1,2,3,4,5,6,10,11];
var ar=[];
for (var i = 0; i < a.length; i++) {
if(a[i] % 2 !== 0) {
ar.push(a[i]);
}
}
console.log(ar);
var numberArray=[1,2,3,4,5,6,7,8];
var count = 0;
var result = 0;
for (let i = 0; i <= (numberArray.length-1); i++)
{
if (numberArray[i] % 2 != 0)
{
result += numberArray[i];
count++;
}
}
alert(result / count);
I want to generate a vector of 100 values composed by [1 0]:
This is how I did it in Matlab:
n = 100;
Seq1 = [1 0]; % sequence of 1-0
Vector = repmat(Seq1,(n/2),1); % Creates n/2 sequences of 1-0
The result is a vector like: [1 0 1 0 1 0 1 0...]
Is there a way to get the same result with JavaScript?
You could mimic the function repmat with a while loop.
function repmat(array, count) {
var result = [];
while (count--) {
result = result.concat(array);
}
return result;
}
var nTrials = 100,
Seq1 = [1, 0],
Vector = repmat(Seq1, nTrials / 2);
console.log(Vector);
Assuming you're looking for a way to add a 1 and then a 0, not an array containing 1 and 0:
var myArray = [];
nTrials = 30;
for(i = 1; i<= nTrials/2; i++){
myArray.push(1);
myArray.push(0)
}
document.body.innerHTML = myArray[1];}
https://jsfiddle.net/6seqs6af/1/
FWIW, here is the full repmat implementation in JavaScript.
It uses arrow functions (=>) which isn't available in all browsers.
// Seq1 is an Array (1D vector). We need a Matrix which JavaScript doesn't have
// natively. But we can derive a Matrix type from an Array by adding
// `numberOfRows` and `numberOfColumns` properties as well as a `set` method
function Matrix(numberOfRows, numberOfColumns) {
this.numberOfColumns = numberOfColumns;
this.numberOfRows = numberOfRows;
this.length = numberOfColumns * numberOfRows;
this.fill();
}
Matrix.prototype = Array.prototype;
Matrix.prototype.set = function() {
for (var i = 0; i < arguments.length; i++) {
this[i] = arguments[i];
}
return this;
}
Matrix.prototype.toString = function() {
return this.reduce((acc, x, idx) => acc + (idx % this.numberOfColumns === this.numberOfColumns - 1 ? x + '\n' : x + ', '), '');
}
Matrix.prototype.at = function(row, column) {
return this[row * this.numberOfColumns + column];
}
// Repmap
// ======
function repmat(mat, repeatColumns, repeatRows) {
var numberOfColumns = mat.numberOfColumns * repeatColumns;
var numberOfRows = mat.numberOfRows * repeatRows;
var values = [];
for (var y = 0; y < numberOfRows; y++) {
for (var x = 0; x < numberOfColumns; x++) {
values.push(mat.at(y % mat.numberOfRows, x % mat.numberOfColumns));
}
}
var result = new Matrix(numberOfRows, numberOfColumns);
result.set.apply(result, values);
return result;
}
// Calculation
// ===========
var nTrials = 100;
var seq1 = new Matrix(1, 2);
seq1.set(1, 0);
var vector = repmat(seq1, nTrials / 2, 1);
console.log(vector.toString());