I am trying to push numbers in an array into another array in groups of two.
If I have an array [1,4,3,2]; it should return [[1,4],[3,2]];
var arrayPairSum = function(nums) {
var len = nums.length / 2;
var arr = [];
for(var i = 0; i < len; i ++) {
var newArr = [];
newArr.push(nums[i]);
newArr.push(nums[i + 1]);
arr.push(newArr);
}
console.log(arr); //this should give me [[1,4],[3,2]];
};
arrayPairSum([1,4,3,2]);
can anyone see what I need to do to achieve this? I cannot figure it out.
You can use reduce method to achieve this. reduce method accepts a callback method provided on every item in the array.
In the other words, this method applies a function against an accumulator and each element in the array (from left to right) to reduce it to a single value.
var array=[1,4,3,2,8];
var contor=array.reduce(function(contor,item,i){
if(i%2==0)
contor.push([array[i],array[i+1]].filter(Boolean));
return contor;
},[]);
console.log(contor);
If you really want to iterate over the array, may skip every second index, so i+=2 ( as satpal already pointed out) :
var arrayPairSum = function(nums) {
var len = nums.length - 1;//if nums.length is not even, it would crash as youre doing nums[i+1], so thats why -1
var arr = [];
for (var i = 0; i < len; i += 2) {
var newArr = [];
newArr.push(nums[i]);
newArr.push(nums[i + 1]);
arr.push(newArr);
}
console.log(arr); //this should give me [[1,4],[3,2]];
};
arrayPairSum([1, 4, 3, 2]);
The upper one crops away every non pair at the end. If you want a single [value] at the end, may go with
len=nums.length
And check later before pushing
if(i+1<nums.length) newArr.push(nums[i+1]);
You were pretty close. Simply change the length to nums.length and in the loop increment i by 2.
var arrayPairSum = function(nums) {
var len = nums.length - 1;
var arr = [];
for(var i = 0; i < len; i+=2) {
var newArr = [];
newArr.push(nums[i]);
newArr.push(nums[i + 1]);
arr.push(newArr);
}
console.log(arr); //this should give me [[1,4],[3,2]];
};
arrayPairSum([1,4,3,2]);
Related
Let's say that I'm doing this because of my homework. I would like to develop some kind of schedule for the week to come (array of 6-7 elements - output result). But I have one problem. I need to figure it out how one element be positioned in the array and also his frequency must be exactly what user input is. Elements must be positioned at different index in the array.
I'm having that kind of input from user (just an example);
var arrayOfElements = ["el1","el2","el3"];
var el1Frequency = 3;
var el2Frequency = 2;
var el3Frequency = 1;
//output array of schedule (this is just an example)
var finaloutPutArray = ["el1","el2","el3","el1","el2","el1"];
Index of elements el1 is 0, 3 and 5, basically, I don't want elements to be repeated like this;
["el1","el1","el2","el3"...];
["el2","el1","el1","el3"];
Can you please give me some ideas on how to solve this problem.
I started like this;
var finalSchedule = [];
var totalDaysPerWeek = 6;
for(var i =0; i < totalDaysPerWeek; i++) {
...
}
This is one pattern, check my working snippet:
var arrayOfElements = ["el1","el2","el3"];
var obj = { el1: 3,
el2: 2,
el3: 1};
// First determine the max recurring of an element, this will be the number of cycles fo your loop
// Check key values
var arr = Object.keys(obj).map(function ( key ) { return obj[key]; });
// Get max value
var max = Math.max.apply( null, arr );
var finalArray = [];
// Iterate from 0 to max val
for(i = 0; i < max; i += 1){
// Iterate on array of elements
for(k = 0; k < arrayOfElements.length; k += 1) {
// If config of recurring
if( obj[arrayOfElements[k]] >= i+1 ) {
// Push into array
finalArray.push(arrayOfElements[k]);
}
}
}
console.log(finalArray);
I am trying to rotate an array from the center position. Say for example:
For an array [1,2,3,4,5] if I select 5 as the current element, the new array should be [3,4,5,1,2]. If I select 4, it should be [2,3,4,5,1]
I tried the below code and it works for some extent but it positions the selected at the beginning instead of center.
Any help with the approach will be really appreciated
var arr = [0,1,2,3,4];
function rot(arr, ind)
{
var narr = [];
var len = arr.length;
for(var i=0; i<arr.length; i++)
{
narr.push((i+ind<len?arr[i+ind]:arr[len-i-1]));
}
return narr;
}
console.log(rot(arr,0))
Try
let rot= (a,v,i=a.indexOf(v)) => a.map((x,j)=> a[(i+1+j+a.length/2)%a.length|0])
where n|0 cast float to integer. We use arrow function, indexOf and map
let arr=[1,2,3,4,5];
let rot=(a,v,i=a.indexOf(v))=>a.map((x,j)=>a[(i+1+j+a.length/2)%a.length|0]);
console.log( rot(arr,4) );
console.log( rot(arr,5) );
First, find how much you are going rotate the array (delta) then you splice as much items as its magnitude from the front if it is negative or the back if it is positive.
Put the spliced items on the opposite end.
function rot(arr, center) {
const index = arr.indexOf(center)
if (index === -1) {
throw new Error('')
}
if (arr.length % 2 === 0) {
throw new Error('')
}
const cIndex = Math.floor(arr.length/2)
const delta = cIndex - index
let narr = [...arr]
if (delta > 0) {
let temp = narr.splice(-delta)
narr = [...temp, ...narr]
}
else if (delta < 0) {
let temp = narr.splice(0, -delta)
narr = [...narr, ...temp]
}
return narr
}
let arr = [1,2,3,4,5]
console.log(rot(arr, 1))
console.log(rot(arr, 2))
console.log(rot(arr, 3))
console.log(rot(arr, 4))
console.log(rot(arr, 5))
this is basically the same as what #Mohammad posted, but instead of using ind as the index, you use arr.indexOf(ind) to get the relevant index.
I'd also add something to deal with the case you can't find the value...
let arrr = [1,2,3,4,5];
function rot(arr, ind) {
let narr = [...arr]; //copy and de-reference our array
let indexFound = narr.indexOf(ind); //find out where ind is
if(indexFound < 0 ) return narr; //ind was not found in your array
let len = Math.floor((narr.length/2) - indexFound); //find out how far your target is from the center
let doShift = (len < 0); //if len is negative, then we need elements from the beginning moved, otherwise move the elements off the end with pop()
len = Math.abs(len); //make sure len is always positive so our loop can run
for(var i=0; i<len; i++) {
if(doShift) narr.push(narr.shift());
else narr.unshift(narr.pop());
}
return narr;
}
console.log(rot(arrr,3));
console.log(rot(arrr,1));
console.log(rot(arrr,5));
Edit: modified to handle shifting from the end or beginning
So I have this function where I've need to take out the evens and odds and put them into separate arrays but I need the evens array to print first rather than the odds.
var numbersArray = [1,2,34,54,55,34,32,11,19,17,54,66,13];
function divider(numbersArray) {
var evensOdds = [[], []];
for (var i = 0; i < numbersArray.length; i++) {
evensOdds[i & 1].push(numbersArray[i]);
}
return evensOdds;
}
If you want to split the number by their even and odd values, instead of using the index (i), determine the sub array to push into using the value - numbersArray[i] % 2.
var numbersArray = [1,2,34,54,55,34,32,11,19,17,54,66,13];
function divider(numbersArray) {
var evensOdds = [[], []];
for (var i = 0; i < numbersArray.length; i++) {
evensOdds[numbersArray[i] % 2].push(numbersArray[i]);
}
return evensOdds;
}
console.log(divider(numbersArray));
If you want to split them by even and odd indexes use (i + 1) % 2 to determine the sub array to push into:
var numbersArray = [1,2,34,54,55,34,32,11,19,17,54,66,13];
function divider(numbersArray) {
var evensOdds = [[], []];
for (var i = 0; i < numbersArray.length; i++) {
evensOdds[(i + 1) % 2].push(numbersArray[i]);
}
return evensOdds;
}
console.log(divider(numbersArray));
Just for fun, a forEach version of the accepted answer.
var numbersArray = [1,2,34,54,55,34,32,11,19,17,54,66,13];
var even_odd = [ [], [] ];
numbersArray.forEach( e => even_odd[e%2].push(e) );
console.log(even_odd);
I am working on a code where I need to reverse certain no of elements in an array and rest should remain same. For example is an array has values of 1,2,3,4,5,6 and I have to reverse 4 elements of it then output should be 4,3,2,1,5,6. I am using below code to achieve this but getting error, please suggest.
function reverseArray(n, a) {
var interimArray1 = [];
//var interimArray2=[];
//var finalArray=[];
for (var i < n; i >= 0; i--) {
interimArray1.push[a[i]];
}
for (var i = n; i < a.length; i++) {
interimArray1.push[a[i]];
}
for (var i = 0; i < interimArray1.length; i++) {
console.log(interimArray1[i]);
}
}
var arr = [1, 2, 3, 4, 5, 6];
var num = 4;
reverseArray(num, arr);
The error in your code is that you intend to call the push method on a[i] like so:
interimArray1.push(a[i]);
but instead you write:
interimArray1.push[a[i]];
You make that mistake twice. To give arguments to the push method, you must use round parenthesis ().
With that fixed, you will see that your code works perfectly.
You can use Array#slice, Array#splice as follow.
function partialReverse(arr, num, from = 0) {
var slicedArr = arr.slice(from, num + from);
arr.splice(from, num); // Remove `num` items from array
arr.splice(from, 0, ...slicedArr.reverse()); // Add `num` reversed items
return arr;
}
var arr = [1, 2, 3, 4, 5, 6];
console.log(partialReverse(arr, 4, 0)); // Reverse four items from `arr` starting from 0th index
console.log(partialReverse(arr, 4, 1)); // Reverse four items from `arr` starting from 1st index
Lots of hints but you seem to be missing them. ;-)
You need to assign an initial value to i, so:
for (var i = n; ... )
===========^
Also, you need to use () to call functions, not [], so:
interimArray1.push(a[i]);
==================^====^
Same in the following for block. Otherwise, the code works though it's more verbose than it needs to be.
This is working :
I'm sure there are faster ways of doing it. Also, it will only work for elements at the beginning of the array but you can adjust the function for what you want to achieve.
var reverseArray = function(arr,elementsToReverse) {
var tempArrayRev = [];
var tempArray = [];
for (var i=0;i<arr.length;i++) {
if (i < elementsToReverse) {
tempArrayRev[i] = arr[i];
} else {
tempArray.push(arr[i]);
}
}
return tempArrayRev.reverse().concat(tempArray);
}
var array = [1,2,3,4,5,6];
document.getElementById('arrayOutput').innerHTML += reverseArray(array,4);
<div id="arrayOutput">Array :<br></div>
This is the answer you can test it.
function reverseArray(n, a) {
var interimArray1 = [];
for (var i = 0; i < a.length; i++) {
interimArray1.push(a[i]);
}
for (var i = num; i >=0; i--) {
interimArray1[i-1] = a[n - i];
}
for (var i = 0; i < interimArray1.length; i++) {
console.log(interimArray1[i]);
}
}
var arr = [1, 2, 3, 4, 5, 6];
var num = 4;
reverseArray(num, arr);
You could use something like this.
function reverseArray(n, arrIn) {
// Splice splits the array in 2 starting at 0 index going n long
var arrOut = arrIn.splice(0,n);
// reverse is pretty straight forward
arrOut = arrOut.reverse();
// Concat joins the two together
return arrOut.concat(arrIn);
}
I want to create an array like this:
s1 = [[[2011-12-02, 3],[2011-12-05,3],[5,13.1],[2011-12-07,2]]];
How to create it using a for loop? I have another array that contains the values as
2011-12-02,3,2011-12-05,3,2011-12-07,2
One of possible solutions:
var input = ['2011-12-02',3,'2011-12-05',3,'2011-12-07',2]
//or: var input = '2011-12-02,3,2011-12-05,3,2011-12-07,2'.split(",");
var output = [];
for(i = 0; i < input.length; i += 2) {
output.push([t[i], t[i + 1]])
}
If your values always come in pairs:
var str = '2011-12-02,3,2011-12-05,3,2011-12-07,2',//if you start with a string then you can split it into an array by the commas
arr = str.split(','),
len = arr.length,
out = [];
for (var i = 0; i < len; i+=2) {
out.push([[arr[i]], arr[(i + 1)]]);
}
The out variable is an array in the format you requested.
Here is a jsfiddle: http://jsfiddle.net/Hj6Eh/
var s1 = [];
for (x = 0, y = something.length; x < y; x++) {
var arr = [];
arr[0] = something[x].date;
arr[1] = something[x].otherVal;
s1.push(arr);
}
I've guessed here that the date and the other numerical value are properties of some other object, but that needn't be the case...
I think you want to create an array which holds a set of arrays.
var myArray = [];
for(var i=0; i<100;i++){
myArray.push([2011-12-02, 3]); // The values inside push should be dynamic as per your requirement
}