Trying to leave input parameters unchanged - javascript

I am trying to write a function that performs operations on an array and returns a different copy of the array and leaves the original one unchanged. I thought I could do this by declaring var array2 = array and then proceeding with the array operations. What am I doing wrong?
Here is my sample function:
var partition = function(array, p){
var pivot = array[p];
var length = array.length;
// make a copy and move pivot to the front
var array2 = array;
array2[p] = array2[0];
array2[0] = pivot;
// partition the array
var i = 1;
for (var j = 1; j < length; j++){
//console.log('i='+i+', j='+j)
if (array2[j] < pivot) {
var temp = array2[j];
array2[j] = array2[i];
array2[i] = temp;
i++;
}
}
//console.log('array after partitioning: ' + array)
// swap pivot
array2[0] = array2[i-1];
array2[i-1] = pivot;
var answer = {array: array2, p: i-1}
return answer;
};
And my sample call:
var a = [3, 2, 1];
partition(a, 0);
console.log(a); // prints [1,2,3] but I want [3,2,1]

add this line instead of array2=array because it's only create a reference to array not the new array2.
var array2 =array.slice(0); //it will create a new array not the reference.
Fiddle http://jsfiddle.net/N64w3/

var array2 = array.slice(0) or just array.slice() will clone your array
Here you have reference to that

Related

Group items of two in one array

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]);

immutable js flat zip two lists

If I have two immutable lists,
const a = [1,2,3];
const b = [a,b,c,d];
is there an easy way to merge/zip them to result in:
const c = [1,a,2,b,3,c,d];
Is interleave what you are looking for: https://facebook.github.io/immutable-js/docs/#/List/interleave
const a = List([1, 2, 3]);
const b = List([a, b, c, d]);
a.interleave(b)
const a = [1, 2, 3] don't make this array immutable at all. What const does is to guarantee that you're not reassigning a. Pushing or removing items to a don't constitute a reassignment.
This is a real immutable list using immutable.js:
const a = List([1, 2, 3]);
Now.. back to zip.
If you don't want another reference, you can just use a for in the smallest list and build the zip yourself. If you want a ready-made function, you might consider using Underscore's Zip, which does exactly what you want.
If you want to develop a function from scratch then you can use the code below. Basically, the function merges the two array as interleaving the two. Consider 2 arrays, array1 = [1,2,3,4] and array2 = [a,b,c,d,e,f]. Then the below code will output a new array [1,a,2,b,3,c,4,d,e,f]
var i = 0;
var j = 0;
var k = 0;
var len1 = array1.length;
var len2 = array2.length;
var flag = true;
var newArr = [];
//Merge the 2 arrays till the smaller sized array
while (i < len1 && j < len2)
{
if(flag){
newArr[k] = array1[i];
i++; k++;
flag = false;
}
else{
newArr[k] = array2[j];
j++; k++;
flag = true;
}
}
/* Copy the remaining elements of array1[], if there are any */
while (i < len1)
{
newArr[k] = array1[i];
i++;
k++;
}
/* Copy the remaining elements of array2[], if there are any */
while (j < len2)
{
newArr[k] = array2[j];
j++;
k++;
}
console.log(newArr)

JavaScript -- search for array values in object

I have an array and an array of objects. I want to search each value in the array in the array of objects and if it doesnt exist, I want to remove the value from the array.
var arr= [1,2,3,4];
var recs = [{id:1},{id:2},{id:3}]; //4 doesnt exist in recs, remove from arr
//arr = [1,2,3];
Heres my attempt. Obviously does not work. I am not sure how I can compare each arr index with all the values in recs before moving on the next arr index:
var arr= [1, 2, 3, 4], index;
var recs = [{a:1},{a:2},{a:3}];
for(var i = 0; i<arr.length; i++){
for(var val in recs[i]){
if(arr[i] != recs[i][val]){
index = arr.indexOf(arr[i]);
arr.splice(index, 1);
}
}
}
thank you!!
If you are okay with leaving your original array instance alone and creating a new one (essentially treating it as immutable)
var newArr = arr.filter(function(num) {
return !recs.every(function(obj) {
return obj.a !== num;
});
});
Detail of the methods used: Array.filter is passed a function, runs that function on each element inside, and then returns a new array with only the elements that returned true in the function.
Array.every works a little similar, but returns a Boolean, true or false. It returns true if all of the elements of the array returned true inside of the function.
var arr= [1, 2, 3, 4];
var recs = [{a:1},{a:2},{a:3}];
// don't use var in for loops.
// Variables declared with var have function scope, so declare them at the top of your function
var i;
var j;
var value;
var found;
// iterate over the array
for (i = 0; i < arr.length; i++)
{
value = arr[i];
found = false;
// iterate over the other array
for (j = 0 ; j < recs.length ; j++)
{
// if we found what we were looking for, make a note and exit loop
if (recs[j].a == value)
{
found = true;
break;
}
}
if (!found)
{
arr.splice(i, 1);
// To compensate the loop's i++, to avoid skipping the next entry
i--;
}
}
alert(arr.join(','));

Rearranging Elements For Array of Array

I am trying to rearrange elements in an array of arrays but have been unsucessful. Can anyone offer suggestions? Here are two options I have tried. I want to swap/switch places for the first and second elements.
arr1 is an array of arrays (i.e. arr[][]) so I created arr2 to be an updated arr1
var arr2 = [];
for (var n = 0; n <arr1.length; n++){
arr2[n][0] = arr1[n][1];
arr2[n][1] = arr1[n][0];
}
The other thing I tried was:
function order(arr[]){
return [arr[n][1],arr[n][0], arr[n][2], arr[n][3]];
}
var arr2 = order(arr1);
You also need to create a new array for each item:
var arr2 = [];
for(var n = 0; n < arr1.length; n++) {
arr2[n] = [arr1[n][1], arr1[n][0]];
}
it's quite easy:
var a = arr1[n][0];
arr2[n][0] = arr1[n][1];
arr2[n][1] = a;
you need to save the first value as a variable, because if you do as you did(arr2[n][0] = arr1[n][1];), your two array indexes will have the same value.
You did:
a = 1, b = 2;
a = b;
b = a;
Which resolves in a = 2, b = 2
Also, your code as it is now, doesn't work. You need to create a new array for the simulation of multidimensional arrays in javascript.
for(i = 0; i < (yourdesiredamountofarrays); i++)
{
yourarray[i] = new Array();
}
The first example you need to use a temporary variable for the switch:
var arr2 = [];
for (var n = 0; n <arr1.length; n++){
var tempVal = arr1[n][1];
arr2[n][1] = arr1[n][0];
arr2[n][0] = tempArr;
}
The second example, in JS the variable shouldn't have brackets next to it, as it's just a loosely typed variable name.
function order(arr){
return [arr[n][1],arr[n][0], arr[n][2], arr[n][3], arr[n][4]];
}
var arr2 = order(arr1);
Next time, before asking you should check the console. The stackoverflow wiki page on JS has lots of great resources for learning to debug JS.

How to create an array like this in JavaScript?

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
}

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