Why loop translate variable value to another variable that was equated to one which loop should only change?
let arrr = [1,1,1,1,1];
let preArrr = [0,0,0,0,0];
preArrr = arrr;
for (let i=0; i < arrr.length; i++) arrr[i] = i;
console.log(arrr, preArrr) // arrr = [0,1,2,3,4], preArrr = [0,1,2,3,4]
forEach loop give me the same result
arrr.forEach(function(e, i) { arrr[i] = i })
console.log(arrr, preArrr) // arrr = [0,1,2,3,4], preArrr = [0,1,2,3,4]
but if i change array by instant it wont connect
let arrr = [1,1,1,1,1];
let preArrr = [0,0,0,0,0];
preArrr = arrr;
arrr = [0,1,2,3,4]
console.log(arrr, preArrr) // arrr1 = [0,1,2,3,4], preArrr1 = [1,1,1,1,1];
so, how to avoid this connection, but still use loop? I`m trying to save previous state of the array
Because you changed preArr to point at same array as arr
That doesn't make a copy when you do preArr=arr. Instead both variables are references to the exact same array
If you want to copy the original into preArr use Array#slice()
let arrr = [1, 1, 1, 1, 1];
let preArrr = arrr.slice();
for (let i = 0; i < arrr.length; i++) arrr[i] = i;
console.log('arrr', JSON.stringify(arrr))
console.log('preArrr', JSON.stringify(preArrr))
i noticed that slice() method doesnt prevent reference to the same array, if an array you want to copy has 2 dimentions
to avoid that:
function slice2D(arr) {
let store = [];
for (let i = 0; i < arr.length; i++) {
store.push(arr[i].slice());
}
return store;
}
let pasteArr = slice2D(copyArr);
Related
For some reason, the manipulated doubleArray below is not shown in the console. Any variables that I declare after the for loop won't show to the console on both cases. Consider that in the first algorithm, there is only one for loop with x being incremented everytime. Whereas, in the second algorithm, it's a nested for loop. Can someone help me fix my error in both algorithms?
First Algorithm:
var isDuplicate = function() {
var helloWorld = [1,2,3,4,3];
var doubleValue = [];
var x = 0;
for (i = 0; i < helloWorld.length; i++) {
x = x + 1;
if (helloWorld[i] === helloWorld[x] && i !== x) {
doubleValue.push(helloWorld[i])
console.log(helloWorld[i]);
} else {
continue;
}
}
console.log(doubleValue);
};
The second Algorithm:
var isDuplicate = function() {
var helloWorld = [1,2,3,4,3];
var doubleValue = [];
for (i = 0; i < helloWorld.length; i++) {
for (x = 1; x < helloWorld.length; i++) {
if (helloWorld[i] === helloWorld[x] && i !== x) {
doubleValue.push(helloWorld[x]);
}
}
}
console.log(doubleValue);
};
In first algorithm, you are only checking if the number at current index is equal to the number at the next index, meaning you are only comparing numbers at consecutive indexes. First algorithm will work only if you have duplicate numbers on consecutive indexes.
In second algorithm, you are incrementing i in both loops, increment x in nested loop, change x = 1 to x = i + 1 and your error will be fixed.
Here's the fixed second code snippet
var isDuplicate = function() {
var helloWorld = [1,2,3,4,3, 1, 2];
var doubleValue = [];
for (let i = 0; i < helloWorld.length; i++) {
for (let x = i + 1; x < helloWorld.length; x++) {
if (helloWorld[i] === helloWorld[x] && i !== x) {
doubleValue.push(helloWorld[x]);
}
}
}
console.log(doubleValue);
};
isDuplicate();
Heres's another way to find the duplicates in an array, using an object. Loop over the array, if current number is present as key in the object, push the current number in the doubleValue array otherwise add the current number as key-value pair in the object.
const isDuplicate = function() {
const helloWorld = [1,2,3,4,3, 1, 2];
const doubleValue = [];
const obj = {};
helloWorld.forEach(n => obj[n] ? doubleValue.push(n): obj[n] = n);
console.log(doubleValue);
};
isDuplicate();
Not entirely sure what you are trying to do. If you are only looking for a method to remove duplicates you can do the following:
const hello_world = [1, 2, 2, 3, 4, 5, 5];
const duplicates_removed = Array.from(new Set(hello_world));
A set is a data object that only allows you to store unique values so, when converting an array to a set it will automatically remove all duplicate values. In the example above we are creating a set from hello_world and converting it back to an array.
If you are looking for a function that can identify all the duplicates in an array you can try the following:
const hello_world = [1, 2, 2, 3, 4, 5, 5];
const duplicates_found = hello_world.filter((item, index) => hello_world.indexOf(item) != index);
The main problem by finding duplicates is to have nested loop to compare each element of the array with any other element exept the element at the same position.
By using the second algorithm, you can iterate from the known position to reduce the iteration count.
var isDuplicate = function(array) {
var doubleValue = [];
outer: for (var i = 0; i < array.length - 1; i++) { // add label,
// declare variable i
// no need to check last element
for (var j = i + 1; j < array.length; j++) { // start from i + 1,
// increment j
if (array[i] === array[j]) { // compare values, not indices
doubleValue.push(array[i]);
continue outer; // prevent looping
}
}
}
return doubleValue;
};
console.log(isDuplicate([1, 2, 3, 4, 3])); // [3]
You could take an object for storing seen values and use a single loop for getting duplicate values.
const
getDuplicates = array => {
const
seen = {}
duplicates = [];
for (let value of array) {
if (seen[value]) duplicates.push(value);
else seen[value] = true;
}
return duplicates;
};
console.log(getDuplicates([1, 2, 3, 4, 3])); // [3]
Your first algorithm doesn't work because it only looks for duplicates next to each other. You can fix it by first sorting the array, then finding the duplicates. You can also remove the x and replace it by ++i in the loop.
var isDuplicate = function() {
var helloWorld = [1,2,3,4,3,6];
var doubleValue = [];
helloWorld = helloWorld.sort((a, b) => { return a - b });
for (i = 0; i < helloWorld.length; i++) {
if (helloWorld[i] === helloWorld[++i]) {
doubleValue.push(helloWorld[i])
console.log(helloWorld[i]);
} else {
continue;
}
}
console.log(doubleValue);
};
isDuplicate();
For the second algorithm loop, you probably meant x++ instead of i++ in the second loop. This would fix the problem.
var isDuplicate = function() {
var helloWorld = [1,2,3,4,3,4];
var doubleValue = [];
for (i = 0; i < helloWorld.length; i++) {
for (x = i + 1; x < helloWorld.length; x++) {
if (helloWorld[i] === helloWorld[x]) {
doubleValue.push(helloWorld[x]);
}
}
}
console.log(doubleValue);
};
isDuplicate()
The first algorithm can't be fixed, it can only detect consecutive duplicates,
in the second algorithm you increment i in both loops.
To avoid the duplicates beeing listed too often, you should start the second loop with i + 1
const a = 3
const b = 4
const arr = new Array(a).fill([]);
for (let i = 0; i < a; i++) {
for (let j = 0; j < b; j++) {
arr[i][j] = i;
}
}
console.log(JSON.stringify(arr))
I would expect to return
[[0,0,0,0],[1,1,1,1],[2,2,2,2]]
but it returns
[[2,2,2,2],[2,2,2,2],[2,2,2,2]]
I think this is not because of the closure things but please let me know what is the problem of this code not working as expected.
* edited -> sorry, I forgot to paste initializing arr code *
Don't assign a value directly in a 2-dimension array. Initialize the first dimension first.
var arr = []
const a = 3
const b = 4
for(let i=0; i<a; i++){
arr.push([])
for(let j=0; j<b; j++){
arr[i][j] = i;
}
}
console.log(arr)
Don't initialize your array like this:
const arr = new Array(a).fill([]);
By doing this, you pass references and not values, as you could do with integers. The difference between references and values is important.
I found and post another question about Array.fill() for the other people having the same issue.
Array.fill(Array) creates copies by references not by value
I create multiple objects and push them to the array objArr:
var objArr = [];
var obj = {};
var height = [9,8,7,3,6,5,2,4];
for (var i = 0; i < 8; i++) {
debugger;
var mountainH = height[i];
obj.h = mountainH;
obj.index = i;
objArr.push(obj);
}
for (var i = 0; i < objArr.length; i++) {
alert(objArr[i].h);
}
But as you can see, each object has the same values. Why?
Put the initialization of obj within your for-loop.
You were re-assigning new values to a global variable obj.
var objArr = [];
var height = [9,8,7,3,6,5,2,4];
for (var i = 0; i < 8; i++) {
debugger;
var obj = {};
var mountainH = height[i];
obj.h = mountainH;
obj.index = i;
objArr.push(obj);
}
for (var i = 0; i < objArr.length; i++) {
console.log(objArr[i].h);
}
Because the scope of obj in your code is global and it should rather be contained in the for loop.
If you will not declare it inside the loop then the value will get overwritten of the same obj on each iteration instead of a new memory allocation.
var objArr = [];
var height = [9, 8, 7, 3, 6, 5, 2, 4];
for (var i = 0; i < 8; i++) {
debugger;
var mountainH = height[i];
var obj = {};
obj.h = mountainH;
obj.index = i;
objArr.push(obj);
}
console.log(obj);
As noted, you need to initialize a new object in each iteration of the loop, otherwise all your array members simply share the same object reference.
Additionally, your code can be greatly reduced by building the array using .map(), and fully using the object literal initializer to declare the properties.
var height = [9,8,7,3,6,5,2,4];
var objArr = height.map((n, i) => ({h: n, index: i}));
console.log(objArr);
This is shorter and clearer. For every number in height, it creates a new object and adds it to a new array, which is returned from .map().
It can be even a little shorter with the newer features for object literals.
var height = [9,8,7,3,6,5,2,4];
var objArr = height.map((h, index) => ({h, index}));
console.log(objArr);
Why won't this function reverseArrayInPlace work? I want to do simply what the function says - reverse the order of elements so that the results end up in the same array arr. I am choosing to do this by using two arrays in the function. So far it just returns the elements back in order...
var arr = ["a","b","c","d","e","f"]
var arr2 = []
var reverseArrayInPlace = function(array){
var arrLength = array.length
for (i = 0; i < arrLength; i++) {
arr2.push(array.pop())
array.push(arr2.shift())
}
}
reverseArrayInPlace(arr)
Here's a simpler way of reversing an array, using an in-place algorithm
function reverse (array) {
var i = 0,
n = array.length,
middle = Math.floor(n / 2),
temp = null;
for (; i < middle; i += 1) {
temp = array[i];
array[i] = array[n - 1 - i];
array[n - 1 - i] = temp;
}
}
You "split" the array in half. Well, not really, you just iterate over the first half. Then, you find the index which is symmetric to the current index relative to the middle, using the formula n - 1 - i, where i is the current index. Then you swap the elements using a temp variable.
The formula is correct, because it will swap:
0 <-> n - 1
1 <-> n - 2
and so on. If the number of elements is odd, the middle position will not be affected.
pop() will remove the last element of the array, and push() will append an item to the end of the array. So you're repeatedly popping and pushing just the last element of the array.
Rather than using push, you can use splice, which lets you insert an item at a specific position in an array:
var reverseArrayInPlace = function (array) {
var arrLength = array.length;
for (i = 0; i < arrLength; i++) {
array.splice(i, 0, array.pop());
}
}
(Note that you don't need the intermediate array to do this. Using an intermediate array isn't actually an in-place reverse. Just pop and insert at the current index.)
Also, interesting comment -- you can skip the last iteration since the first element will always end up in the last position after length - 1 iterations. So you can iterate up to arrLength - 1 times safely.
I'd also like to add that Javascript has a built in reverse() method on arrays. So ["a", "b", "c"].reverse() will yield ["c", "b", "a"].
A truly in-place algorithm will perform a swap up to the middle of the array with the corresponding element on the other side:
var reverseArrayInPlace = function (array) {
var arrLength = array.length;
for (var i = 0; i < arrLength/2; i++) {
var temp = array[i];
array[i] = array[arrLength - 1 - i];
array[arrLength - 1 - i] = temp;
}
}
If you are doing Eloquent Javascript, the exercise clearly states to not use a new array for temporary value storage. The clues in the back of the book present the structure of the solution, which are like Stefan Baiu's answer.
My answer posted here uses less lines than Stefan's since I think it's redundant to store values like array.length in variables inside a function. It also makes it easier to read for us beginners.
function reverseArrayInPlace(array) {
for (var z = 0; z < Math.floor(array.length / 2); z++) {
var temp = array[z];
array[z] = array[array.length-1-z];
array[array.length-1-z] = temp;
}
return array;
}
You are calling the function with arr as parameter, so both arr and array refer to the same array inside the function. That means that the code does the same as:
var arr = ["a","b","c","d","e","f"]
var arr2 = []
var arrLength = arr.length;
for (i = 0; i < arrLength; i++) {
arr2.push(arr.pop())
arr.push(arr2.shift())
}
The first statements get the last item from arr and places it last in arr2. Now you have:
arr = ["a","b","c","d","e"]
arr2 = ["f"]
The second statement gets the first (and only) item from arr2 and puts it last in arr:
arr = ["a","b","c","d","e","f"]
arr2 = []
Now you are back where you started, and the same thing happens for all iterations in the loop. The end result is that nothing has changed.
To use pop and push to place the items reversed in the other array, you can simply move the items until the array is empty:
while (arr.length > 0) {
arr2.push(arr.pop());
}
If you want to move them back (instead of just using the new array), you use shift to get items from the beginning of arr2 and push to put them at the end of arr:
while (arr2.length > 0) {
arr.push(arr2.shift());
}
Doing a reversal in place is not normally done using stack/queue operations, you just swap the items from the beginning with the items from the end. This is a lot faster, and you don't need another array as a buffer:
for (var i = 0, j = arr.length - 1; i < j; i++, j--) {
var temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
This swaps the pairs like this:
["a","b","c","d","e"]
| | | |
| +-------+ |
+---------------+
I think you want a simple way to reverse an array. Hope it will help you
var yourArray = ["first", "second", "third", "...", "etc"]
var reverseArray = yourArray.slice().reverse()
console.log(reverseArray)
You will get
["etc", "...", "third", "second", "first"]
With the constraints I had for this assignment, this is the way I figured out how to solve the problem:
var arr = ["a","b","c","d","e","f"]
var arr2 = []
var reverseArrayInPlace = function(array){
var arrLength = array.length
for (i = 0; i < arrLength; i++) {
arr2.push(array.pop())
}
for (i = 0; i < arrLength; i++) {
array[i] = arr2.shift()
}
}
reverseArrayInPlace(arr)
Thank you for all your help!
***** edit ******
For all of you still interested, I rewrote it using some help from this thread and from my own mental devices... which are limited at this point. Here is it:
arr = [1,2,3,4,5,6,7,8,9,10,11,12,13]
arr2 = ["a","b","c","d","e","f"]
arr3 = [1,2,3]
arr4 = [1,2,3,4]
arr5 = [1,2,3,4,5]
var reverseArrayInPlace2 = function(array) {
var arrLength = array.length
var n = arrLength - 1
var i = 0
var middleTop = Math.ceil(arrLength/2)
var middleBottom = Math.floor(arrLength/2)
while (i < Math.floor(arrLength/2)) {
array[-1] = array[i]
array[i] = array[n]
array[n] = array[-1]
// console.log(array)
i++
n--
}
return array
}
console.log(reverseArrayInPlace2(arr))
console.log(reverseArrayInPlace2(arr2))
console.log(reverseArrayInPlace2(arr3))
console.log(reverseArrayInPlace2(arr4))
console.log(reverseArrayInPlace2(arr5))
P.S. what is wrong with changing global variables? What would the alternative be?
Here is my solution with no temp array. Nothing groundbreaking, just shorter version of some proposed solutions.
let array = [1, 2, 3, 4, 5];
for(let i = 0; i<Math.floor((array.length)/2); i++){
var pointer = array[i];
array[i] = array[ (array.length-1) - i];
array[(array.length-1) - i] = pointer;
}
console.log(array);
//[ 5, 4, 3, 2, 1 ]
I know this is a old question, but I came up with an answer I do not see above. It is similar to the approved answer above, but I use array destructuring instead of a temporary variable to swap the elements in the array.
const reverseArrayInPlace = array => {
for (let i = 0; i < array.length / 2; i++) {
[array[i], array[array.length - 1 - i]] = [array[array.length - 1 - i], array[i]]
}
return array
}
const myArray = [1,2,3,4,5,6,7,8,9];
console.log(reverseArrayInPlace(myArray))
This solution uses a shorthand for the while
var arr = ["a","b","c","d","e","f"]
const reverseInPlace = (array) => {
let end = array.length;
while(end--)
array.unshift(array.pop());
return array;
}
reverseInPlace(arr)
function reverseArrayInPlace (arr) {
var tempArr = [];
for (var i = 0; i < arr.length; i++) {
// Temporarily store last element of original array
var holdingPot = arr.pop();
// Add last element into tempArr from the back
tempArr.push(holdingPot);
// Add back value popped off from the front
// to keep the same arr.length
// which ensures we loop thru original arr length
arr.unshift(holdingPot);
}
// Assign arr with tempArr value which is the reversed
// array of the original array
arr = tempArr;
return arr;
}
I'm working on a codecademy.com lesson on arrays. I'm supposed to write nested loops to put each card of every suit in a deck of cards in an array.
I'm really messing this up. This is one combination that I've tried that doesn't work. The only indication that I have that I'm sort of close is that it returns "52" so at least 52 objects are going into the array. can anyone point out what I'm doing wrong?
//array 1: the suits
var suits = ["clubs","hearts","diamonds","spades"];
//array 2: the ranks
var ranks = [2,3,4,5,6,7,8,9,10,"J","Q","K","A"];
//using for loops, modify the "deck" array so that it becomes a
//two-dimensional array that stores every card in the deck;
//e.g. [1, "clubs"], [2, "clubs"],etc...["A", "spades"]
var deck = [];
for(i = 0; i < suits.length; i++){
for (j = 0; j < ranks.length; j++){
var cardArray = [];
cardArray[0] = suits[i];
cardArray[0][0] = ranks[j];
deck.push(cardArray);
}
}
Each iteration, a new array is being added to deck that looks like the following:
cardArray: [ [ ranks[j] ] ]
When cardArray[0][0] is set, it is overwriting cardArray[0] as an array with index 0 containing ranks[j]. Instead, set cardArray[0] to suits, and cardArray[1] to ranks.
cardArray[0] = suits[i];
cardArray[1] = ranks[j];
deck.push(cardArray);
This results in:
for (var i = 0; i < suits.length; i++){
for (var j = 0; j < ranks.length; j++){
var cardArray = [];
cardArray[0] = suits[i];
cardArray[1] = ranks[j];
deck.push(cardArray);
}
}
You should use a var declaration for your loop counter. Your problematic code is
cardArray[0] = suits[i];
cardArray[0][0] = ranks[j];
which does things like
var foo = "clubs";
foo[0] = "J";
which obviously does not work. I think what you want is
var deck = [];
for(var i = 0; i < suits.length; i++){
var cardArray = [];
for (j = 0; j < ranks.length; j++){
var twoCards = [];
twoCards[0] = suits[i];
twoCards[1] = ranks[j];
cardArray.push(twoCards);
// // or the same thing as this loop body, shorter:
// cardArray.push([suits[i], ranks[j]]);
}
deck.push(cardArray);
}