Related
The output that I want to have is newArray = [4, 9, 16, 25]. But I don't get it. Where did I make errors? Please help me.
var array = [2, 3, 4, 5];
var result = [];
function multiply(a) {
return a * a;
}
function newArray (a) {
for (i=0; i<a.lenght; i++){
result.push(multiply(a.value));
}
}
var newArray = newArray(array);
var array = [2, 3, 4, 5];
function multiply(a) {
return a * a;
}
function newArray (a) {
return a.map(multiply)
}
var result = newArray(array);
console.log(result)
This is another way to do it
const array = [2,3,4,5];
function multipliedArray(arr) {
return arr.map(x => x * x);
}
console.log(multipliedArray(array));
Keeping your logic as it is.
You misspelled a.length.
And you missed the index with array a.
var array = [2, 3, 4, 5];
var result = [];
function multiply(a) {
return a * a;
}
function newArray (a) {
for (i=0; i<a.length; i++){ //spelling mistake
result.push(multiply(a[i])); // index should be used
}
return result;
}
console.log(newArray(array));
I found that if I use ES6, I can change the codes like the following.
const arrayPast = [2, 3, 4, 5];
const result = [];
const appendArray = arrayPast.map(x => x * x);
result.push(appendArray);
Another thought, using forEach
const oldArray = [3, 4, 5, 6];
const square = [];
const newArray = oldArray.forEach((x) => square.push(x * x));
I have 10 different arrays. Each array has different numbers.
array1 = [1,2,3,4,5]
array2 = [6,7,8,9,10]
...
array 10 = [51,52,53,54]
let's say I pass in 7. Then I want to know which array it is from and want to return array number. So in this case it is going to be 2.
Should I write a switch statement for each array? Appreciate it in javascript.
try:
var arrays = [array1, array2, ..., array10];
for(var i=0; i<arrays.length; ++i) {
if (arrays[i].indexOf(value) != -1) {
console.log('found in array' + (i+1));
}
}
You cannot directly retrieve the name of array.The reason is this variable is only storing a reference to the object.
Instead you can have a key inside the same array which represent its name. Then indexOf can be used to find the array which contain the number , & if it is so, then get the array name
var array1 = [1,2,3,4,5];
array1.name ="array1";
var array2 = [6,7,8,9,10];
array2.name ="array2";
var array10 = [51,52,53,54]
array10.name ="array10";
var parArray = [array1,array2,array10]
function _getArrayName(number){
for(var o=0;o<parArray.length;o++){
var _tem = parArray[o];
if(parArray[o].indexOf(number) !==-1){
console.log(parArray[o].name);
}
}
}
_getArrayName(6) //prints array2
jsfiddle
One fast method should be using hash tables or as i would like to call LUT. Accordingly this job boils down to a single liner as follows;
var arrs = {
arr1 : [1,2,3,4,5],
arr2 : [6,7,8,9,10],
arr3 : [12,14,16,17],
arr4 : [21,23,24,25,27,20],
arr5 : [31,34,35,39],
arr6 : [45,46,44],
arr7 : [58,59],
arr8 : [66,67,69,61],
arr9 : [72,73,75,79,71],
arr0 : [81,85,98,99,90,80]
},
lut = Object.keys(arrs).reduce((p,c) => {arrs[c].forEach(n => p[n]=c); return p},{}),
findar = n => lut[n];
document.write("<pre>" + findar(12) + "</pre>");
One way to do this is have the arrays in an object and iterate over the keys/values. This method doesn't presume the arrays (and therefore their names) are in sequential order.
Note: this will always return a the first match from the function and terminate the search.
var obj = {
array1: [1, 2, 3, 4, 5],
array2: [6, 7, 8, 9, 10],
array3: [51, 52, 53, 54],
array4: [51, 52, 53, 54, 7]
}
function finder(obj, test) {
var keys = Object.keys(obj);
for (var i = 0; i < keys.length; i++) {
var key = keys[i];
if (obj[key].indexOf(test) > -1) {
return key.match(/\d+/)[0];
}
}
return false;
}
finder(obj, 7); // '2'
DEMO
If you want to find all instances of a value in all arrays the function needs to be altered slightly.
function finder(obj, test) {
var keys = Object.keys(obj);
var out = [];
for (var i = 0; i < keys.length; i++) {
var key = keys[i];
if (obj[key].indexOf(test) > -1) {
out.push(key.match(/\d+/)[0]);
}
}
return out;
}
finder(obj, 7); // ['2', '4']
DEMO
Let’s assume we have this two arrays:
x = [1, 2, 3];
y = ['a', 'b'];
What would be the best way to combine them and get the following result:
newArray = ['1a', '1b', '2a', '2b', '3a', '3b'];
Here is one way of doing that:
x.reduce(function(arr, x) {
return arr.concat(y.map(function(y) {
return x + y;
}));
}, []);
//=> ["1a", "1b", "2a", "2b", "3a", "3b"]
Try this:
var x = [1, 2, 3];
var y = ['a', 'b'];
var output = [];
for (var i = 0; i < x.length; i++) {
for (var j = 0; j < y.length; j++) {
output.push(x[i]+y[j]);
}
}
document.getElementById('output').innerHTML = JSON.stringify(output);
<div id="output"></div>
Try this..
var x = [1, 2, 3];
var y = ['a', 'b'];
var newarr = [];
for(var i=0;i<x.length;i++){
for(var j=0;j<y.length;j++){
newarr.push(x[i]+y[j]);
}
}
//alert(newarr);
DEMO
If arrow functions are supported you obtain the desired result like this:
[].concat.apply([],
x.map(x => y.map(y => x+y))
);
If not, you have to write it like this
[].concat.apply([],
x.map(function(x) { return y.map(function(y) {return x+y })})
);
Explanation:
The middle line yields the following result:
[ ["1a", "1b"], ["2a", "2b"], ["3a", "3b"] ]
Then the Array.prototype.concat method is used to concatenate the inner arrays.
You could simply create a array to be returned and do a simple loop for the array that contains numbers. Inside of that loop, you create another loop for the array of combinations to the numbers (var b=0,e=comb.length;e>b;b++). Using the i from the first loop (for(var i=0,l=array.length;l>i;i++)) you push the array at it (a[i]) with the array of combinations at the position b (c[b]) (inside of the loop that's inside of the first loop) to the new array. Finally, return the new array.
function CombineExample(a,c){
var New=[];
for(var i=0,l=a.length;l>i;i++){
for(var b=0,e=c.length;e>b;b++){
New.push(a[i]+c[b])
}
}
return New
}
Clean! And do this to use:
CombineExample([1,2,3],['a','b'])
/* returns ["1a", "1b", "2a", "2b", "3a", "3b"] */
Use nested loops to iterate all elements of the participating arrays. Populate new array elements inside the inner loop:
var x = [1, 2, 3];
var y = ['a', 'b'];
var newArray = [];
x.forEach(function(xItem) {
y.forEach(function(yItem) {
newArray.push(xItem.toString().concat(yItem));
});
});
console.log(newArray);
The simplest approach:
var x = ["a", "b", "c"];
var y = [1, 2, 3];
var newArray = [];
var i = 0;
for (;i < x.length;++i) {
var j = 0;
for (;j < y.length;++j) {
newArray.push(x[i] + y[j]);
}
}
;
Please do note that if both arrays are numeric, this will actually add the numbers, not concatenate. You'd need to do some string conversion.
var x = [1, 2, 3];
var y = ['a', 'b'];
var z = [];
for(var i=0;i<x.length;i++){
for(var j=0;j<y.length;j++){
z.push(x[i]+y[j]);
}
}
Are you seriously asking for that?
for hours i've been trying to figure out how to sort 2 array dependently.
Let's say I have 2 arrays.
First one:
array1 = ['zzzzz', 'aaaaaa', 'ccccc'];
and the second one:
array2 = [3, 7, 1];
I sort the first one with array1.sort(); and it becomes [aaaaaa, cccccc, zzzzzz]
now what I want is that the second one becomes [7, 1, 3]
I think it's quite simple but i'm trying to implement this in something a little more complex, im new and i keep mixing up things.
Thanks
I would "zip" them into one array of objects, then sort that with a custom sort callback, then "unzip" them back into the two arrays you wanted:
var array1 = ['zzzzz', 'aaaaaa', 'ccccc'],
array2 = [3, 7, 1],
zipped = [],
i;
for(i=0; i<array1.length; ++i) {
zipped.push({
array1elem: array1[i],
array2elem: array2[i]
});
}
zipped.sort(function(left, right) {
var leftArray1elem = left.array1elem,
rightArray1elem = right.array1elem;
return leftArray1elem === rightArray1elem ? 0 : (leftArray1elem < rightArray1elem ? -1 : 1);
});
array1 = [];
array2 = [];
for(i=0; i<zipped.length; ++i) {
array1.push(zipped[i].array1elem);
array2.push(zipped[i].array2elem);
}
alert('Sorted arrays:\n\narray1: ' + array1 + '\n\narray2: ' + array2);
Here's a working fiddle.
Here's a simple function that will do the trick:
function sortTogether(array1, array2) {
var merged = [];
for(var i=0; i<array1.length; i++) { merged.push({'a1': array1[i], 'a2': array2[i]}); }
merged.sort(function(o1, o2) { return ((o1.a1 < o2.a1) ? -1 : ((o1.a1 == o2.a1) ? 0 : 1)); });
for(var i=0; i<merged.length; i++) { array1[i] = merged[i].a1; array2[i] = merged[i].a2; }
}
Usage demo (fiddle here):
var array1 = ['zzzzz', 'aaaaaa', 'ccccc'];
var array2 = [3, 7, 1];
console.log('Before..: ',array1,array2);
sortTogether(array1, array2); // simply call the function
console.log('After...: ',array1,array2);
Output:
Before..: ["zzzzz", "aaaaaa", "ccccc"] [3, 7, 1]
After...: ["aaaaaa", "ccccc", "zzzzz"] [7, 1, 3]
Instead of two arrays of primitive types (strings, numbers) you can make an array of objects where one property of the object is string (containing "aaaaa", "cccccc", "zzzzzz") and another is number (7,1,3). This way you will have one array only, which you can sort by any property and the other property will remain in sync.
It just so happens I had some old code lying around that might do the trick:
function arrVirtualSortGetIndices(array,fnCompare){
var index=array.map(function(e,i,a){return i;});
fnCompare=fnCompare || defaultStringCompare;
var idxCompare=function (aa,bb){return fnCompare(array[aa],array[bb]);};
index.sort(idxCompare);
return index;
function defaultStringCompare(aa,bb){
if(aa<bb)return -1;
if(bb<aa)return 1;
return 0;
}
function defaultNumericalCompare(aa,bb){
return aa-bb;
}
}
function arrReorderByIndices(array,indices){
return array.map(
function(el,ix,ar){
return ar[indices[ix]];
}
);
}
var array1 = ['zzzzz', 'aaaaaa', 'ccccc'];
var array2 = [3, 7, 1];
var indices=arrVirtualSortGetIndices(array1);
var array2sorted=arrReorderByIndices(array2,indices);
array2sorted;
/*
7,1,3
*/
Sorry, I don't do 'fors'. At least not when I don't have to.
And fiddle.
Also, an alternative fiddle that sorts the results when given an array of objects like this:
given:
var list = [
{str:'zzzzz',value:3},
{str:'aaaaa',value:7},
{str:'ccccc',value:1}
];
outputs:
[
{str: "aaaaa", value: 7},
{str: "ccccc", value: 1},
{str: "zzzzz", value: 3}
]
Assumption:
The arrays are the same length (this is implied by your question)
the contents can be compared with > and < (true in your example, but I wanted to make it clear that it was assumed here)
So then we can use an insertion sort.
var value,len = array1.length;
for (i=0; i < len; i++) {
value = array1[i];
for (j=i-1; j > -1 && array1[j] > value; j--) {
array1[j+1] = array1[j];
array2[j+1] = array2[j];
}
items[j+1] = value;
}
Using a solution found here to find the new indices after sorting an array, you can apply those indices to array2 like so.
function sortWithIndices(toSort) {
for (var i = 0; i < toSort.length; i++) {
toSort[i] = [toSort[i], i];
}
toSort.sort(function(left, right) {
return left[0] < right[0] ? -1 : 1;
});
toSort.sortIndices = [];
for (var j = 0; j < toSort.length; j++) {
toSort.sortIndices.push(toSort[j][2]);
toSort[j] = toSort[j][0];
}
return toSort;
}
var array1 = ['zzzz', 'aaaa', 'cccc'];
var array2 = [3, 7, 1];
// calculate the indices of array1 after sorting. (attached to array1.sortIndices)
sortWithIndices(array1);
// the final array after applying the sorted indices from array1 to array2
var final = [];
// apply sorted indices to array2
for(var i = 0; i < array1.sortIndices.length; i++)
final[i] = array2[array1.sortIndices[i]];
// output results
alert(final.join(","));
JSFiddle Demo
I am saving some data in order using arrays, and I want to add a function that the user can reverse the list. I can't think of any possible method, so if anybody knows how, please help.
Javascript has a reverse() method that you can call in an array
var a = [3,5,7,8];
a.reverse(); // 8 7 5 3
Not sure if that's what you mean by 'libraries you can't use', I'm guessing something to do with practice. If that's the case, you can implement your own version of .reverse()
function reverseArr(input) {
var ret = new Array;
for(var i = input.length-1; i >= 0; i--) {
ret.push(input[i]);
}
return ret;
}
var a = [3,5,7,8]
var b = reverseArr(a);
Do note that the built-in .reverse() method operates on the original array, thus you don't need to reassign a.
Array.prototype.reverse() is all you need to do this work. See compatibility table.
var myArray = [20, 40, 80, 100];
var revMyArr = [].concat(myArray).reverse();
console.log(revMyArr);
// [100, 80, 40, 20]
Heres a functional way to do it.
const array = [1,2,3,4,5,6,"taco"];
function reverse(array){
return array.map((item,idx) => array[array.length-1-idx])
}
20 bytes
let reverse=a=>[...a].map(a.pop,a)
const original = [1, 2, 3, 4];
const reversed = [...original].reverse(); // 4 3 2 1
Concise and leaves the original unchanged.
reveresed = [...array].reverse()
The shortest reverse method I've seen is this one:
let reverse = a=>a.sort(a=>1)
**
Shortest reverse array method without using reverse method:
**
var a = [0, 1, 4, 1, 3, 9, 3, 7, 8544, 4, 2, 1, 2, 3];
a.map(a.pop,[...a]);
// returns [3, 2, 1, 2, 4, 8544, 7, 3, 9, 3, 1, 4, 1, 0]
a.pop method takes an last element off and puts upfront with spread operator ()
MDN links for reference:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/pop
two ways:
counter loop
function reverseArray(a) {
var rA = []
for (var i = a.length; i > 0; i--) {
rA.push(a[i - 1])
}
return rA;
}
Using .reverse()
function reverseArray(a) {
return a.reverse()
}
This is what you want:
array.reverse();
DEMO
Here is a version which does not require temp array.
function inplaceReverse(arr) {
var i = 0;
while (i < arr.length - 1) {
arr.splice(i, 0, arr.pop());
i++;
}
return arr;
}
// Useage:
var arr = [1, 2, 3];
console.log(inplaceReverse(arr)); // [3, 2, 1]
I've made some test of solutions that not only reverse array but also makes its copy. Here is test code. The reverse2 method is the fastest one in Chrome but in Firefox the reverse method is the fastest.
var array = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
var reverse1 = function() {
var reversed = array.slice().reverse();
};
var reverse2 = function() {
var reversed = [];
for (var i = array.length - 1; i >= 0; i--) {
reversed.push(array[i]);
}
};
var reverse3 = function() {
var reversed = [];
array.forEach(function(v) {
reversed.unshift(v);
});
};
console.time('reverse1');
for (var x = 0; x < 1000000; x++) {
reverse1();
}
console.timeEnd('reverse1'); // Around 184ms on my computer in Chrome
console.time('reverse2');
for (var x = 0; x < 1000000; x++) {
reverse2();
}
console.timeEnd('reverse2'); // Around 78ms on my computer in Chrome
console.time('reverse3');
for (var x = 0; x < 1000000; x++) {
reverse3();
}
console.timeEnd('reverse3'); // Around 1114ms on my computer in Chrome
53 bytes
function reverse(a){
for(i=0,j=a.length-1;i<j;)a[i]=a[j]+(a[j--]=a[i++],0)
}
Just for fun, here's an alternative implementation that is faster than the native .reverse method.
You can do
var yourArray = ["first", "second", "third", "...", "etc"]
var reverseArray = yourArray.slice().reverse()
console.log(reverseArray)
You will get
["etc", "...", "third", "second", "first"]
> var arr = [1,2,3,4,5,6];
> arr.reverse();
[6, 5, 4, 3, 2, 1]
array.reverse()
Above will reverse your array but modifying the original.
If you don't want to modify the original array then you can do this:
var arrayOne = [1,2,3,4,5];
var reverse = function(array){
var arrayOne = array
var array2 = [];
for (var i = arrayOne.length-1; i >= 0; i--){
array2.push(arrayOne[i])
}
return array2
}
reverse(arrayOne)
function reverseArray(arr) {
let reversed = [];
for (i = 0; i < arr.length; i++) {
reversed.push((arr[arr.length-1-i]))
}
return reversed;
}
Using .pop() method and while loop.
var original = [1,2,3,4];
var reverse = [];
while(original.length){
reverse.push(original.pop());
}
Output: [4,3,2,1]
I'm not sure what is meant by libraries, but here are the best ways I can think of:
// return a new array with .map()
const ReverseArray1 = (array) => {
let len = array.length - 1;
return array.map(() => array[len--]);
}
console.log(ReverseArray1([1,2,3,4,5])) //[5,4,3,2,1]
// initialize and return a new array
const ReverseArray2 = (array) => {
const newArray = [];
let len = array.length;
while (len--) {
newArray.push(array[len]);
}
return newArray;
}
console.log(ReverseArray2([1,2,3,4,5]))//[5,4,3,2,1]
// use swapping and return original array
const ReverseArray3 = (array) => {
let i = 0;
let j = array.length - 1;
while (i < j) {
const swap = array[i];
array[i++] = array[j];
array[j--] = swap;
}
return array;
}
console.log(ReverseArray3([1,2,3,4,5]))//[5,4,3,2,1]
// use .pop() and .length
const ReverseArray4 = (array) => {
const newArray = [];
while (array.length) {
newArray.push(array.pop());
}
return newArray;
}
console.log(ReverseArray4([1,2,3,4,5]))//[5,4,3,2,1]
As others mentioned, you can use .reverse() on the array object.
However if you care about preserving the original object, you may use reduce instead:
const original = ['a', 'b', 'c'];
const reversed = original.reduce( (a, b) => [b].concat(a) );
// ^
// |
// +-- prepend b to previous accumulation
// original: ['a', 'b', 'c'];
// reversed: ['c', 'b', 'a'];
Pure functions to reverse an array using functional programming:
var a = [3,5,7,8];
// ES2015
function immutableReverse(arr) {
return [ ...a ].reverse();
}
// ES5
function immutableReverse(arr) {
return a.concat().reverse()
}
It can also be achieved using map method.
[1, 2, 3].map((value, index, arr) => arr[arr.length - index - 1])); // [3, 2, 1]
Or using reduce (little longer approach)
[1, 2, 3].reduce((acc, curr, index, arr) => {
acc[arr.length - index - 1] = curr;
return acc;
}, []);
reverse in place with variable swapping (mutative)
const myArr = ["a", "b", "c", "d"];
for (let i = 0; i < (myArr.length - 1) / 2; i++) {
const lastIndex = myArr.length - 1 - i;
[myArr[i], myArr[lastIndex]] = [myArr[lastIndex], myArr[i]]
}
Reverse by using the sort method
This is a much more succinct method.
const resultN = document.querySelector('.resultN');
const resultL = document.querySelector('.resultL');
const dataNum = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
const dataLetters = ['a', 'b', 'c', 'd', 'e'];
const revBySort = (array) => array.sort((a, b) => a < b);
resultN.innerHTML = revBySort(dataNum);
resultL.innerHTML = revBySort(dataLetters);
<div class="resultN"></div>
<div class="resultL"></div>
Using ES6 rest operator and arrow function.
const reverse = ([x, ...s]) => x ? [...reverse(s), x] : [];
reverse([1,2,3,4,5]) //[5, 4, 3, 2, 1]
Use swapping and return the original array.
const reverseString = (s) => {
let start = 0, end = s.length - 1;
while (start < end) {
[s[start], s[end]] = [s[end], s[start]]; // swap
start++, end--;
}
return s;
};
console.log(reverseString(["s", "t", "r", "e", "s", "s", "e", "d"]));
Infact the reverse() may not work in some cases, so you have to make an affectation first as the following
let a = [1, 2, 3, 4];
console.log(a); // [1,2,3,4]
a = a.reverse();
console.log(a); // [4,3,2,1]
or use concat
let a = [1, 2, 3, 4];
console.log(a, a.concat([]).reverse()); // [1,2,3,4], [4,3,2,1]
What about without using push() !
Solution using XOR !
var myARray = [1,2,3,4,5,6,7,8];
function rver(x){
var l = x.length;
for(var i=0; i<Math.floor(l/2); i++){
var a = x[i];
var b = x[l-1-i];
a = a^b;
b = b^a;
a = a^b;
x[i] = a;
x[l-1-i] = b;
}
return x;
}
console.log(rver(myARray));
JavaScript already has reverse() method on Array, so you don't need to do that much!
Imagine you have the array below:
var arr = [1, 2, 3, 4, 5];
Now simply just do this:
arr.reverse();
and you get this as the result:
[5, 4, 3, 2, 1];
But this basically change the original array, you can write a function and use it to return a new array instead, something like this:
function reverse(arr) {
var i = arr.length, reversed = [];
while(i) {
i--;
reversed.push(arr[i]);
}
return reversed;
}
Or simply chaning JavaScript built-in methods for Array like this:
function reverse(arr) {
return arr.slice().reverse();
}
and you can call it like this:
reverse(arr); //return [5, 4, 3, 2, 1];
Just as mentioned, the main difference is in the second way, you don't touch the original array...
How about this?:
function reverse(arr) {
function doReverse(a, left, right) {
if (left >= right) {
return a;
}
const temp = a[left];
a[left] = a[right];
a[right] = temp;
left++;
right--;
return doReverse(a, left, right);
}
return doReverse(arr, 0, arr.length - 1);
}
console.log(reverse([1,2,3,4]));
https://jsfiddle.net/ygpnt593/8/