I want to access data of var a so it is: 245 but instead it only accesses the last one. so if i print it out it says 5
var A = [1, 2, 3, 4, 5];
var B = A[[1], [3], [4]];
console.log(B)
When accessing an object using square bracket notation — object[expression] — the expression resolves to the string name of the property.
The expression [1], [3], [4] consists of three array literals separated by comma operators. So it becomes [4]. Then it gets converted to a string: "4". Hence your result.
JavaScript doesn't have any syntax for picking non-contiguous members of an array in a single operation. (For contiguous members you have the slice method.)
You need to get the values one by one.
var A = [1, 2, 3, 4, 5];
var B = [A[1], A[3], A[4]];
console.log(B.join(""))
var A = [1, 2, 3, 4, 5];
var B = [A[1], A[3], A[4]];
console.log(B)
You'll need to access A multiple times for each index.
var A = [1, 2, 3, 4, 5];
var B = A[1];
console.log(A[1], A[3], A[4])
You can access them directly like that.
If you want to access index 2 for example, you should do console.log(A[1]);
You can't access multiple indices at the same time.
A variable can have only one value.
#Quentin solution resolve the problem, I wrote this solution to recommend you to create an array of index, and iterate over it.
Note: You are getting the last index, because you are using the comma operator. The comma operator allows you to put multiple expressions. The resulting will be the value of the last comma separated expression.
const A = [1, 2, 3, 4, 5];
const indexes = [1,3,4];
const B = indexes.map(i => A[i]).join``;
console.log(B);
Related
Is there a way to use regex to check if an array contains exactly one occurence of each number in a range ?
myArr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
I have tried this :
let regex = /[1-9]{1}/;
But this only checks that the array contains at least one occurence in the range : )
The described validation is not a particularly good use case for regex.
One alternative way to find the answer you seek is to:
Create a Set with the array items. (A Set by default only retains unique values.)
Convert the Set back to array.
Compare the lengths of the original array and the new array. If they mismatch, the difference is the number of array items that exist in duplicate.
// return TRUE if myArr only has unique values
[...new Set(myArr)].length === myArr.length
You can just filter for duplicates and compare the original array with the filtered to see if it had any duplicates. Upside here is that you can use the filtered array if you need it
let myArr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 5]
let uniques = myArr.filter((v, i, a) => a.indexOf(v) === i)
let hasDupes = myArr.length != uniques.length
console.log("hasdupes?", hasDupes);
console.log(uniques)
I have an initial array,
I've been trying to change values (orders) by using pop, splice methods inside a for loop and finally I push this array to the container array.
However every time initial array is values are pushed. When I wrote console.log(initial) before push method, I can see initial array has been changed but it is not pushed to the container.
I also tried to slow down the process by using settimeout for push method but this didnt work. It is not slowing down. I guess this code is invoked immediately
I would like to learn what is going on here ? Why I have this kind of problem and what is the solution to get rid of that.
function trial(){
let schedulePattern = [];
let initial = [1,3,4,2];
for(let i = 0; i < 3; i++){
let temp = initial.pop();
initial.splice(1,0,temp);
console.log(initial);
schedulePattern.push(initial);
}
return schedulePattern;
}
**Console.log**
(4) [1, 2, 3, 4]
(4) [1, 4, 2, 3]
(4) [1, 3, 4, 2]
(3) [Array(4), Array(4), Array(4)]
0 : (4) [1, 3, 4, 2]
1 : (4) [1, 3, 4, 2]
2 : (4) [1, 3, 4, 2]
length : 3
When you push initial into schedulePattern, it's going to be a bunch of references to the same Array object. You can push a copy of the array instead if you want to preserve its current contents:
schedulePattern.push(initial.slice(0));
Good answer on reference types versus value types here: https://stackoverflow.com/a/13266769/119549
When you push the array to schedulepattern, you are passing a reference to it.
you have to "clone" the array.
use the slice function.
function trial(){
let schedulePattern = [];
let initial = [1,3,4,2];
for(let i = 0; i < 3; i++){
let temp = initial.pop();
initial.splice(1,0,temp);
console.log(initial);
schedulePattern.push(initial.slice());
}
return schedulePattern;
}
You have to know that arrays are mutable objects. What does it mean? It means what is happening to you, you are copying the reference of the object and modifying it.
const array = [1,2,3]
const copy = array;
copy.push(4);
console.log(array); // [1, 2, 3, 4]
console.log(copy); // [1, 2, 3, 4]
There are a lot of methods in Javascript which provide you the way you are looking for. In other words, create a new array copy to work properly without modify the root.
const array = [1,2,3]
const copy = Array.from(array);
copy.push(4);
console.log(array); // [1, 2, 3]
console.log(copy); // [1, 2, 3, 4]
I encourage you to take a look at Array methods to increase your knowledge to take the best decision about using the different options you have.
I am new to Javascript and is confused why the following won't work?
var array = [1, 2, 3, 4]
var spread = ...array;
I was expecting it would become 1, 2, 3, 4. Instead, it gave an error message Unexpected token .... Can anyone explain this to me?
Thank you so much!
This is the correct way, however you're not gaining anything doing that.
var array = [1, 2, 3, 4]
var spread = [...array];
console.log(spread);
If you really want to destructure that array, you need destructuring assignment:
var array = [1, 2, 3, 4]
var [one, two, three, four] = array;
console.log(one, two, three, four);
The correct way of doing what you want is:
var array = [1, 2, 3, 4]
var spread = [...array];
The syntax for using spread is:
For function calls:
myFunction(...iterableObj);
For array literals or strings:
[...iterableObj, '4', 'five', 6];
For object literals (new in ECMAScript 2018):
let objClone = { ...obj };
So, based on the syntax, for an array by using spread you are missing the square brackets []:
var array = [1, 2, 3, 4]
var spread = [...array];
console.log(spread);
This question already has answers here:
Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array
(97 answers)
Closed 6 years ago.
There is a javascript array
var arr = [0, 1, 2, 2, 3, 3, 5];
I want to choose elements that repeats twice. In this case its 2 and 3. and i want attach them into a variable.
var a = 2, b = 3;
As far as i know there is no built-in function to do that job. How can i do that. Thanks.
You can use filter to get the values that occur twice.
var arr = [0, 1, 2, 2, 3, 3, 5];
var dups = arr.filter ( (v,i,a) => a.indexOf(v) < i );
console.log(dups);
In comments you stated you would only have doubles, but no values that occur more than twice. Note that the above would return a value more than once, if the latter would be the case.
This returns the values in an array, which is how you should work. To put them in separate values can be done as follows:
var [a, b, ...others] = dups;
...but you would have to know how many variables to reserve for that, and it does not make your further program any easier. JavaScript has many nice functions (methods) for arrays, so you should in fact leave them in an array.
There is no built in function to do that indeed.
You will have to loop thought the array and keeping track of the number of occurrences of the elements, while building a response array.
You could filter a sorted array.
var arr = [0, 1, 2, 2, 3, 3, 5],
repeats = arr.filter(function (a, i, aa) {
return aa[i - 1] === a;
});
console.log(repeats);
Most simple way to do this is the following:
var dups = [];
var arr = [0, 1, 2, 2, 3, 3, 5];
arr.forEach(function (v, i, a){
delete arr[i];
if (arr.indexOf(v) !== -1){
dups.push(v);
}
});
console.log(dups);
It's destructive however.
I want to replace elements in some array from 0 element, with elements of another array with variable length. Like:
var arr = new Array(10), anotherArr = [1, 2, 3], result;
result = anotherArr.concat(arr);
result.splice(10, anotherArr.length);
Is there some better way?
You can use the splice method to replace part of an array with items from another array, but you have to call it in a special way as it expects the items as parameters, not the array.
The splice method expects parameters like (0, anotherArr.Length, 1, 2, 3), so you need to create an array with the parameters and use the apply method to call the splice method with the parameters:
Array.prototype.splice.apply(arr, [0, anotherArr.length].concat(anotherArr));
Example:
var arr = [ 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j'];
var anotherArr = [ 1, 2, 3 ];
Array.prototype.splice.apply(arr, [0, anotherArr.length].concat(anotherArr));
console.log(arr);
Output:
[ 1, 2, 3, 'd', 'e', 'f', 'g', 'h', 'i', 'j']
Demo: http://jsfiddle.net/Guffa/bB7Ey/
In ES6 with a single operation, you can do this to replace the first b.length elements of a with elements of b:
let a = [1, 2, 3, 4, 5]
let b = [10, 20, 30]
a.splice(0, b.length, ...b)
console.log(a) // -> [10, 20, 30, 4, 5]
It could be also useful to replace the entire content of an array, using a.length (or Infinity) in the splice length:
let a = [1, 2, 3, 4, 5]
let b = [10, 20, 30]
a.splice(0, a.length, ...b)
// or
// a.splice(0, Infinity, ...b)
console.log(a) // -> [10, 20, 30], which is the content of b
The a array's content will be entirely replaced by b content.
Note 1: in my opinion the array mutation should only be used in performance-critical applications, such as high FPS animations, to avoid creating new arrays. Normally I would create a new array maintaining immutability.
Note 2: if b is a very large array, this method is discouraged, because ...b is being spread in the arguments of splice, and there's a limit on the number of parameters a JS function can accept. In that case I encourage to use another method (or create a new array, if possible!).
In ES6, TypeScript, Babel or similar you can just do:
arr1.length = 0; // Clear your array
arr1.push(...arr2); // Push the second array using the spread opperator
Simple.
For anyone looking for a way to replace the entire contents of one array with entire contents of another array while preserving the original array:
Array.prototype.replaceContents = function (array2) {
//make a clone of the 2nd array to avoid any referential weirdness
var newContent = array2.slice(0);
//empty the array
this.length = 0;
//push in the 2nd array
this.push.apply(this, newContent);
};
The prototype function takes an array as a parameter which will serve as the new array content, clones it to avoid any weird referential stuff, empties the original array, and then pushes in the passed in array as the content. This preserves the original array and any references.
Now you can simply do this:
var arr1 = [1, 2, 3];
var arr2 = [3, 4, 5];
arr1.replaceContents(arr2);
I know this is not strictly what the initial question was asking, but this question comes up first when you search in google, and I figured someone else may find this helpful as it was the answer I needed.
You can just use splice, can add new elements while removing old ones:
var arr = new Array(10), anotherArr = [1, 2, 3];
arr.splice.apply(arr, [0, anotherArr.length].concat(anotherArr))
If you don't want to modify the arr array, you can use slice that returns a shallow copy of the array:
var arr = new Array(10), anotherArr = [1, 2, 3], result = arr.slice(0);
result.splice.apply(result, [0, anotherArr.length].concat(anotherArr));
Alternatively, you can use slice to cut off the first elements and adding the anotherArr on top:
result = anotherArr.concat(arr.slice(anotherArr.length));
I'm not sure if it's a "better" way, but at least it allows you to choose the starting index (whereas your solution only works starting at index 0). Here's a fiddle.
// Clone the original array
var result = arr.slice(0);
// If original array is no longer needed, you can do with:
// var result = arr;
// Remove (anotherArr.length) elements starting from index 0
// and insert the elements from anotherArr into it
Array.prototype.splice.apply(result, [0, anotherArr.length].concat(anotherArr));
(Damnit, so many ninjas. :-P)
You can just set the length of the array in this case. For more complex cases see #Guffa's answer.
var a = [1,2,3];
a.length = 10;
a; // [1, 2, 3, undefined x 7]