What I am trying to do is to return the [1, 4] array, however, I do not understand what's the mistake which ends up returning [1]. Any clues? Thank you!
const removeFromArray = function(arr) {
for (let i = arr.length - 1; i >= 0; i--) {
arr.splice(arr[i], 2);
}
return arr;
};
console.log(
removeFromArray([1, 2, 3, 4], 3, 2)
)
It's not exactly clear to me what you want to achieve.
You define a function which only takes one argument:
const removeFromArray = function(arr) {...}
But then you call the function with 3 arguments, an array and two numbers:
removeFromArray([1, 2, 3, 4], 3, 2)
Now your function only takes the first input (the array) and removes all elements instead the first one.
Please consider the syntax: splice(start, deleteCount)
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/splice
Maybe this rm() does what you want?
const rm=(arr, ...rem)=>arr.filter(a=>!rem.includes(a));
console.log(rm([1, 2, 3, 4], 3, 2));
It treats the first argument as the array arr that is to be filtered. The following arguments then make up the array rem, containing all the elements that are to be taken out of array arr.
You should consider using the built in filter method for arrays.
removeFromArray = (array, unwanted, otherUnwanted) => {
const filtered = array.filter((number) => {
return number !== unwanted && number !== otherUnwanted
});
return filtered;
};
console.log(removeFromArray[1,2,3,4], 3, 2]
To make the function more scalable for future use the second parameter could be an array.
betterRemoveFromArray = (array, unwantedNumbers) => {
const filtered = array.filter((number) => {
return !unwantedNumbers.includes(number)
});
return filtered;
};
console.log(removeFromArray3([1, 2, 3, 4], [2, 3]));
You need to write a variadic function which means it accepts a variable number of arguments. To represent it, use the rest parameter syntax which allows the function to accept an indefinite number of arguments as an array.
Then, use the filter method on the 1st argument, like so:
const removeFromArray = function(arr, ...theArgs) {
arr = arr.filter(arg => !theArgs.includes(arg));
return arr;
};
console.log(
removeFromArray([1, 2, 3, 4], 3, 2)
)
Related
map() can't mutate the calling array, instead it returns a new Array with modified values.
But, the following code mutating the original Array, is there any wrong in my understanding?
const arr = [1, 2, 3, 4, 5];
arr.map((num, index, arr1) => {
return arr1[index] = num * 2;
});
console.log(arr); // [2, 4, 6, 8, 10]
Well, you're mutating the original array by passing its reference into the callback function inside map() (arr1) and then manually accessing the indices. It will create a new array if you just return the value from that function.
const arr = [1, 2, 3, 4, 5];
const arr1 = arr.map((num) => {
return num * 2;
});
console.log(arr); // [1, 2, 3, 4, 5]
console.log(arr1); // [2, 4, 6, 8, 10]
The third argument to the callback function of map is the
original/source array on which the map is called upon
The arr and arr1 are both same i.e both are referencing on the same array, You can see it by using console.log(arr === arr1). So what ever you operation perform on the arr1, it gonna affect the arr.
const arr = [1, 2, 3, 4, 5];
arr.map((num, index, arr1) => {
console.log(arr1 === arr);
return num * 2;
});
You can just return num * 2 from the callback function. map internally creates a new array and return it. So you don't have to assign it as
arr1[index] = num * 2
You can also make it one-liner as:
arr.map((num, index, arr1) => num * 2)
const arr = [1, 2, 3, 4, 5];
const result = arr.map((num, index, arr1) => {
return num * 2;
});
console.log(arr); // [2, 4, 6, 8, 10]
console.log(result); // [2, 4, 6, 8, 10]
Array.map creates a new array populated with the results of calling a provided function on every element in the calling array.
Here its specifed that you must call or execute a function on every element of calling array.
What is the issue with your code?
You are not actually calling a function, you are instead updating the original array. If you are looking to create a new array by multiplying each node of the element with 2, you should do something like below.
Working Example
const arr = [1, 2, 3, 4, 5];
const newArray = arr.map((nodeFromOriginalArray, indexOfCurrentElement, arrayFromMapCalled) => {
return nodeFromOriginalArray * 2;
});
console.log(arr);
console.log(newArray);
Lets debug the paremeters inside the map function.
Here we have provided three arguments.
First argument nodeFromOriginalArray: The current element being processed in the array. This will be each node from your calling array.
Second argument indexOfCurrentElement: The index of the current element being processed in the array. Which means, the index of current element in calling array.
Third argument arrayFromMapCalled: The array map was called upon. This is the array on which the map function is getting executed. Please note, this is the original array. Updating properties inside this array results in updating your calling array. This is what happened in your case.
You should not modify your original array, which is the third parameter. Instead, you should return your node multipled by 2 inside map and assign this to a new array. Updating the third paramater inside the map function will mutate your calling array.
When calling map on an array, you provide a mapper with three arguments, an item in the array, it's index and the array itself (as you've represented in your snippet).
map takes the value returned by the function mapper as the element at the index in a new array returned by the operation.
const arr = [1,2,3,4,5]
const doubled = arr.map(x => x * 2) // [2,4,6,8, 10]
A over simplified implementation of map (without the index and originalArray params) might look like this. Let's assume that instead of being a method on the array instance, it's a function that takes an array and a mapper function.
I would not recommend re-implementing in production code, there's the native implementation as well as several libraries such as lodash and underscore that implement it.
function map(arr, mapper) {
const result = [];
for (const item of arr) {
const resultItem = mapper(item);
result.push(resultItem);
}
return result;
}
function double(x) {
return x * 2;
}
const doubled = map([1,2,3,4,5,6], double); // [2, 4, 6, 8 ,10, 12]
I am trying to compare two given parameters of a function. The exact problem is as follows:
You will be provided with an initial array (the first argument in the destroyer function), followed by one or more arguments. Remove all elements from the initial array that are of the same value as these arguments.
Note
You have to use the arguments object.
console.log(destroyer([1, 2, 3, 1, 2, 3], 2, 3)); // expected output: [1,1]
I am using filter method to iterate over the array but I couldn't compare the args with the elements of the array inside the callback of the filter.
function destroyer(arr, ...args) {
let result = arr.filter(num => {
for (let i = 0; i<=args.length; i++ ){
num !== args[i]
}
});
return result;
}
console.log(destroyer([1, 2, 3, 1, 2, 3], 2, 3));
I can iterate with for loop but I cannot use the output of for loop to do filter iteration.
Any ideas?
Probably an easier way to achieve the goal using .filter() with .includes(). Additionally you can use ...rest so called rest parameters for you function, see form the documentation:
The rest parameter syntax allows us to represent an indefinite number of arguments as an array.
Try as the following:
const destroyer = (arr, ...rest) => {
return arr.filter(num => !rest.includes(num));
}
console.log(destroyer([1, 2, 3, 1, 2, 3], 2, 3));
I hope this helps!
The filter() method creates a new array with all elements that pass the test implemented by the provided function.
Example:
const words = ['spray', 'limit', 'elite', 'exuberant', 'destruction', 'present'];
const result = words.filter(word => word.length > 6);
console.log(result);
// expected output: Array ["exuberant", "destruction", "present"]
via MDN
Filter iterates over all elements of some array and returns a new array. It puts an element in the new array only if callback (your function invoked as a parameter of filter) return true otherwise it's omitted.
Next it's worth to use rest parameters to achieve two arrays (initial and values to exclude).
The rest parameter syntax allows us to represent an indefinite number of arguments as an array.
function sum(...theArgs) {
return theArgs.reduce((previous, current) => {
return previous + current;
});
}
console.log(sum(1, 2, 3));
// expected output: 6
console.log(sum(1, 2, 3, 4));
// expected output: 10
Solution with explanation:
//Declare your function, first parameter is initial array, the second one is also array created by using rest parameters
function destroyer(initialArray = [], ...toExclude) {
// filter initialArray, if el (single element) is NOT included in "toExclude" it returns true
// and add this particular element to the result array
let result = initialArray.filter(el => toExclude.includes(el) == false);
//return result
return result;
}
It's a bit fuzzy for me, but I'm trying to create a simple function that takes an array and some arguments and removes the arguments from the array.
For example if I have an array such as [1,2,3,4] and the arguments 2,3 it would return [1,4] as a result.
This is what I have so far:
const removeFromArray = (arr) => {
let args = Array.from(arguments);
return arr.filter(function (item) {
!args.includes(item);
});
}
It doesn't work though. It works if I want to remove all the items from the array, but doesn't if I only go for specific ones.
How can I make this work so that it works even if I'm supplying an argument that is not part of the array (I want it to ignore those) and also if I have strings in the array as well?
Thanks in advance!
You could take rest parameters ... for the items to exclude the values.
const removeFromArray = (array, ...items) => array.filter(item => !items.includes(item));
console.log(removeFromArray([1, 2, 3, 4], 1, 4))
To use you style, you need to exclude the first item of arguments, because this is the array with all items.
let args = Array.from(arguments).slice(1);
// ^^^^^^^^^
In arrow functions the implicit object arguments doesn't exist, so declare the function using the keyword function, likewise, you need to return the result of the function includes within the handler of the function filter.
const removeFromArray = function() {
let [arr, ...args] = Array.from(arguments);
return arr.filter(function (item) {
return !args.includes(item);
});
}
console.log(removeFromArray([1,2,3,4], 2, 3));
Add a second parameter for the items you want to have removed.
const removeFromArray = (arr, toBeRemoved) => {
return arr.filter(item => !toBeRemoved.includes(item));
}
Example call:
const array = [1, 2, 3, 4]
const newArray = removeFromArray(arr, [2, 3])
console.log(newArray) // [1, 4]
Stick it to the Array.prototype, might use it somewhere else
Array.prototype.remove = function(...args) {
return this.filter(v => !args.includes(v));
}
In Lisps that I've seen there usually exists a 'rest' operator which returns all elements of a list in the same order, without the 0th element of that list. Without using iteration (no for or while loops), and without iteration happening via some other function or method does there exist a way to create a similar rest operator in Javascript?
For example, rest([1, 2, 3]) should return [2, 3], and rest([10, 2, 6]) should return [2, 6], and in general where 'arr' is the passed in array rest([arr[0], ..., arr[n]]) should return [arr[1], ..., arr[n]].
I had the idea of having a default value of 0 for a hidden 'counter' parameter of rest, but what I have below doesn't quite work:
function rest(arr, counter = 0)
{
if (counter === arr.length)
{
return [];
}
else
{
rest(arr, counter + 1)[0] = arr[counter]
}
}
In ES6, you can use rest syntax of Array destructuring. As from Docs:
When destructuring an array, you can unpack and assign the remaining part of it to a variable using the rest pattern:
Demo:
let rest = (arr) => {
let [first, ...rest] = arr;
return rest;
};
console.log(rest([1, 2, 3]));
console.log(rest([10, 5, 6]));
Instructions:
Write a function called getAllElementsButLast.
Given an array, getAllElementsButLast returns an array with all the elements but the last.
Below is my code that will not pass the requirements for the question. I am not sure why this is not correct even though I am getting back all the elements besides the last.
var arr = [1, 2, 3, 4]
function getAllElementsButLast(array) {
return arr.splice(0, arr.length - 1)
}
getAllElementsButLast(arr) // [1, 2, 3]
I think the reason why it's not accepted is because with splice() you change the input array. And that's not what you want. Instead use slice(). This method doesn't change the input array.
var arr = [1, 2, 3, 4]
function getAllElementsButLast(array) {
var newArr = array.slice(0, array.length - 1);
return newArr;
}
var r = getAllElementsButLast(arr);
console.log(r);
console.log(arr);