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function dropElements(arr, func) {
let output = arr.reduce((acc=[], elem) => {
if (func(elem)){
acc.push(arr.slice(arr.indexOf(elem)))
return acc
}
}, [])
return output
}
let tester = dropElements([1, 2, 3, 4,5,6,3,2,1], function(n) {return n >= 3;})
console.log(tester)
I want it to output [3,4,5,6,3,2,1]. But is printing copies of size decreasing arrays.
The statement of the problem: "Given the array arr, iterate through and remove each element starting from the first element (the 0 index) until the function func returns true when the iterated element is passed through it."
You can use Array#slice along with Array#findIndex to find the first index at which the callback function returns true.
function dropElements(arr, func) {
const idx = arr.findIndex(func);
return idx >= 0 ? arr.slice(idx) : [];
}
console.log(dropElements([1, 2, 3, 4, 5, 6, 3, 2, 1], n => n >= 3));
If you specifically want to use Array#reduce, you could use a variable to store whether or not the callback has returned true for any element so far.
function dropElements(arr, func) {
let found = false;
return arr.reduce((acc, elem) => {
found = found || func(elem);
if (found) acc.push(elem);
return acc;
}, []);
}
console.log(dropElements([1, 2, 3, 4, 5, 6, 3, 2, 1], n => n >= 3));
Use the shift method to remove the first element from the array until the array is empty or func(arr[0]) returns a truthy value. This method is inefficient as it could be, but it matches the problem statement. Particularly:
It removes each element singularly, rather than all at once.
It removes elements while iterating.
It operates on the array itself, rather than a copy.
function dropElements(arr, func) {
while (arr.length > 0 && !func(arr[0])) {
arr.shift();
}
return arr;
}
let tester = dropElements([1, 2, 3, 4,5,6,3,2,1], function(n) {return n >= 3;});
console.log(tester);
I have a number array [2, 1, 3, 4, 5, 1] and want to remove the smallest number in the list. But somehow my IF statement gets skipped.
I checked and by itself "numbers[i + 1]" and "numbers[i]" do work, but "numbers[i + 1] < numbers[i]" doesn't...
function removeSmallest(numbers) {
var smallestNumberKEY = 0;
for (i = 0; i <= numbers.lenths; i++) {
if (numbers[i + 1] < numbers[i]) {
smallestNumberKEY = i + 1;
}
}
numbers.splice(smallestNumberKEY, 1);
return numbers;
}
document.write(removeSmallest([2, 1, 3, 4, 5, 1]));
You have a typo in your code, array doesn't have lenths property
function removeSmallest(numbers) {
var smallestNumberKEY = 0;
for (var i = 0; i < numbers.length - 1; i++) {
if (numbers[i + 1] < numbers[i]) {
smallestNumberKEY = i + 1;
numbers.splice(smallestNumberKEY, 1);
}
}
return numbers;
}
document.write(removeSmallest([2, 1, 3, 4, 5, 1]));
But your algorithm wont work for another array, e.g [5, 3, 1, 4, 1], it will remove a value 3 too.
You can find the min value with Math.min function and then filter an array
function removeSmallest(arr) {
var min = Math.min(...arr);
return arr.filter(e => e != min);
}
You can use Array#filter instead
function removeSmallest(arr) {
var min = Math.min.apply(null, arr);
return arr.filter((e) => {return e != min});
}
console.log(removeSmallest([2, 1, 3, 4, 5, 1]))
Short one liner. If the smallest value exist multiple times it will only remove ONE. This may or may not be what you want.
const result = [6,1,3,1].sort().filter((_,i) => i) // result = [1,3,6]
It works by sorting and then creating a new array from the items where indeces are truthy(anything but 0)
another solution with splice and indexOf:
array = [2, 1, 3, 4, 5, 1];
function replace(arr){
arr = arr.slice(); //copy the array
arr.splice( arr.indexOf(Math.min.apply(null, arr)),1)
return arr;
}
document.write( replace(array) ,'<br> original array : ', array)
edit : making a copy of the array will avoid the original array from being modified
"Short" solution using Array.forEach and Array.splice methods:
function removeSmallest(numbers) {
var min = Math.min.apply(null, numbers);
numbers.forEach((v, k, arr) => v !== min || arr.splice(k,1));
return numbers;
}
console.log(removeSmallest([2, 1, 3, 4, 5, 1])); // [2, 3, 4, 5]
This is a proposal with a single loop of Array#reduce and without Math.min.
The algorithm sets in the first loop min with the value of the element and returns an empty array, because the actual element is the smallest value and the result set should not contain the smallest value.
The next loop can have
a value smaller than min, then assign a to min and return a copy of the original array until the previous element, because a new minimum is found and all other previous elements are greater than the actual value and belongs to the result array.
a value greater then min, then the actual value is pushed to the result set.
a value equal to min, then the vaue is skipped.
'use strict';
var removeSmallest = function () {
var min;
return function (r, a, i, aa) {
if (!i || a < min) {
min = a;
return aa.slice(0, i);
}
if (a > min) {
r.push(a);
}
return r;
}
}();
document.write('<pre>' + JSON.stringify([2, 1, 3, 2, 4, 5, 1].reduce(removeSmallest, []), 0, 4) + '</pre>');
I like this oneliner: list.filter(function(n) { return n != Math.min.apply( Math, list ) })
check it out here: https://jsfiddle.net/rz2n4rsd/1/
function remove_smallest(list) {
return list.filter(function(n) { return n != Math.min.apply( Math, list ) })
}
var list = [2, 1, 0, 4, 5, 1]
console.log(list) // [2, 1, 0, 4, 5, 1]
list = remove_smallest(list)
console.log(list) // [2, 1, 4, 5, 1]
list = remove_smallest(list)
console.log(list) // [2, 4, 5]
I had to do this but I needed a solution that did not mutate the input array numbers and ran in O(n) time. If that's what you're looking for, try this one:
const removeSmallest = (numbers) => {
const minValIndex = numbers.reduce((finalIndex, currentVal, currentIndex, array) => {
return array[currentIndex] <= array[finalIndex] ? currentIndex : finalIndex
}, 0)
return numbers.slice(0, minValIndex).concat(numbers.slice(minValIndex + 1))
}
function sumOfPaiars(ints){
var array = [];
var min = Math.min(...ints)
console.log(min)
for(var i=0;i<ints.length;i++){
if(ints[i]>min){
array.push(ints[i])
}
}
return array
}
If you only wish to remove a single instance of the smallest value (which was my use-case, not clear from the op).
arr.sort().shift()
Here is a piece of code that is work properly but is not accepted from codewars:
let numbers = [5, 3, 2, 1, 4];
numbers.sort(function numbers(a, b) {
return a - b;
});
const firstElement = numbers.shift();
I have a function which takes an array and further arguments like this:
function arrayandmore([1, 2, 3], 2, 3)
I need to return the array ([1, 2, 3]) without those elements which equals the arguments coming behind the array. So in this case, the returned array would be:
([1]).
One of my approaches is:
function destroyer(arr) {
var args = Array.from(arguments);
var i = 0;
while (i < args.length) {
var result = args[0].filter(word => word !== args[i]);
i++;
}
console.log(result);
}
destroyer([1, 1, 3, 4], 1, 3);
Console returns:
[ 1, 1, 4 ]
I don't understand, why it returns one too - I don't understand, why it does not work.
It is the same with using splice.
function destroyer(arr) {
var args = Array.from(arguments);
var quant = args.length - 1;
for (var i = 1; i <= quant; i++) {
if (arr.indexOf(args[i]) !== -1) {
arr = arr.splice(arr.indexOf(args[i]));
}
console.log(arr);
}
}
destroyer([1, 1, 3, 4], 1, 3);
I think, both ways should work. But I don't figure out why they don't.
Your while won't work because result is being overwritten in every loop. So, it only ever removes the last parameter to the destroyer function
You can use the rest parameter syntax to separate the array and the items to be removed.
Then use filter and includes like this:
function destroyer(arr, ...toRemove) {
return arr.filter(a => !toRemove.includes(a))
}
console.log(destroyer([1, 1, 3, 4, 5], 1, 3))
I am trying to use the filter method on an array to loop through the array based on a variable number of arguments.
Below is my attempt at this:
function destroyer(arr) {
var argArr = arr.slice.call(arguments, 1);
var filteredArray = arr.filter(function(val) {
for (var i = 0; i < argArr.length; i++) {
return val != argArr[i];
};
});
console.log(filteredArray);
}
destroyer([1, 2, 3, 1, 2, 3], 2, 3);
When I do this, only the first element of the arguments array is disposed of. This therefore returns:
[1, 3, 1, 3]
I have found a few examples online of possible ways to resolve this but they are vastly different from what I understand just yet. Is there any way to get mine to work, or even understand why the additional elements of the arguments array are not being called.
If you use ES6 you can do it with rest operator and Array#includes function
function destroyer(arr, ...params){
return arr.filter(item => !params.includes(item));
}
console.log(destroyer([1, 2, 3, 1, 2, 3], 2, 3));
With your logic you can do like this. If val is equal to the current argArr's item then return false, if nothing was found after the loop: return true
function destroyer(arr) {
var argArr = Array.prototype.slice.call(arguments, 1);
var filteredArray = arr.filter(function(val) {
for (var i = 0; i < argArr.length; i++) {
if(val === argArr[i]){
return false;
}
};
return true;
});
console.log(filteredArray);
}
destroyer([1, 2, 3, 1, 2, 3], 2, 3);
Because with your code you always test if current element in filter is equal or not equal to second parameter in function which is 2 and return true/false. Instead you can use indexOf to test if current element in filter is inside arguments array.
function destroyer(arr) {
var argArr = arr.slice.call(arguments, 1);
var filteredArray = arr.filter(function(val) {
return argArr.indexOf(val) == -1
});
console.log(filteredArray);
}
destroyer([1, 2, 3, 1, 2, 3], 2, 3);
Problem is in your this line
return val != argArr[i];
Change logic like this , it will avoid to do extra looping also .
function destroyer(arr) {
var argArr = arr.slice.call(arguments, 1); debugger
var filteredArray = arr.filter(function(val) {
return !(argArr.indexOf(val) >= 0);
});
console.log(filteredArray);
}
destroyer([1, 2, 3, 1, 2, 3], 2, 3);
I tried to remove some words according to the passing variable.
However, I wrote two versions of code which have minor different!
And they resulted in different kinds of output which I don't understand why!
So I need u guys help, and big big thanks for u guys!
This function will be accepting the different numbers of variables,
which might be ( [arr],1,2 ) or ( [arr],1,2,8,9...) etc,
and remove the the variable in the first array according to the passing numbers.
For example: destroyer ( [1, 2, 3, 4], 2, 3 ) --> output should be [1,4]
And here is my code. ( Notice the minor difference with bold fonts! )
function destroyer(arr) {
for ( var i = 1; i < arguments.length; i++ ){
arr = arguments[0].filter(function(value){
if( value == arguments[i]){
return false;
}else{
return true;
}
});
}
return arr;
}
destroyer([1, 2, 3, 1, 2, 3], 2, 3);
The output will be [1,2,3,1,2,3], which means value == arguments[i] doesn't work. However,
function destroyer(arr) {
for ( var i = 1; i < arguments.length; i++ ){
filter = arguments[i];
arr = arguments[0].filter(function(value){
if( value == filter){
return false;
}else{
return true;
}
});
}
return arr;
}
destroyer([1, 2, 3, 1, 2, 3], 2, 3);
This version works perfectly showing me [1,1].
So what's going wrong with the first version??? Thank you!!
The problem with the first one is that arguments applies to the .filter() callback function (the closest function scope, not the parent function scope) so arguments[i] is not what you want it to be.
You could copy those arguments into an actual array which you could then refer to from the .filter() callback.
function destroyer(arr) {
var args = [].slice.call(arguments, 1);
for ( var i = 0; i < args.length; i++ ){
arr = arr.filter(function(value){
if( value == args[i]){
return false;
}else{
return true;
}
});
}
return arr;
}
var a = destroyer([1, 2, 3, 1, 2, 3], 2, 3);
// show output
document.write(JSON.stringify(a));
Personally, I'd suggest a bit simpler version:
function destroyer(arr) {
var args = [].slice.call(arguments, 1);
return arr.filter(function(value) {
return args.indexOf(value) === -1;
});
}
var a = destroyer([1, 2, 3, 1, 2, 3], 2, 3);
// show output
document.write(JSON.stringify(a));
Which, can be written simpler in ES6 using the spread operator:
function destroyer(arr, ...args) {
return arr.filter(function(value) {
return args.indexOf(value) === -1;
});
}
var a = destroyer([1, 2, 3, 1, 2, 3], 2, 3);
// show output
document.write(JSON.stringify(a));
Or, if you prefer even shorter ES6 notation and want to use the new Array.prototype.includes():
function destroyer(arr, ...args) {
return arr.filter(value => !args.includes(value));
}
var a = destroyer([1, 2, 3, 1, 2, 3], 2, 3);
document.write(JSON.stringify(a));
I would write it like that in an es6 version:
function destroyer( arr, ...elem ) {
return arr.reduce( ( prev, next, index ) => elem.includes( arr[index] ) ? prev : prev.concat(next) , [])
}