Identifying non-matches in two javascript arrays [duplicate] - javascript

I was wondering how I'd go about implementing a method in javascript that removes all elements of an array that clear a certain condition. (Preferably without using jQuery)
Ex.
ar = [ 1, 2, 3, 4 ];
ar.removeIf( function(item, idx) {
return item > 3;
});
The above would go through each item in the array and remove all those that return true for the condition (in the example, item > 3).
I'm just starting out in javascript and was wondering if anyone knew of a short efficient way to get this done.
--update--
It would also be great if the condition could work on object properties as well.
Ex.
ar = [ {num:1, str:"a"}, {num:2, str:"b"}, {num:3, str:"c"} ];
ar.removeIf( function(item, idx) {
return item.str == "c";
});
Where the item would be removed if item.str == "c"
--update2--
It would be nice if index conditions could work as well.
Ex.
ar = [ {num:1, str:"a"}, {num:2, str:"b"}, {num:3, str:"c"} ];
ar.removeIf( function(item, idx) {
return idx == 2;
});


You can use Array filter method.
The code would look like this:
ar = [1, 2, 3, 4];
ar = ar.filter(item => !(item > 3));
console.log(ar) // [1, 2, 3]


You could add your own method to Array that does something similar, if filter does not work for you.
Array.prototype.removeIf = function(callback) {
var i = 0;
while (i < this.length) {
if (callback(this[i], i)) {
this.splice(i, 1);
}
else {
++i;
}
}
};
To me, that's one of the coolest features of JavaScript. Ian pointed out a more efficient way to do the same thing. Considering that it's JavaScript, every bit helps:
Array.prototype.removeIf = function(callback) {
var i = this.length;
while (i--) {
if (callback(this[i], i)) {
this.splice(i, 1);
}
}
};
This avoids the need to even worry about the updating length or catching the next item, as you work your way left rather than right.


You can use Array.filter(), which does the opposite:
ar.filter(function(item, idx) {
return item <= 3;
});


You can use lodash.remove
var array = [1, 2, 3, 4];
var evens = _.remove(array, function(n) {
return n % 2 == 0;
});
console.log(array);
// => [1, 3]
console.log(evens);
// => [2, 4]


Make it a one-liner with arrow function:
ar = ar.filter(i => i > 3);


simply write the following example if condition could work on object properties as well
var ar = [ {num:1, str:"a"}, {num:2, str:"b"}, {num:3, str:"c"} ];
var newArray = [];
for (var i = 0, len = ar.length; i<len; i++) {
if (ar[i].str == "b")
{newArray.push(ar[i]);};
};
console.log(newArray);
See the example Live Example


if you need to remove exactly one item, and you know for sure that the item exists, you can use this one-liner:
ar.splice(ar.findIndex(el => el.id === ID_TO_REMOVE), 1);
// or with custom method:
let ar = [ {id:1, str:"a"}, {id:2, str:"b"}, {id:3, str:"c"}, {id:4,str:"d"} ];
ar.removeById = id => ar.splice(ar.findIndex(el => el.id === id), 1);
ar.removeById(ID_TO_REMOVE);
http://jsfiddle.net/oriadam/72kgprw5/
ES6 only


I love these kinds of questions and just a different version from me too... :)
Array.prototype.removeIf = function(expression) {
var res = [];
for(var idx=0; idx<this.length; idx++)
{
var currentItem = this[idx];
if(!expression(currentItem))
{
res.push(currentItem);
}
}
return res;
}
ar = [ 1, 2, 3, 4 ];
var result = ar.removeIf(expCallBack);
console.log(result);
function expCallBack(item)
{
return item > 3;
}


My solution for an array of numbers would be:
ar = ar.filter(item => item < 4);


For the in-place remove, my solution is
ar.filter(item => !(item > 3))
.forEach(obsoleteItem => ar.splice(ar.indexOf(obsoleteItem), 1));


Incorrect way
First of all, any answer that suggests to use filter does not actually remove the item. Here is a quick test:
var numbers = [1, 2, 2, 3];
numbers.filter(x => x === 2);
console.log(numbers.length);
In the above, the numbers array will stay intact (nothing will be removed). The filter method returns a new array with all the elements that satisfy the condition x === 2 but the original array is left intact.
Sure you can do this:
var numbers = [1, 2, 2, 3];
numbers = numbers.filter(x => x === 2);
console.log(numbers.length);
But that is simply assigning a new array to numbers.
Correct way to remove items from array
One of the correct ways, there are more than 1, is to do it as following. Please keep in mind, the example here intentionally has duplicated items so the removal of duplicates can be taken into consideration.
var numbers = [1, 2, 2, 3];
// Find all items you wish to remove
// If array has objects, then change condition to x.someProperty === someValue
var numbersToRemove = numbers.filter(x => x === 2);
// Now remove them
numbersToRemove.forEach(x => numbers.splice(numbers.findIndex(n => n === x), 1));
// Now check (this is obviously just to test)
console.log(numbers.length);
console.log(numbers);
Now you will notice length returns 2 indicating only numbers 1 and 3 are remaining in the array.

Related

Split array into arrays of numbers where the sum is equal to a specific target

I need to create a function that take as parameter an array and a target. It should return an array of arrays where the sum of these numbers equals to the target
sumPairs(array, target) {
}
For example:
sumPairs([1, 2, 3, 4, 5], 7) // output : [[2, 5], [3, 4]]
I know I have to use map(), and probably reduce(), set(), or filter() maybe (I read their documentation in MDN but still cant find out). I tried some ways but I can't get it.
If you guys could help me to find out how to dynamically create arrays and push them into a new array..
I read there some solutions (Split array into arrays of matching values) but I hate to just use created functions without knowing what they really do or how they work.

Some very basic code for achieving it, Just run all over combinations and conditionally add the items you want.
function sumPairs(array, target) {
var res = [];
for(var i = 0; i < array.length; i++){
for(var j = 0; j < array.length; j++){
if(i!=j && array[i]+array[j]==target &&
res.filter((x)=> x[0] == array[j] && x[1] == array[i]).length == 0 )
res.push([array[i], array[j]]);
}
}
return res;
}
var result = sumPairs([1, 2, 3, 4, 5], 7);
console.log(result);
Option 2 - see this answer for more options (like using reduce)
function sumPairs(array, target) {
return array.flatMap(
(v, i) => array.slice(i+1).filter(w => (v!=w && v+w==target)).map(w=> [w,v])
);
}
var result = sumPairs([1, 2, 3, 4, 5], 7);
console.log(result);

"The exercise says that it sould be arrays of pairs that sum the
target value so I think only 2 items"
If you need a pair that matches a sum and you pick any number from the list, you are left with
the following equation to solve num + x = sum where we want to find x. E.g. if you picked 7 and the target sum is 10 then you know you are looking for a 3.
Therefore, we can first construct a counting map of the numbers available in our list linear (O(n)) time and then search for matches in linear time as well rather than brute forcing with a quadratic algorithm.
const nums = [1, 2, 3, 4, 5];
console.log(findSumPairs(nums, 7));
function findSumPairs(nums, sum) {
const countByNum = countGroupByNum(nums);
return nums.reduce((pairs, num) => {
countByNum[num]--;
const target = sum - num;
if (countByNum[target] > 0) {
countByNum[target]--;
pairs.push([num, target]);
} else {
countByNum[num]++;
}
return pairs;
}, []);
}
function countGroupByNum(nums) {
return nums.reduce((acc, n) => (acc[n] = (acc[n] || 0) + 1, acc), {});
}
Here's another implementation with more standard paradigms (e.g. no reduce):
const nums = [1, 2, 3, 4, 5];
console.log(findSumPairs(nums, 7));
function findSumPairs(nums, sum) {
const countByNum = countGroupByNum(nums);
const pairs = [];
for (const num of nums) {
const target = sum - num; //Calculate the target to make the sum
countByNum[num]--; //Make sure we dont pick the same num instance
if (countByNum[target] > 0) { //If we found the target
countByNum[target]--;
pairs.push([num, target]);
} else {
countByNum[target]++; //Didin't find a match, return the deducted num
}
}
return pairs;
}
function countGroupByNum(nums) {
const countByNum = {};
for (const num of nums) {
countByNum[num] = (countByNum[num] || 0) + 1;
}
return countByNum;
}

You can also sort your array and find all the pairs with given sum by using two pointer method. Place the first pointer to the start of the array and the second pointer to the end.
if the sum of the values at the two places is :
More than target: Decrement your second pointer by 1
Less than target: Increment your first pointer by 1
Equal to target: This is one possible answer, push them to your answer array and increment your first pointer by 1 and decrement your second pointer by 1.
This is more performant solution with complexity O(n*log(n))

How to get the sum of all duplicates in an array?

I am trying to calculate the sum of all duplicates in an array. For example:
duplicate([1,1,2,3,3]) --> should return 8.
I have written the following function to calculate the sum of duplicates in an array using JavaScript. Currently it is returning an array with duplicates one less than what they are present in the array.
function duplicate(arr) {
var sum = 0;
arr.sort();
var stack = [];
for(var i = 0; i < arr.length; i++){
if (arr[i] === arr[i+1]) {
stack.push(arr[i])
}
}
return stack;
}
console.log(duplicate([1,2,1,2,2,3,3]))
This is returning [ 1, 2, 2, 3 ]
How do I get the correct array and calculate the correct sum? I have to use Object for that?

To make the logic easier, you might filter out the non-duplicates by checking whether their indexOf in the array is equal to their lastIndexOf in the array:
function duplicate(arr) {
const duplicates = arr.filter(elm => arr.indexOf(elm) !== arr.lastIndexOf(elm));
return duplicates.reduce((a, b) => a + b);
}
console.log(duplicate([1,1,2,3,3])); // --> should return 8.
console.log(duplicate([1,2,1,2,2,3,3]));

Initially create an object where the keys will be the integer and their value will be the number of occurrence. Then if the number of occurrence is more than 1 , multiply the number with number of occurrence.
function duplicate(arr) {
let dupVal = 0;
let k = arr.reduce((acc, curr, index) => {
if (acc[curr] === undefined) {
acc[curr] = 1
} else {
acc[curr] += 1;
}
return acc
}, {});
for (let keys in k) {
if (k[keys] > 1) {
dupVal += parseInt(keys, 10) * k[keys]
}
}
return dupVal;
}
console.log(duplicate([1, 2, 1, 2, 2, 3, 3]))

Try This one
const arr = [1,1,2,3,3]
let dup = arr.filter((value, index)=>{
// creating a copy of main array
let copyarr = [].concat(arr)
// removing present value
copyarr.splice(index,1)
// after removing present value, if you still
// get the value in copied array that means
// it has duplicates
if(copyarr.indexOf(value)>-1){
return true
}
return false
})
// now add it using reduce
let sum = dup.reduce((acc, value)=> acc+value,0)
console.log(sum)
Copy above code and paste into chrome devTool. You will get the answer.

The problem is that you are matching value with immediate next value in array, in array that is sorted already it will work, but not on unsorted one. So try to sort the array first and then run your code.
Edit :
Looks like sorting is added in code,
But another condition => if there is number that is repeated more than twice it should be handled and only appear once in stack, if that is required.
This will : console.log(duplicate([1,2,1,2,2,3,3]))
Result this : [1,2,3]
function duplicate(arr) {
var sum = 0;
arr.sort();
var stack = [];
for(var i = 0; i < arr.length; i++){
if (arr[i] === arr[i+1]) {
if(stack.length == 0 || (arr[i] != stack[stack.length-1])){
stack.push(arr[i])
}
}
}
return stack;
}

you can use JS Array.reduce method to accomplish your requirement in a shorter way
function sumDplicates(arr) {
return arr.reduce(function(tot, val, index, _arr) {
if (_arr.lastIndexOf(val) > index || _arr.indexOf(val) != index)
return tot + val;
return tot
}, 0)
}
console.log(sumDplicates([1, 1, 2, 3, 3]));
console.log(sumDplicates([1, 2, 1, 2, 2, 3, 3]));

You can pursue your original sorting approach with a slight modification:
if (arr[i] === arr[i + 1] || arr[i] === arr[i - 1])
That is, check if the previous or the next element in the sorted array is equal to the current element for it to qualify as a duplicate.
The following solution accomplishes this with filter and reduce:
function duplicate(array) {
return array
.sort((a, b) => a - b)
.filter((a, i, arr) => (arr[i] === arr[i + 1] || arr[i] === arr[i - 1]))
.reduce((a, b) => a + b, 0);
}
console.log(duplicate([1, 1, 2, 3, 3]));
console.log(duplicate([1, 2, 1, 2, 3, 3]));

Array.reduce() and Array.lastIndexOf() will simply solve your problem.
function sum(arr)
{
return arr.reduce(function(sum, item){
return arr.lastIndexOf(item)!==arr.indexOf(item) ? sum+=item : sum;
},0)
}
console.log(sum([1,1,2,3,3]));
console.log(sum([1,2,3,4]));
console.log(sum([1,2,2,3,4]));
console.log(sum([1,1,2,2,3,3,4,4]));

Though I don't know much about JavaScript, If I were you, I would have simply kept a temporary array, which copies all the duplicate variables and then use that array for sum.
Also, if you want to add the particular number as many times as it appears, I will suggest creating a table like the one in sparse matrices
and then referring to it during addition.
This logic, though not space efficient, is very easy to implement.

Here is an approach with a single Array.reduce and nothing else. No Array.indexOf or Array.lastIndexOf. Although it might not be as concise it does not traverse the array looking for indexes multiple times nor does any Array.filter:
const sumDubs = arr => arr.reduce((r,c) => {
if(r[c]) {
r[c] += 1
r.sum += r[c] > 2 ? (r[c]*c) - ((r[c]-1)*c) : r[c]*c
} else r[c] = 1
return r
}, { sum: 0 }).sum
console.log(sumDubs([1, 1, 2, 3, 3])) // 8
console.log(sumDubs([1, 2, 1, 2, 2, 3, 3])) // 14
console.log(sumDubs([1, 2, 1, 2, 2, 3, 3, 3, 1, 2, 4, 4])) // 28
The idea is to keep track of the on-going sum via a property in the accumulator of the Array.reduce and simply keep calculating the sum based on which number is duplicated and more importantly how many times.

How to compare to different arrays in javascript/jquery and get back the columns with difference in an array

Since I don't know how this works very well and other questions such as:
How to compare arrays in JavaScript?
or, JavaScript array difference
are completely different questions and after 1 and a half hours of scouring the web no results I am going to go and ask this question simply. Say you have two arrays [1,2,3] and [1,4,1]. Then how would you get a program to make our variable (say x) to be [1,2] (because column 1 and 2 are different if you count the first column as column 0)?

Please see this:
var array1 = [1, 2, 3];
var array2 = [1, 4, 1];
var columndiff = [];
for (i = 0; i < array1.length; i++) {
if (array1[i] != array2[i])
columndiff.push(i);
}
console.log(columndiff);
Note: Here we are assuming that array1 and array2 has equal length

ES2015 code:
const a = [1, 2, 3];
const b = [1, 4, 1];
const result = a.reduce((res, val, index) => {
if (val !== b[index]) res.push(index);
return res;
}, []);
console.log(result); // [1, 2]

You can simply do like this;
var arr = [1,3,4,5,2,3],
brr = [1,3,5,5,1,2],
res = arr.reduce((p,c,i) => c !== brr[i] ? p.concat(i) : p,[]);
console.log(res);

remove trailing elements from array that are equal to zero - better way

I have very long array containing numbers. I need to remove trailing zeros from that array.
if my array will look like this:
var arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
I want to remove everything except [1, 2, 0, 1, 0, 1].
I have created function that is doing what is expected, but I'm wondering if there is a build in function I could use.
var arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
for(i=arr.length-1;i>=0;i--)
{
if(arr[i]==0)
{
arr.pop();
} else {
break;
}
}
console.log(arr);
Can this be done better/faster?

Assuming:
var arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
You can use this shorter code:
while(arr[arr.length-1] === 0){ // While the last element is a 0,
arr.pop(); // Remove that last element
}
Result:
arr == [1,2,0,1,0,1]

var arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
var copy = arr.slice(0);
var len = Number(copy.reverse().join('')).toString().length;
arr.length = len;
arr -> [1, 2, 0, 1, 0, 1]
how it works
copy.reverse().join('') becomes "00000000000000000101021"
when you convert a numerical string to number all the preceding zeroes are kicked off
var len = Number(copy.reverse().join('')) becomes 101021
now by just counting the number i know from where i have to remove the trailing zeroes and the fastest way to delete traling elements is by resetting the length of the array.
arr.length = len;
DEMO

const arr = [0,0,0,1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
My solution is:
arr.join('').replace(/0+$/g,'').split('').map(Number);
It will remove trailing zeros in the given array.
Result is [0,0,0,1,2,0,1,0,1];
If you also needed to remove leading zeros, you can adjust the regex like this:
arr.join('').replace(/^0+|0+$/g,'').split('').map(Number);
Now It will remove not only trailing zeros, but leading zeros too.
Result is [1,2,0,1,0,1];

Accepted answer is perfectly well. Just for fun here is a reducing approach.
var a = [1,0,1,0,1,0,1,2,3,4,5,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0];
f = a => { var b = true;
return a.reduceRight((r,n,i) => ( b ? n && ( b = false
, r[i] = n
)
: r[i] = n
, r
)
, []
);
};
console.log(f(a));

Here's a one-liner representing javascript at both its best and worst at the same time.
var arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
arr.slice(0,arr.reduceRight(([d,l],c) => [d||c,l-!(d||c)], [false,arr.length])[1])
Output:
[ 1, 2, 0, 1, 0, 1 ]

A simpler immutable approach using reduceRight
const arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
const prunedArray = arr.reduceRight((acc, item) => {
if(item === 0 && acc.length === 0) {
return acc;
}
return acc.concat(item);
}, []);
console.log(prunedArray); // [1, 0, 1, 0, 2, 1]

Remove multiple elements from array in Javascript/jQuery

I have two arrays. The first array contains some values while the second array contains indices of the values which should be removed from the first array. For example:
var valuesArr = new Array("v1","v2","v3","v4","v5");
var removeValFromIndex = new Array(0,2,4);
I want to remove the values present at indices 0,2,4 from valuesArr. I thought the native splice method might help so I came up with:
$.each(removeValFromIndex,function(index,value){
valuesArr.splice(value,1);
});
But it didn't work because after each splice, the indices of the values in valuesArr were different. I could solve this problem by using a temporary array and copying all values to the second array, but I was wondering if there are any native methods to which we can pass multiple indices at which to remove values from an array.
I would prefer a jQuery solution. (Not sure if I can use grep here)

There's always the plain old for loop:
var valuesArr = ["v1","v2","v3","v4","v5"],
removeValFromIndex = [0,2,4];
for (var i = removeValFromIndex.length -1; i >= 0; i--)
valuesArr.splice(removeValFromIndex[i],1);
Go through removeValFromIndex in reverse order and you can .splice() without messing up the indexes of the yet-to-be-removed items.
Note in the above I've used the array-literal syntax with square brackets to declare the two arrays. This is the recommended syntax because new Array() use is potentially confusing given that it responds differently depending on how many parameters you pass in.
EDIT: Just saw your comment on another answer about the array of indexes not necessarily being in any particular order. If that's the case just sort it into descending order before you start:
removeValFromIndex.sort(function(a,b){ return b - a; });
And follow that with whatever looping / $.each() / etc. method you like.

I suggest you use Array.prototype.filter
var valuesArr = ["v1","v2","v3","v4","v5"];
var removeValFrom = [0, 2, 4];
valuesArr = valuesArr.filter(function(value, index) {
return removeValFrom.indexOf(index) == -1;
})

Here is one that I use when not going with lodash/underscore:
while(IndexesToBeRemoved.length) {
elements.splice(IndexesToBeRemoved.pop(), 1);
}

Not in-place but can be done using grep and inArray functions of jQuery.
var arr = $.grep(valuesArr, function(n, i) {
return $.inArray(i, removeValFromIndex) ==-1;
});
alert(arr);//arr contains V2, V4
check this fiddle.

A simple and efficient (linear complexity) solution using filter and Set:
const valuesArr = ['v1', 'v2', 'v3', 'v4', 'v5'];
const removeValFromIndex = [0, 2, 4];
const indexSet = new Set(removeValFromIndex);
const arrayWithValuesRemoved = valuesArr.filter((value, i) => !indexSet.has(i));
console.log(arrayWithValuesRemoved);
The great advantage of that implementation is that the Set lookup operation (has function) takes a constant time, being faster than nevace's answer, for example.

This works well for me and work when deleting from an array of objects too:
var array = [
{ id: 1, name: 'bob', faveColor: 'blue' },
{ id: 2, name: 'jane', faveColor: 'red' },
{ id: 3, name: 'sam', faveColor: 'blue' }
];
// remove people that like blue
array.filter(x => x.faveColor === 'blue').forEach(x => array.splice(array.indexOf(x), 1));
There might be a shorter more effecient way to write this but this does work.

It feels necessary to post an answer with O(n) time :). The problem with the splice solution is that due to the underlying implementation of array being literally an array, each splice call will take O(n) time. This is most pronounced when we setup an example to exploit this behavior:
var n = 100
var xs = []
for(var i=0; i<n;i++)
xs.push(i)
var is = []
for(var i=n/2-1; i>=0;i--)
is.push(i)
This removes elements starting from the middle to the start, hence each remove forces the js engine to copy n/2 elements, we have (n/2)^2 copy operations in total which is quadratic.
The splice solution (assuming is is already sorted in decreasing order to get rid of overheads) goes like this:
for(var i=0; i<is.length; i++)
xs.splice(is[i], 1)
However, it is not hard to implement a linear time solution, by re-constructing the array from scratch, using a mask to see if we copy elements or not (sort will push this to O(n)log(n)). The following is such an implementation (not that mask is boolean inverted for speed):
var mask = new Array(xs.length)
for(var i=is.length - 1; i>=0; i--)
mask[is[i]] = true
var offset = 0
for(var i=0; i<xs.length; i++){
if(mask[i] === undefined){
xs[offset] = xs[i]
offset++
}
}
xs.length = offset
I ran this on jsperf.com and for even n=100 the splice method is a full 90% slower. For larger n this difference will be much greater.

I find this the most elegant solution:
const oldArray = [1, 2, 3, 4, 5]
const removeItems = [1, 3, 5]
const newArray = oldArray.filter((value) => {
return !removeItems.includes(value)
})
console.log(newArray)
output:
[2, 4]
or even shorter:
const newArray = oldArray.filter(v => !removeItems.includes(v))

function filtermethod(element, index, array) {
return removeValFromIndex.find(index)
}
var result = valuesArr.filter(filtermethod);
MDN reference is here

In pure JS you can loop through the array backwards, so splice() will not mess up indices of the elements next in the loop:
for (var i = arr.length - 1; i >= 0; i--) {
if ( yuck(arr[i]) ) {
arr.splice(i, 1);
}
}

A simple solution using ES5. This seems more appropriate for most applications nowadays, since many do no longer want to rely on jQuery etc.
When the indexes to be removed are sorted in ascending order:
var valuesArr = ["v1", "v2", "v3", "v4", "v5"];
var removeValFromIndex = [0, 2, 4]; // ascending
removeValFromIndex.reverse().forEach(function(index) {
valuesArr.splice(index, 1);
});
When the indexes to be removed are not sorted:
var valuesArr = ["v1", "v2", "v3", "v4", "v5"];
var removeValFromIndex = [2, 4, 0]; // unsorted
removeValFromIndex.sort(function(a, b) { return b - a; }).forEach(function(index) {
valuesArr.splice(index, 1);
});

Quick ES6 one liner:
const valuesArr = new Array("v1","v2","v3","v4","v5");
const removeValFromIndex = new Array(0,2,4);
const arrayWithValuesRemoved = valuesArr.filter((value, i) => removeValFromIndex.includes(i))

If you are using underscore.js, you can use _.filter() to solve your problem.
var valuesArr = new Array("v1","v2","v3","v4","v5");
var removeValFromIndex = new Array(0,2,4);
var filteredArr = _.filter(valuesArr, function(item, index){
return !_.contains(removeValFromIndex, index);
});
Additionally, if you are trying to remove items using a list of items instead of indexes, you can simply use _.without(), like so:
var valuesArr = new Array("v1","v2","v3","v4","v5");
var filteredArr = _.without(valuesArr, "V1", "V3");
Now filteredArr should be ["V2", "V4", "V5"]

You can correct your code by replacing removeValFromIndex with removeValFromIndex.reverse(). If that array is not guaranteed to use ascending order, you can instead use removeValFromIndex.sort(function(a, b) { return b - a }).

Here's one possibility:
valuesArr = removeValFromIndex.reduceRight(function (arr, it) {
arr.splice(it, 1);
return arr;
}, valuesArr.sort(function (a, b) { return b - a }));
Example on jsFiddle
MDN on Array.prototype.reduceRight

filter + indexOf (IE9+):
function removeMany(array, indexes) {
return array.filter(function(_, idx) {
return indexes.indexOf(idx) === -1;
});
});
Or with ES6 filter + find (Edge+):
function removeMany(array, indexes = []) {
return array.filter((_, idx) => indexes.indexOf(idx) === -1)
}

Here's a quickie.
function removeFromArray(arr, toRemove){
return arr.filter(item => toRemove.indexOf(item) === -1)
}
const arr1 = [1, 2, 3, 4, 5, 6, 7]
const arr2 = removeFromArray(arr1, [2, 4, 6]) // [1,3,5,7]

Try this
var valuesArr = new Array("v1", "v2", "v3", "v4", "v5");
console.info("Before valuesArr = " + valuesArr);
var removeValFromIndex = new Array(0, 2, 4);
valuesArr = valuesArr.filter((val, index) => {
return !removeValFromIndex.includes(index);
})
console.info("After valuesArr = " + valuesArr);

Sounds like Apply could be what you are looking for.
maybe something like this would work?
Array.prototype.splice.apply(valuesArray, removeValFromIndexes );

var valuesArr = new Array("v1","v2","v3","v4","v5");
var removeValFromIndex = new Array(0,2,4);
console.log(valuesArr)
let arr2 = [];
for (let i = 0; i < valuesArr.length; i++){
if ( //could also just imput this below instead of index value
valuesArr[i] !== valuesArr[0] && // "v1" <--
valuesArr[i] !== valuesArr[2] && // "v3" <--
valuesArr[i] !== valuesArr[4] // "v5" <--
){
arr2.push(valuesArr[i]);
}
}
console.log(arr2);
This works. However, you would make a new array in the process. Not sure if thats would you want or not, but technically it would be an array containing only the values you wanted.

You can try Lodash js library functions (_.forEach(), _.remove()). I was using this technique to remove multiple rows from the table.
let valuesArr = [
{id: 1, name: "dog"},
{id: 2, name: "cat"},
{id: 3, name: "rat"},
{id: 4, name: "bat"},
{id: 5, name: "pig"},
];
let removeValFromIndex = [
{id: 2, name: "cat"},
{id: 5, name: "pig"},
];
_.forEach(removeValFromIndex, (indi) => {
_.remove(valuesArr, (item) => {
return item.id === indi.id;
});
})
console.log(valuesArr)
/*[
{id: 1, name: "dog"},
{id: 3, name: "rat"},
{id: 4, name: "bat"},
];*/
Don't forget to clone (_.clone(valuesArr) or [...valuesArr]) before mutate your array

You could try and use delete array[index] This won't completely remove the element but rather sets the value to undefined.

removeValFromIndex.forEach(function(toRemoveIndex){
valuesArr.splice(toRemoveIndex,1);
});

For Multiple items or unique item:
I suggest you use Array.prototype.filter
Don't ever use indexOf if you already know the index!:
var valuesArr = ["v1","v2","v3","v4","v5"];
var removeValFrom = [0, 2, 4];
valuesArr = valuesArr.filter(function(value, index) {
return removeValFrom.indexOf(index) == -1;
}); // BIG O(N*m) where N is length of valuesArr and m is length removeValFrom
Do:
with Hashes... using Array.prototype.map
var valuesArr = ["v1","v2","v3","v4","v5"];
var removeValFrom = {};
([0, 2, 4]).map(x=>removeValFrom[x]=1); //bild the hash.
valuesArr = valuesArr.filter(function(value, index) {
return removeValFrom[index] == 1;
}); // BIG O(N) where N is valuesArr;

You could construct a Set from the array and then create an array from the set.
const array = [1, 1, 2, 3, 5, 5, 1];
const uniqueArray = [...new Set(array)];
console.log(uniqueArray); // Result: [1, 2, 3, 5]

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