Removing specific items from a javascript array based on index values

Question

Suppose I have a simple javascript array like such:

var test_array  = ["18081163__,0,0.15,15238", "34035", "Somerset", "Local", "31221", "29640", "42575", "749", "1957", "45809", "17597", "43903", "1841", "1", "Norfolk Road", "Other"]

It has a length = 16. I want to remove all items based on index except [0, 2, 3, 14]. I know I could use splice to do it piece by piece like such:

test_array.splice(1,1);
test_array.splice(3, 10);
test_array.splice(4, 1);

How could this be done in a single line of code to remove items with index [1, 4,5,6,7,8,9,10,11,12,13,15]?

do you want to keep the object reference to test_array? what array of indices do you have? the one to keep or the one to delete — Nina Scholz
– Nina Scholz, Commented May 12, 2021 at 18:34
I wanted to remove indices [1, 4,5,6,7,8,9,10,11,12,13,15]. Don't really care about keeping the original test_array. — gwydion93
– gwydion93, Commented May 12, 2021 at 19:07

Mamun · Accepted Answer · 2021-05-12 18:38:44Z

2

Consider you have an array of indexes based on which you want to remove items, then you can try using Array.prototype.filter()

The filter() method creates a new array with all elements that pass the test implemented by the provided function.

and Array.prototype.includes()

The includes() method determines whether an array includes a certain value among its entries, returning true or false as appropriate.

Demo:

var test_array  = ["18081163__,0,0.15,15238", "34035", "Somerset", "Local", "31221", "29640", "42575", "749", "1957", "45809", "17597", "43903", "1841", "1", "Norfolk Road", "Other"];
var index = [1,4,5,6,7,8,9,10,11,12,13,15];
test_array = test_array.filter((item, idx) => !index.includes(idx)); //remove the items whose index does not include the index array

console.log(test_array);

edited May 12, 2021 at 18:38

answered May 12, 2021 at 18:33

Mamun

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2 Comments

gwydion93 Over a year ago

Great solution using .filter! Short and sweet; this is exactly what I was looking for.

Mamun Over a year ago

@gwydion93, glad that it helps:)

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