The best way to remove array element by value
JavascriptArraysJavascript Problem Overview
I have an array like this
arr = ["orange","red","black","white"]
I want to augment the array object defining a deleteElem()
method which acts like this:
arr2 = arr.deleteElem("red"); // ["orange","black","white"] (with no hole)
What is the best way to accomplish this task using just the value parameter (no index)?
Javascript Solutions
Solution 1 - Javascript
Here's how it's done:
var arr = ["orange","red","black","white"];
var index = arr.indexOf("red");
if (index >= 0) {
arr.splice( index, 1 );
}
This code will remove 1 occurency of "red" in your Array.
Solution 2 - Javascript
Back when I was new to coding I could hardly tell what splice
was doing, and even today it feels less readable.
But readability counts.
I would rather use the filter method like so:
arr = ["orange","red","black","white","red"]
arr = arr.filter(val => val !== "red");
console.log(arr) // ["orange","black","white"]
Note how all occurrences of "red" are removed from the array.
From there, you can easily work with more complex data such as array of objects.
arr = arr.filter(obj => obj.prop !== "red");
Solution 3 - Javascript
There is an underscore method for this, http://underscorejs.org/#without
arr = ["orange","red","black","white"];
arr = _.without(arr, "red");
Solution 4 - Javascript
The trick is to go through the array from end to beginning, so you don't mess up the indices while removing elements.
var deleteMe = function( arr, me ){
var i = arr.length;
while( i-- ) if(arr[i] === me ) arr.splice(i,1);
}
var arr = ["orange","red","black", "orange", "white" , "orange" ];
deleteMe( arr , "orange");
arr is now ["red", "black", "white"]
Solution 5 - Javascript
My approach, let's see what others have to say. It supports an "equals" method as well.
// Remove array value
// @param {Object} val
Array.prototype.removeByValue = function (val) {
for (var i = 0; i < this.length; i++) {
var c = this[i];
if (c == val || (val.equals && val.equals(c))) {
this.splice(i, 1);
break;
}
}
};
Read https://stackoverflow.com/a/3010848/356726 for the impact on iterations when using prototype with Array.
Solution 6 - Javascript
Array.prototype.deleteElem = function(val) {
var index = this.indexOf(val);
if (index >= 0) this.splice(index, 1);
return this;
};
var arr = ["orange","red","black","white"];
var arr2 = arr.deleteElem("red");
Solution 7 - Javascript
Or simply check all items, create a new array with non equal and return it.
var arr = ['orange', 'red', 'black', 'white'];
console.info('before: ' + JSON.stringify(arr));
var deleteElem = function ( val ) {
var new_arr = [];
for ( var i = 0; i < this.length; i++ ) {
if ( this[i] !== val ) {
new_arr.push(this[i]);
}
}
return new_arr;
};
arr = deleteElem('red');
console.info('after: ' + JSON.stringify(arr));
Solution 8 - Javascript
The best way is to use splice and rebuild new array, because after splice, the length of array does't change.
Check out my answer:
function remove_array_value(array, value) {
var index = array.indexOf(value);
if (index >= 0) {
array.splice(index, 1);
reindex_array(array);
}
}
function reindex_array(array) {
var result = [];
for (var key in array) {
result.push(array[key]);
}
return result;
}
example:
var example_arr = ['apple', 'banana', 'lemon']; // length = 3
remove_array_value(example_arr, 'banana');
banana is deleted and array length = 2
Solution 9 - Javascript
If order the array (changing positions) won't be a problem you can solve like:
var arr = ["orange","red","black","white"];
arr.remove = function ( item ) {
delete arr[item];
arr.sort();
arr.pop();
console.log(arr);
}
arr.remove('red');
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
Solution 10 - Javascript
Here you go:
arr.deleteElem = function ( val ) {
for ( var i = 0; i < this.length; i++ ) {
if ( this[i] === val ) {
this.splice( i, 1 );
return i;
}
}
};
Live demo: http://jsfiddle.net/4vaE2/3/
The deleteElem
method returns the index of the removed element.
var idx = arr.deleteElem( 'red' ); // idx is 1