从数组中删除匹配特定字符串的所有元素

Question

What is the easiest way to remove all elements from array that match specific string? For example:

array = [1,2,'deleted',4,5,'deleted',6,7];

I want to remove all 'deleted' from the array.

Answer 1

Simply use the Array.prototype.filter() function for obtain elements of a condition

var array = [1,2,'deleted',4,5,'deleted',6,7];
var newarr = array.filter(function(a){return a !== 'deleted'})

Update: ES6 Syntax

let array = [1,2,'deleted',4,5,'deleted',6,7]
let newarr = array.filter(a => a !== 'deleted')

Answer 2

If you have multiple strings to remove from main array, You can try this

// Your main array 
var arr = [ '8','abc','b','c'];

// This array contains strings that needs to be removed from main array
var removeStr = [ 'abc' , '8'];

arr = arr.filter(function(val){
  return (removeStr.indexOf(val) == -1 ? true : false)
})

console.log(arr);

// 'arr' Outputs to :
[ 'b', 'c' ]

OR

Better Performance(Using hash) , If strict type equality not required

// Your main array 
var arr = [ '8','deleted','b','c'];

// This array contains strings that needs to be removed from main array
var removeStr = [ 'deleted' , '8'];
var removeObj = {};  // Use of hash will boost performance for larger arrays
removeStr.forEach( e => removeObj[e] = true);

var res = arr.filter(function(val){
  return !removeObj[val]
})

console.log(res);

// 'arr' Outputs to :
[ 'b', 'c' ]

Answer 3

array = array.filter(function(s) {
    return s !== 'deleted';
});

Answer 4

If you want the same array then you can use

var array = [1,2,'deleted',4,5,'deleted',6,7];
var index = "deleted";
for(var i = array.length - 1; i >= 0; i--) {
    if(array[i] === index) {
       array.splice(i, 1);
    }
}

EXAMPLE 1

else you can use Array.prototype.filter which creates a new array with all elements that pass the test implemented by the provided function.

 var arrayVal = [1,2,'deleted',4,5,'deleted',6,7];
function filterVal(value) {
  return value !== 'deleted';
}
var filtered = arrayVal.filter(filterVal);

EXAMPLE 2

Answer 5

A canonical answer would probably look like this:

[10, 'deleted', 20, 'deleted'].filter(x => x !== 'deleted');
//=> [10, 20]

There's nothing unexpected here; any developers can read, understand and maintain this code. From that perspective this solution is great. I just want to offer some different perspectives.

Firstly I sometimes struggle with the semantic of filter when the condition is "reversed":

[2, 3, 2, 3].filter(x => x === 2);
[2, 3, 2, 3].filter(x => x !== 2);

This is a contrived example but I bet a few readers did pause for a nanosecond. These small cognitive bumps can be exhausting in the long run.

I personally wish there would be a reject method:

[2, 3, 2, 3].filter(x => x === 2);
[2, 3, 2, 3].reject(x => x === 2);

Secondly there's a lot of "machinery" in this expression x => x === 2 : a function expression, a parameter and an equality check.

This could be abstracted away by using a curried function:

const eq =
  x => y =>
    x === y;

[2, 3, 2, 3].filter(eq(2));
//=> [2, 2]

We can see that eq(2) is the same as x => x === 2 just shorter and with added semantic.

Now let's build a reject function and use eq :

const reject =
  (pred, xs) =>
    xs.filter(x =>
      pred(x) === false);

reject(eq(2), [2, 3, 2, 3]);
//=> [3, 3]

But what if we need to reject other things? Well we can build an either function that uses eq :

const either =
  (...xs) => y =>
    xs.some(eq(y));

reject(either(1, 2), [1, 3, 2, 3]);
//=> [3, 3]

Finally to answer your question:

reject(eq('deleted'), [1, 'deleted', 3]);
//=> [1, 3]

reject(either('deleted', 'removed'), [1, 'deleted', 3, 'removed', 5]);
//=> [1, 3, 5]

We could go further and remove based on different predicates eg remove if matches the string "delete" or is 0 .

Let's build a eitherfn function that takes a list of predicates:

const eitherfn =
  (...fn) => y =>
    fn.some(f =>
      f(y));

And now let's build a match function:

const match =
  x => y =>
    typeof y === 'string'
      ? y.includes(x)
      : false;

Then:

reject(eitherfn(match('delete'), eq(0)), [0, 1, 'deleted', 3, 'will delete', 5])
// [1, 3, 5]