how to return nothing ( empty array) with array.map function

Question

Right now, if 'Everything' in the list is detected, the output becomes [""].
Expected output: []

Copy.names = rule.names.map(function(x) {                                
    if (x.name ==='Everything') {                                   
        return '';
    } else {
        return x.name;
    }
});

Lawtun

just write return; Also, consider using Array.prototype.filter for this. – Lutz

if i do that, it returns 'null' which I don't want – Lawtun

You probably want to use a method such as filter. map returns an element for every element you iterate over. – Thorn

@Lawtun Change map to filter, return '' to false and return x.name to true – Scenography

Or you could simply put return x.name === 'Everything' instead of the whole if...else – Scenography

Map will always return SOMETHING, you will need to change it to filter or, using a regular for loop to get the desired behavior. – Lutz

guys, my backend expects an empty list [] if "everything" is selected. Otherwise, it expects a list of the other names ['apple', 'orange'] So what's the best way? – Lawtun

Possible duplicate of Removing elements with Array.map in JavaScript – Chicory

Answer 1

Use Array.prototype.filter:

Copy.names = rule.names.filter(function(x) {                                
    return x.name !=='Everything';
}).map(function (x) {
    return x.name;
});

Kaiser

I did this, it's actually returning 'Everything' , which is wrong – Lawtun

If you are using the code in your original post, the problem is that Array.prototype.map applies a transformation to every element in your list. You must instead use some logic to remove or ignore elements you don't want. Array.prototype.map does not have the ability to ignore elements alone, so it must be combined with another function, like Array.prototype.filter: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… – Kaiser

Shouldn't it be x.name !== 'Everything'? – Chicory

Answer 2

Another solution with Array.filter():

names.map(
  (x) => x.name === 'Everything' && x.name
).filter(Boolean)

Plasterboard

Answer 3

If you can use Lodash (which I highly recommend), you can deal with it in elegant way by using _.flatMap:

Copy.names = _.flatMap(rule.names, function(x) {
    if (x.name ==='Everything') {                                   
        return [];
    } else {
        return [x.name];
    }
})

As you can see, it's similiar to map, except that you return array of items instead of item.

Chicory

One caveat to this approach is that you will create a temporary array for every element in rule.names. This isn't a big deal for just a few elements, but can become costly with large lists. – Kaiser

please dont ever import a heavy dependency for just one function... – Oryx

@samthet I think the package gets reduced in the build package to include only the used functions – Vaniavanilla

Answer 4

If you can use ES6, you can use generator for that:

Copy.names = Array.from(function* () {
    for (var x of rule.names) {
       if (x.name ==='Everything') {                                   
            // do nothing
       } else {
            yield x.name;
       }
    }
})

If not... you can always go for imperative way:

Copy.names = []

for (var x of rule.names) {
   if (x.name ==='Everything') {                                   
        // do nothing
   } else {
        Copy.names.push(x.name);
   }
}

Chicory

I don't think that the first approach (Using Array.from with generator) works. I tried it in different ES6 enabled environments and it always returns an empty array. – Polyzoarium

Answer 5

An answer I wrote elsewhere covers this, but if you want to be able to accomplish the transform of Array.map() but also to change the output length, you would need to use Array.reduce().

Usually, though, it will make more sense to filter--preferably before you map, but if needed, then after.

Damson

Answer 6

For this you should use reduce instead of map, here an example:

Copy.names = rule.names.reduce(function (arr, x) {
  x.name !== 'Everything' && arr.push(x.name)
  return arr
}, [])

Reinold