915

How can I skip an array element in .map?

My code:

var sources = images.map(function (img) {
    if(img.src.split('.').pop() === "json"){ // if extension is .json
        return null; // skip
    }
    else{
        return img.src;
    }
});

This will return:

["img.png", null, "img.png"]
5
  • 57
    You can't, but you could filter out all null values afterwards. Commented Jul 17, 2014 at 14:50
  • 1
    Why not? I know using continue doesn't work but it would be good to know why (also would avoid double looping) - edit - for your case couldn't you just invert the if condition and only return img.src if the result of the split pop !== json? Commented Apr 25, 2018 at 13:35
  • 2
    @GrayedFox Then implicit undefined would be put into the array, instead of null. Not that better... Commented Apr 7, 2020 at 17:27
  • @GrayedFox For each element in the initial array, the map function will insert the return value of the sub function, hence there is no way to avoid returning a value. To avoid double looping use a for loop or a forEach to fill a new array. Commented Nov 21, 2023 at 14:39
  • The map's inner mapping function can just return() (ie. no return val) rather than the invalid continue() , which will insert an undefined value as that element. That's consistent with the map() function's design: map the processed array to a returned array 1:1 by values. Otherwise use flatMap() as described in an answer below. Commented Apr 11, 2024 at 18:46

20 Answers 20

1287

Just .filter() it first:

var sources = images.filter(function(img) {
  if (img.src.split('.').pop() === "json") {
    return false; // skip
  }
  return true;
}).map(function(img) { return img.src; });

If you don't want to do that, which is not unreasonable since it has some cost, you can use the more general .reduce(). You can generally express .map() in terms of .reduce:

someArray.map(function(element) {
  return transform(element);
});

can be written as

someArray.reduce(function(result, element) {
  result.push(transform(element));
  return result;
}, []);

So if you need to skip elements, you can do that easily with .reduce():

var sources = images.reduce(function(result, img) {
  if (img.src.split('.').pop() !== "json") {
    result.push(img.src);
  }
  return result;
}, []);

In that version, the code in the .filter() from the first sample is part of the .reduce() callback. The image source is only pushed onto the result array in the case where the filter operation would have kept it.

update — This question gets a lot of attention, and I'd like to add the following clarifying remark. The purpose of .map(), as a concept, is to do exactly what "map" means: transform a list of values into another list of values according to certain rules. Just as a paper map of some country would seem weird if a couple of cities were completely missing, a mapping from one list to another only really makes sense when there's a 1 to 1 set of result values.

I'm not saying that it doesn't make sense to create a new list from an old list with some values excluded. I'm just trying to make clear that .map() has a single simple intention, which is to create a new array of the same length as an old array, only with values formed by a transformation of the old values.

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17 Comments

Doesn't this require you loop over the entire array twice? Is there any way to avoid that?
@AlexMcMillan you could use .reduce() and do it all in one pass, though performance-wise I doubt it'd make a significant difference.
With all these negative, "empty"-style values (null, undefined, NaN etc) it would be good if we could utilise one inside a map() as an indicator that this object maps to nothing and should be skipped. I often come across arrays I want to map 98% of (eg: String.split() leaving a single, empty string at the end, which I don't care about). Thanks for your answer :)
@AlexMcMillan well .reduce() is sort-of the baseline "do whatever you want" function, because you have complete control over the return value. You might be interested in the excellent work by Rich Hickey in Clojure concerning the concept of transducers.
@vsync you can't skip an element with .map(). You can however use .reduce() instead, so I'll add that.
|
381

Since 2019, Array.prototype.flatMap is a good option.

images.flatMap(({src}) => src.endsWith('.json') ? [] : src);

From MDN:

flatMap can be used as a way to add and remove items (modify the number of items) during a map. In other words, it allows you to map many items to many items (by handling each input item separately), rather than always one-to-one. In this sense, it works like the opposite of filter. Simply return a 1-element array to keep the item, a multiple-element array to add items, or a 0-element array to remove the item.

9 Comments

this is the really answer, simple and strong enough. we learn this is better than filter and reduce.
great, but I wish there was a native method without the need to return the empty array - maybe a method that would put all truthy values in the new array and skip falsy ones.
First, props to MDN for providing this kind of commentary. It is not common that documentation has this sort of practical use case examples. Second, I do wish it was more specific about the slightly more efficient part. How much more efficient than map followed by flat?
Wonderful function! Shorter modern syntax with nullish coalescing: foo.flatMap(bar => bar.baz ?? [ ])
|
42

I think the most simple way to skip some elements from an array is by using the filter() method.

By using this method (ES5) and the ES6 syntax you can write your code in one line, and this will return what you want:

let images = [{src: 'img.png'}, {src: 'j1.json'}, {src: 'img.png'}, {src: 'j2.json'}];

let sources = images.filter(img => img.src.slice(-4) != 'json').map(img => img.src);

console.log(sources);

4 Comments

that's exactly what .filter() was made for
Is this better than forEach and completing it in one pass instead of two?
As you wish @wuliwong. But please take into account that this will be still O(n) in complexity meassure and please look at least at these two articles too: frontendcollisionblog.com/javascript/2015/08/15/… and coderwall.com/p/kvzbpa/don-t-use-array-foreach-use-for-instead All the best!
Thank you @simhumileco! Exactly because of that, I am here (and probably many others as well). The question is probably how to combine .filter and .map by only iterating once.
27

TLDR: You can first filter your array and then perform your map but this would require two passes on the array (filter returns an array to map). Since this array is small, it is a very small performance cost. You can also do a simple reduce. However if you want to re-imagine how this can be done with a single pass over the array (or any datatype), you can use an idea called "transducers" made popular by Rich Hickey.

Answer:

We should not require increasing dot chaining and operating on the array [].map(fn1).filter(f2)... since this approach creates intermediate arrays in memory on every reducing function.

The best approach operates on the actual reducing function so there is only one pass of data and no extra arrays.

The reducing function is the function passed into reduce and takes an accumulator and input from the source and returns something that looks like the accumulator

// 1. create a concat reducing function that can be passed into `reduce`
const concat = (acc, input) => acc.concat([input])

// note that [1,2,3].reduce(concat, []) would return [1,2,3]

// transforming your reducing function by mapping
// 2. create a generic mapping function that can take a reducing function and return another reducing function
const mapping = (changeInput) => (reducing) => (acc, input) => reducing(acc, changeInput(input))

// 3. create your map function that operates on an input
const getSrc = (x) => x.src
const mappingSrc = mapping(getSrc)

// 4. now we can use our `mapSrc` function to transform our original function `concat` to get another reducing function
const inputSources = [{src:'one.html'}, {src:'two.txt'}, {src:'three.json'}]
inputSources.reduce(mappingSrc(concat), [])
// -> ['one.html', 'two.txt', 'three.json']

// remember this is really essentially just
// inputSources.reduce((acc, x) => acc.concat([x.src]), [])


// transforming your reducing function by filtering
// 5. create a generic filtering function that can take a reducing function and return another reducing function
const filtering = (predicate) => (reducing) => (acc, input) => (predicate(input) ? reducing(acc, input): acc)

// 6. create your filter function that operate on an input
const filterJsonAndLoad = (img) => {
  console.log(img)
  if(img.src.split('.').pop() === 'json') {
    // game.loadSprite(...);
    return false;
  } else {
    return true;
  }
}
const filteringJson = filtering(filterJsonAndLoad)

// 7. notice the type of input and output of these functions
// concat is a reducing function,
// mapSrc transforms and returns a reducing function
// filterJsonAndLoad transforms and returns a reducing function
// these functions that transform reducing functions are "transducers", termed by Rich Hickey
// source: http://clojure.com/blog/2012/05/15/anatomy-of-reducer.html
// we can pass this all into reduce! and without any intermediate arrays

const sources = inputSources.reduce(filteringJson(mappingSrc(concat)), []);
// [ 'one.html', 'two.txt' ]

// ==================================
// 8. BONUS: compose all the functions
// You can decide to create a composing function which takes an infinite number of transducers to
// operate on your reducing function to compose a computed accumulator without ever creating that
// intermediate array
const composeAll = (...args) => (x) => {
  const fns = args
  var i = fns.length
  while (i--) {
    x = fns[i].call(this, x);
  }
  return x
}

const doABunchOfStuff = composeAll(
    filtering((x) => x.src.split('.').pop() !== 'json'),
    mapping((x) => x.src),
    mapping((x) => x.toUpperCase()),
    mapping((x) => x + '!!!')
)

const sources2 = inputSources.reduce(doABunchOfStuff(concat), [])
// ['ONE.HTML!!!', 'TWO.TXT!!!']

Resources: rich hickey transducers post

2 Comments

but isn't .concat() going to be repeatedly copying acc, making it O(n^2)?
Yeah, it's more of a brain exercise in javascript where clojure (from the linked Rich Hickey post) would support functional operations like this.
26

Here's a fun solution:

/**
 * Filter-map. Like map, but skips undefined values.
 *
 * @param callback
 */
function fmap(callback) {
    return this.reduce((accum, ...args) => {
        const x = callback(...args);
        if(x !== undefined) {
            accum.push(x);
        }
        return accum;
    }, []);
}

Use with the bind operator:

[1,2,-1,3]::fmap(x => x > 0 ? x * 2 : undefined); // [2,4,6]

1 Comment

This method saved me from having to use separate map, filter and concat calls.
22

Why not just use a forEach loop?

let arr = ['a', 'b', 'c', 'd', 'e'];
let filtered = [];

arr.forEach(x => {
  if (!x.includes('b')) filtered.push(x);
});

console.log(filtered)   // filtered === ['a','c','d','e'];

Or even simpler use filter:

const arr = ['a', 'b', 'c', 'd', 'e'];
const filtered = arr.filter(x => !x.includes('b')); // ['a','c','d','e'];

2 Comments

Best would be a simple for loop that filters & creates an new array, but for the context of using map lets keep it like it's now. (was 4yrs ago I asked this question, when I knew nothing about coding)
Fair enough, given that there is no direct way to the above with map and all the solutions used an alternative method I thought I would chip in the simplest way I could think of to do the same.
15

Answer sans superfluous edge cases:

const thingsWithoutNulls = things.reduce((acc, thing) => {
  if (thing !== null) {
    acc.push(thing);
  }
  return acc;
}, [])

Comments

12
var sources = images.map(function (img) {
    if(img.src.split('.').pop() === "json"){ // if extension is .json
        return null; // skip
    }
    else{
        return img.src;
    }
}).filter(Boolean);

The .filter(Boolean) will filter out any falsey values in a given array, which in your case is the null.

Comments

11

To extrapolate on Felix Kling's comment, you can use .filter() like this:

var sources = images.map(function (img) {
  if(img.src.split('.').pop() === "json") { // if extension is .json
    return null; // skip
  } else {
    return img.src;
  }
}).filter(Boolean);

That will remove falsey values from the array that is returned by .map()

You could simplify it further like this:

var sources = images.map(function (img) {
  if(img.src.split('.').pop() !== "json") { // if extension is .json
    return img.src;
  }
}).filter(Boolean);

Or even as a one-liner using an arrow function, object destructuring and the && operator:

var sources = images.map(({ src }) => src.split('.').pop() !== "json" && src).filter(Boolean);

1 Comment

Thanks, the .filter(Boolean) is a genius solution!
5

const arr = [0, 1, '', undefined, false, 2, undefined, null, , 3, NaN];
const filtered = arr.filter(Boolean);
console.log(filtered);

/*
    Output: [ 1, 2, 3 ]
*/

Comments

4

Here's a utility method (ES5 compatible) which only maps non null values (hides the call to reduce):

function mapNonNull(arr, cb) {
    return arr.reduce(function (accumulator, value, index, arr) {
        var result = cb.call(null, value, index, arr);
        if (result != null) {
            accumulator.push(result);
        }

        return accumulator;
    }, []);
}

var result = mapNonNull(["a", "b", "c"], function (value) {
    return value === "b" ? null : value; // exclude "b"
});

console.log(result); // ["a", "c"]

Comments

3

if it null or undefined in one line ES5/ES6

//will return array of src 
images.filter(p=>!p.src).map(p=>p.src);//p = property


//in your condition
images.filter(p=>p.src.split('.').pop() !== "json").map(p=>p.src);

Comments

1

I use .forEach to iterate over , and push result to results array then use it, with this solution I will not loop over array twice

Comments

1

You can use after of you method map(). The method filter() for example in your case:

var sources = images.map(function (img) {
  if(img.src.split('.').pop() === "json"){ // if extension is .json
    return null; // skip
  }
  else {
    return img.src;
  }
});

The method filter:

const sourceFiltered = sources.filter(item => item)

Then, only the existing items are in the new array sourceFiltered.

Comments

1

You can use forEach with the return keyword instead of map like this:

const sources = [];

const images = [{src: 'helloWorld.json'}, {src: 'helloWorld2.png'}]

images.forEach((img) => {
  if(img.src.split('.').pop() === "json") return
    
  sources.push(img.src);
});

console.log(sources)

Comments

0

Here is a updated version of the code provided by @theprtk. It is a cleaned up a little to show the generalized version whilst having an example.

Note: I'd add this as a comment to his post but I don't have enough reputation yet

/**
 * @see http://clojure.com/blog/2012/05/15/anatomy-of-reducer.html
 * @description functions that transform reducing functions
 */
const transduce = {
  /** a generic map() that can take a reducing() & return another reducing() */
  map: changeInput => reducing => (acc, input) =>
    reducing(acc, changeInput(input)),
  /** a generic filter() that can take a reducing() & return */
  filter: predicate => reducing => (acc, input) =>
    predicate(input) ? reducing(acc, input) : acc,
  /**
   * a composing() that can take an infinite # transducers to operate on
   *  reducing functions to compose a computed accumulator without ever creating
   *  that intermediate array
   */
  compose: (...args) => x => {
    const fns = args;
    var i = fns.length;
    while (i--) x = fns[i].call(this, x);
    return x;
  },
};

const example = {
  data: [{ src: 'file.html' }, { src: 'file.txt' }, { src: 'file.json' }],
  /** note: `[1,2,3].reduce(concat, [])` -> `[1,2,3]` */
  concat: (acc, input) => acc.concat([input]),
  getSrc: x => x.src,
  filterJson: x => x.src.split('.').pop() !== 'json',
};

/** step 1: create a reducing() that can be passed into `reduce` */
const reduceFn = example.concat;
/** step 2: transforming your reducing function by mapping */
const mapFn = transduce.map(example.getSrc);
/** step 3: create your filter() that operates on an input */
const filterFn = transduce.filter(example.filterJson);
/** step 4: aggregate your transformations */
const composeFn = transduce.compose(
  filterFn,
  mapFn,
  transduce.map(x => x.toUpperCase() + '!'), // new mapping()
);

/**
 * Expected example output
 *  Note: each is wrapped in `example.data.reduce(x, [])`
 *  1: ['file.html', 'file.txt', 'file.json']
 *  2:  ['file.html', 'file.txt']
 *  3: ['FILE.HTML!', 'FILE.TXT!']
 */
const exampleFns = {
  transducers: [
    mapFn(reduceFn),
    filterFn(mapFn(reduceFn)),
    composeFn(reduceFn),
  ],
  raw: [
    (acc, x) => acc.concat([x.src]),
    (acc, x) => acc.concat(x.src.split('.').pop() !== 'json' ? [x.src] : []),
    (acc, x) => acc.concat(x.src.split('.').pop() !== 'json' ? [x.src.toUpperCase() + '!'] : []),
  ],
};
const execExample = (currentValue, index) =>
  console.log('Example ' + index, example.data.reduce(currentValue, []));

exampleFns.raw.forEach(execExample);
exampleFns.transducers.forEach(execExample);

Comments

0

You can do this

var sources = [];
images.map(function (img) {
    if(img.src.split('.').pop() !== "json"){ // if extension is not .json
        sources.push(img.src); // just push valid value
    }
});

3 Comments

This is not fit with map. Instead you can use forEach or reduce like in the selected answer
The point of Array.map() is that it returns the result. So the above suggestion is not very readable.
In Perl one can do something simple like this: [ "first", "a" eq "ab" ? "second" : () ] I can't find (so far) anything as simple as that in javascript.
0

I use foreach():

var sources = [];

images.forEach(function (img) {
    if(img.src.split('.').pop() !== "json"){ // if extension is .json
        sources.push(img);
    }
});

NOTE: I negated your logic.

Comments

0

you can use map + filter like this :

   var sources = images.map(function (img) {
    if(img.src.split('.').pop() === "json"){ // if extension is .json
        return null; // skip
    }
    else{
        return img.src;
    }})?.filter(x => x !== null);

Comments

-1

Just use .filter() after .map(). This is the smartest way in my opinion.

var sources = images.map(function (img) {
    if(img.src.split('.').pop() === "json"){ // if extension is .json
        return null; // skip
    }
    else{
        return img.src;
    }
}).filter(x => x);

Comments

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