你可以这样做

Var sources = []; 图像。映射(函数(img) { 如果(img.src.split (' . ') .pop () ! = = json){/ /如果扩展不是. json sources.push (img.src);//只推有效值 ｝ })；

下面是一个实用工具方法(ES5兼容)，它只映射非空值(隐藏reduce调用):

函数mapNonNull(arr, cb) { 加勒比海盗。Reduce(函数(累加器，值，索引，arr) { Var结果= cb。调用(null, value, index, arr); If (result != null) { accumulator.push(结果); ｝ 返回蓄电池; },[]); ｝ var result = mapNonNull(["a"， "b"， "c"]，函数(值){ 返回值=== "b" ?Null:值;//排除"b" })； console.log(结果);// ["a"， "c"]

这里有一个有趣的解决方案:

/**
 * 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;
    }, []);
}

与绑定操作符一起使用:

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

TLDR:你可以先过滤你的数组，然后执行你的映射，但这将需要对数组进行两次传递(过滤器返回一个数组映射)。因为这个数组很小，所以它的性能代价非常小。你也可以做简单的减法。然而，如果你想重新想象如何通过对数组(或任何数据类型)的一次传递来实现这一点，你可以使用Rich Hickey流行的“传感器”思想。

答:

我们不应该要求增加点链并对数组[]进行操作。map(fn1).filter(f2)…因为这种方法在内存中为每个约简函数创建中间数组。

最好的方法是对实际的约简函数进行操作，因此只有一次数据传递，没有额外的数组。

reduce函数是传递给reduce的函数，它接受累加器并从源函数输入，然后返回类似累加器的东西

// 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!!!']

资源:丰富的吻痕传感器后

回答无多余的边缘情况:

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

你可以像这样使用map + filter:

   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);