有一个很好的数组方法reduce()从数组中获取一个值。例子:

[0,1,2,3,4].reduce(function(previousValue, currentValue, index, array){
  return previousValue + currentValue;
});

在对象上实现同样效果的最佳方法是什么?我想这样做:

{ 
    a: {value:1}, 
    b: {value:2}, 
    c: {value:3} 
}.reduce(function(previous, current, index, array){
  return previous.value + current.value;
});

但是,Object似乎没有实现任何reduce()方法。


当前回答

一种选择是减少键():

var o = { 
    a: {value:1}, 
    b: {value:2}, 
    c: {value:3} 
};

Object.keys(o).reduce(function (previous, key) {
    return previous + o[key].value;
}, 0);

有了这个,你会想要指定一个初始值,或者第一轮将是'a' + 2。

如果你想将结果作为对象({value:…}),你必须每次初始化并返回对象:

Object.keys(o).reduce(function (previous, key) {
    previous.value += o[key].value;
    return previous;
}, { value: 0 });

其他回答

这并不难自己实现:

function reduceObj(obj, callback, initial) {
    "use strict";
    var key, lastvalue, firstIteration = true;
    if (typeof callback !== 'function') {
        throw new TypeError(callback + 'is not a function');
    }   
    if (arguments.length > 2) {
        // initial value set
        firstIteration = false;
        lastvalue = initial;
    }
    for (key in obj) {
        if (!obj.hasOwnProperty(key)) continue;
        if (firstIteration)
            firstIteration = false;
            lastvalue = obj[key];
            continue;
        }
        lastvalue = callback(lastvalue, obj[key], key, obj);
    }
    if (firstIteration) {
        throw new TypeError('Reduce of empty object with no initial value');
    }
    return lastvalue;
}

在行动:

var o = {a: {value:1}, b: {value:2}, c: {value:3}};
reduceObj(o, function(prev, curr) { prev.value += cur.value; return prev;}, {value:0});
reduceObj(o, function(prev, curr) { return {value: prev.value + curr.value};});
// both == { value: 6 };

reduceObj(o, function(prev, curr) { return prev + curr.value; }, 0);
// == 6

你也可以把它添加到Object原型中:

if (typeof Object.prototype.reduce !== 'function') {
    Object.prototype.reduce = function(callback, initial) {
        "use strict";
        var args = Array.prototype.slice(arguments);
        args.unshift(this);
        return reduceObj.apply(null, args);
    }
}

ES6实现:Object.entries()

const o = {
  a: {value: 1},
  b: {value: 2},
  c: {value: 3}
};

const total = Object.entries(o).reduce(function (total, pair) {
  const [key, value] = pair;
  return total + value.value;
}, 0);

1:

[{value:5}, {value:10}].reduce((previousValue, currentValue) => { return {value: previousValue.value + currentValue.value}})

>> Object {value: 15}

2:

[{value:5}, {value:10}].map(item => item.value).reduce((previousValue, currentValue) => {return previousValue + currentValue })

>> 15

3:

[{value:5}, {value:10}].reduce(function (previousValue, currentValue) {
      return {value: previousValue.value + currentValue.value};
})

>> Object {value: 15}

在这种情况下,你真正想要的是Object.values。下面是一个简明的ES6实现:

const add = {
  a: {value:1},
  b: {value:2},
  c: {value:3}
}

const total = Object.values(add).reduce((t, {value}) => t + value, 0)

console.log(total) // 6

或者仅仅是:

const add = {
  a: 1,
  b: 2,
  c: 3
}

const total = Object.values(add).reduce((t, n) => t + n)

console.log(total) // 6

由于它还没有在答案中得到证实,因此下划线的减少也适用于此。

_.reduce({ 
    a: {value:1}, 
    b: {value:2}, 
    c: {value:3} 
}, function(prev, current){
    //prev is either first object or total value
    var total = prev.value || prev

    return total + current.value
})

注意,_。如果列表对象只有一项,Reduce将返回唯一的值(对象或其他),而不调用迭代器函数。

_.reduce({ 
    a: {value:1} 
}, function(prev, current){
    //not called
})

//returns {value: 1} instead of 1