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努力编写代码......迷失在循环中。

我有 2 个数据集，例如：

var elements = [
        {"id":"21.U2duHWiX.0zu.E0C","amount":"345"},
        {"id":"21.U2duHWiX.A5q.E0C","amount":"344"}
    ]

var elements_in_combination = [
    {"id":"21.U2duHWiX.0zu.E0C","amount":"329","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C"]},
    {"id":"21.U2duHWiX.A5q.E0C","amount":"328","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C"]},
]

我正在寻找使用所有元素的最低数量。

答案是 329 + 328。

在这里，有 3 个元素，例如：

var elements = [
        {"id":"21.U2duHWiX.0zu.E0C","amount":"345"},
        {"id":"21.U2duHWiX.A5q.E0C","amount":"344"},
        {"id":"21.U2duHWiX.P1y.E0C","amount":"343"}
    ]

var elements_in_combination = [
    {"id":"21.U2duHWiX.0zu.E0C","amount":"329","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C"]},
    {"id":"21.U2duHWiX.A5q.E0C","amount":"328","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C"]},
    {"id":"21.U2duHWiX.0zu.E0C","amount":"329","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.P1y.E0C","amount":"327","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.A5q.E0C","amount":"328","combination":["21.U2duHWiX.A5q.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.P1y.E0C","amount":"327","combination":["21.U2duHWiX.A5q.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.0zu.E0C","amount":"314","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.A5q.E0C","amount":"313","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.P1y.E0C","amount":"312","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C","21.U2duHWiX.P1y.E0C"]}
]

这里的答案是 314 + 313 + 312....但我不知道如何使用代码到达那里。

元素越多，事情就越复杂，当它们可能不会全部组合在一起时，例如：

var elements = [
    {"id":"21.U2duHWiX.0zu.E0C","amount":"345"},
    {"id":"21.U2duHWiX.A5q.E0C","amount":"344"},
    {"id":"21.U2duHWiX.J3e.E0C","amount":"342"},
    {"id":"21.U2duHWiX.P1y.E0C","amount":"343"}
]

var elements_in_combination = [
    {"id":"21.U2duHWiX.0zu.E0C","amount":"329","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C"]},
    {"id":"21.U2duHWiX.A5q.E0C","amount":"328","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C"]},
    {"id":"21.U2duHWiX.0zu.E0C","amount":"329","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.J3e.E0C"]},
    {"id":"21.U2duHWiX.J3e.E0C","amount":"326","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.J3e.E0C"]},
    {"id":"21.U2duHWiX.0zu.E0C","amount":"329","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.P1y.E0C","amount":"327","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.A5q.E0C","amount":"328","combination":["21.U2duHWiX.A5q.E0C","21.U2duHWiX.J3e.E0C"]},
    {"id":"21.U2duHWiX.J3e.E0C","amount":"326","combination":["21.U2duHWiX.A5q.E0C","21.U2duHWiX.J3e.E0C"]},
    {"id":"21.U2duHWiX.A5q.E0C","amount":"328","combination":["21.U2duHWiX.A5q.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.P1y.E0C","amount":"327","combination":["21.U2duHWiX.A5q.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.J3e.E0C","amount":"326","combination":["21.U2duHWiX.J3e.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.P1y.E0C","amount":"327","combination":["21.U2duHWiX.J3e.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.0zu.E0C","amount":"314","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C","21.U2duHWiX.J3e.E0C"]},
    {"id":"21.U2duHWiX.A5q.E0C","amount":"313","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C","21.U2duHWiX.J3e.E0C"]},
    {"id":"21.U2duHWiX.J3e.E0C","amount":"311","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C","21.U2duHWiX.J3e.E0C"]},
    {"id":"21.U2duHWiX.0zu.E0C","amount":"314","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.A5q.E0C","amount":"313","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.P1y.E0C","amount":"312","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.A5q.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.0zu.E0C","amount":"314","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.J3e.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.J3e.E0C","amount":"311","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.J3e.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.P1y.E0C","amount":"312","combination":["21.U2duHWiX.0zu.E0C","21.U2duHWiX.J3e.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.A5q.E0C","amount":"313","combination":["21.U2duHWiX.A5q.E0C","21.U2duHWiX.J3e.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.J3e.E0C","amount":"311","combination":["21.U2duHWiX.A5q.E0C","21.U2duHWiX.J3e.E0C","21.U2duHWiX.P1y.E0C"]},
    {"id":"21.U2duHWiX.P1y.E0C","amount":"312","combination":["21.U2duHWiX.A5q.E0C","21.U2duHWiX.J3e.E0C","21.U2duHWiX.P1y.E0C"]}
]

关于如何解决这个问题的任何想法？

（对不起，解释和解决一样困难）

编辑：澄清

这是一个抽象的例子：

var elements = [ 
    { id: A, value: '#' },
    { id: B, value: '#' },
    { id: C, value: '#' }
]

var elements_in_combination = [
    { id: A, value: '#', combinations: [A, B] },
    { id: B, value: '#', combinations: [A, B] },
    { id: A, value: '#', combinations: [A, C] },
    { id: C, value: '#', combinations: [A, C] },
    { id: B, value: '#', combinations: [B, C] }, 
    { id: C, value: '#', combinations: [B, C] },
    { id: A, value: '#', combinations: [A, B, C] },
    { id: B, value: '#', combinations: [A, B, C] },
    { id: C, value: '#', combinations: [A, B, C] },
]

我想知道什么产生了最低值，计算方法是：

[A, B, C] = '##'
or
[A, B] + C = '##'
or
[A, C] + B = '##'
or
A + [B, C] = '##'
or
A + B + C = '##'

然后我需要从元素和具有最佳组合的 elements_in_combination 构建一个数组，例如：

var elements = [ 
    { id: A, value: '#', combinations: [A, B] },
    { id: B, value: '#', combinations: [A, B] },
    { id: C, value: '#' }
]
4

3 回答 3

1

好的。检查这个脚本：

// this part is only needed if your ids are arbitrary, and can contain the join-character
// if not, you could replace this by the identity function
var count = 0, numericids = {};
function getNumericId(id) {
    return id in numericids ? numericids[id] : numericids[id] = count++;
}

// returns the same (reversible) id for all similiar [unsorted] key combinations
function id(keys) {
    return keys.map(getNumericId).sort().join('-');
    // you might remove the getNumericId part if distinct without
}

// now, build a map that holds the summed amount for each single (sub)combination
var amounts = {};
function add(amount, keys) {
    var key = id (keys);
    if (key in amounts)
        amounts[key] += amount;
    else
        amounts[key] = amount;
}
for (var i=0; i<elements.length; i++) // each element is a single combination
    add(Number(elements[i].amount), [elements[i].id]);
for (var i=0; i<elements_in_combination.length; i++)
    add(Number(elements_in_combination[i].amount), elements_in_combination[i].combination);
// so we have the amounts in a good accessible structure now

接下来，我们需要找到一个集合的所有分区。哇。这是一个 NP-hard 问题，不容易解决。三个元素（您问题中的五个组合）很容易变得越来越复杂，对于 6 个元素，您已经有 203 种可能性（贝尔数字。为了进一步阅读，我发现

好的，让我们递归解决这个问题，缓存结果并获取最小值：

// first, get the set for which we want to determine the result:
var initialset = elements.map(function(el){return getNumericId(el.id);}).sort();
// set up a cache for minimum value results:
var cache = {};

function partition(set) {
// returns an array of all partitionings into two parts
    var results = [[[set[0]],[]]];
    for (var i=1; i<set.length; i++)
        for (var j=0, l=results.length; j<l; j++) {
            // repeat the array with duplicates
            results[j+l] = [results[j][0].slice(),results[j][1].slice()];
            // but while we push to the first part in the first half
            results[ j ][0].push(set[i]);
            // we push to the second part in the second half
            results[j+l][1].push(set[i]);
        }
    return results;
}

function getMin(set) {
    var key = set.join('-');
    if (key in cache) // quick escape
        return cache[key];
    var result = {amount:Infinity, set:null};
    if (key in amounts) // there is a combination with this
        result = {amount:amounts[key], set:[key]};
    var divisions = partition(set);
    // for all possibilities to divide the set in two parts
    // (unless the first, which is [set, []])
    for (var i=1; i<divisions.length; i++) {
        // get the minimal amounts of both parts
        var first = getMin(divisions[i][0]);
        var second = getMin(divisions[i][1]);
        var sum = first.amount + second.amount;
        if (sum < result.amount) // and find new minima
            result = {amount:sum, set: first.set.concat(second.set)};
    }
    return cache[key] = result;
}
// And now invoke this monster!
if (!initialset.length) throw new Error("When searching for nothing you would find nothing");
var min = getMin(initialset);
cache = null, amounts = null; // and immediately free the memory

所以，这是你的结果！它包含您在属性中所需的总和以及amount属性中使用的组合键集set。

构建元素数组现在很容易：

var elemArr = [];
function addElem(el, comb) {
    if (min.set.indexOf(id(comb)) >= 0)
         elemArr.push(el);
}
for (var i=0; i<elements.length; i++) // each element is a single combination
    addElem(elements[i], [elements[i].id]);
for (var i=0; i<elements_in_combination.length; i++)
    addElem(elements_in_combination[i], elements_in_combination[i].combination);

return elemArr; // We've done it!

该脚本为您的所有示例返回正确的结果：

  • 329 (21.U2duHWiX.0zu.E0C) + 328 (21.U2duHWiX.A5q.E0C)
  • 314 (21.U2duHWiX.0zu.E0C) + 313 (21.U2duHWiX.A5q.E0C) + 312 (21.U2duHWiX.P1y.E0C)
  • 344 (21.U2duHWiX.A5q.E0C) + 314 (21.U2duHWiX.0zu.E0C) + 311 (21.U2duHWiX.J3e.E0C) + 312 (21.U2duHWiX.P1y.E0C) -[B]-[A,C,D]组合:-)

请注意，这些可能不是唯一的解决方案，因为只找到了许多可能的最小值中的第一个

于 2012-11-08T01:31:41.197 回答
0 
function find_matches(elements, elements_in_combination) {
    var matches = ();
    var element_ids = ();
    for (var i = 0; i < elements.length; i++) {
        element_ids.push(elements[i].id);
    }
    element_ids.sort();
    for (i = 0; i < elements_in_combination.length; i++) {
        combs = elements_in_combination[i].combination.slice(0).sort();
        if (array_equal(element_ids, combs)) {
            matches.push(elements_in_combination[i].amount;
        }
    }
    return matches;
}

请参阅此问题以了解如何实施array_equal()。

于 2012-11-07T21:15:45.833 回答
0

好的..这可能是一个选择，但是因为我不知道“最佳组合”的术语是什么，所以我无法进一步减少它。

以下代码应生成一个对象，其中包含每个元素作为对象。然后，每个元素对象将包含每个唯一数量（从低到高）的另一个对象。然后，金额对象包含该金额的可能组合。

IE。容器对象 ( finalElements ) - 元素 id - 订单和金额 - 组合：

var finalElements = { };

// sort:
elements_in_combination.sort( eic_sortOnAmmount );

function eic_sortOnAmmount( a, b ) {
    return a.amount - b.amount;
}

// parse the elements array and create an object for each element
// add the initial amount as a key:
for( var i in elements ) {
    finalElements[ elements[i].id ] = { order:[] };
    finalElements[ elements[i].id ][ elements[ i ].amount ] = null;
}

// parse the elements_in_combination array
// if the id matches one of the elements in finalElements
// add its amount and combination
for( var i in elements_in_combination ) {
    if( finalElements.hasOwnProperty( elements_in_combination[ i ].id ) ) {
        if( finalElements[ elements_in_combination[ i ].id ].hasOwnProperty( elements_in_combination[ i ].amount ) ) {
            finalElements[ elements_in_combination[ i ].id ][ elements_in_combination[ i ].amount ].push( elements_in_combination[ i ].combination );
        } else {
            finalElements[ elements_in_combination[ i ].id ].order.push(elements_in_combination[ i ].amount);
            finalElements[ elements_in_combination[ i ].id ][ elements_in_combination[ i ].amount] = [ elements_in_combination[ i ].combination ];
        }
    }
}

示例用法：

console.log(finalElements["21.U2duHWiX.0zu.E0C"].order[0]); //produces 314
console.log(finalElements["21.U2duHWiX.0zu.E0C"][finalElements["21.U2duHWiX.0zu.E0C"].order[0]]); // produces the combinations for 314

希望这会有所帮助 - 顺便说一句：空量是原始元素量。

于 2012-11-07T23:58:03.300 回答