Question

我有一个无限范围的集合（可以是0，可以是1000，可以是一百万）。我需要搜索集合中每个模型的属性，并返回相同的属性（及其值）。

例如，如果我在集合中有以下三个模型：

modelOne:
  color: "red"
  age: 10
  size: "large"

modelTwo:
  color: "red"
  age: 11
  size: "medium"

modelThree:
  color: "red"
  age: 9
  size: "large"

我需要应用程序返回color: "red"（或其他一些可以解析的派生），因为它是所有三个模型中唯一相同的属性。

编辑 John Munsch的解决方案非常有效，但现在需求已经改变，因为某些属性现在可以是数组。有没有办法比较常规属性，和属性是数组？

新代码示例：

modelOne:
  color: "red"
  age: 10
  sizes: ["small", "large"]

modelTwo:
  color: "red"
  age: 9
  sizes: ["small", "large"]

Answer 1

这是一个快速的jsFiddle与我的答案版本：http://jsfiddle.net/JohnMunsch/3NMGD/

注意：jsFiddle和下面的代码都已更新，以反映问题的变更要求。

var model = [
    {
      color: "red",
      age: 10,
      size: [ "small", "large" ]
    },
    {
      color: "red",
      age: 11,
      size: [ "small", "large" ]
    },
    {
      color: "red",
      age: 9,
      size: [ "small", "large" ]
    }
];

function findCommonalities(data) {
    if (data.length > 0) {
        // It's safe enough to get the list of keys from the very first
        // element. If the later ones differ, you know that the keys they
        // had but the first element didn't are guaranteed not to be in
        // the common set anyway because the first element didn't
        // have them.
        var keys = _.keys(data[0]);
        var commonalities = { };

        _.each(keys,
            function (key) {
                var values = _.pluck(data, key);

                if (values.length == data.length) {
                    // Sadly calling _.uniq() won't work when the values
                    // plucked out are arrays themselves. It calls ===
                    // and that's not sufficient for arrays.
                    if (_.isArray(values[0])) {
                        // However, we can get a little tricky here.
                        // What if we _.zip() together all of the arrays
                        // (assuming the ordering for each array is the
                        // same) then we'll end up with an array of
                        // arrays where each one can again be tested
                        // with _.uniq() because each one will contain
                        // the same value taken from each array.
                        var zippedValues = _.zip(values);
                        console.log("zippedValues", zippedValues);

                        if (!_.find(zippedValues,
                            function (zippedValue) {
                                var uniqueValues = _.uniq(zippedValue);

                                // Note: This test is the inverse of the
                                // one below. We're trying to find any
                                // array that has more than one value in
                                // it. If we do then there's some
                                // variance.
                                return uniqueValues.length != 1;
                            })) {
                            // We didn't find any arrays that had any
                            // variance so we want this as well.
                            commonalities[key] = values[0];
                        }
                    } else {
                        var uniqueValues = _.uniq(values);

                        if (uniqueValues.length == 1) {
                            commonalities[key] = uniqueValues[0];
                        }
                    }
                }
            }
        );

        return commonalities;
    } else {
        return { };
    }
}

console.log("commonalities: ", findCommonalities(model));

使用少量密钥和少量项目，性能似乎很好，但您需要使用一百万条记录和大量密钥对其进行测试。

Backbone.js：比较集合中的模型属性

1 个答案: