javascript - 如何使用下划线选择最大的数组元素

Question

我有以下数组:

var array = [[{key1: 1, key2: thing}, {key1: 3, key2: thing}],
 [{key1: 4, key2: stuff}, {key1: 2, key2: stuff}],
 [{key1: 3, key2: onlyOne}]]

我需要取出少于两个对象的任何数组，并且每个可能包含“n”个对象的剩余数组应该只包含具有最小 key1 的对象。最后我会:

var updatedArray = [[{key1: 1, key2: thing}],[{key1: 2, key2: stuff}]]

如何使用 Underscore 来获取此内容？

现在我已经完成了:

var soFar = _.filter(_.groupBy(_.map(array, function(n){return _.find(n)}), 
"key2"), function(n){return n.length > 1});

//soFar = [[{key1: 1, key2: thing}, {key1: 3, key2: thing}], 
//[{key1: 4, key2: stuff}, {key1: 2, key2: stuff}]]`

这让我摆脱了只有一个对象的数组，但我无法根据对象的比较找出下一部分来修改每个数组。

Answer 1

这是按照OP要求使用undersore/lodash的答案:

var array = [
    [{
        key1: 1,
        key2: 'thing'
    }, {
        key1: 3,
        key2: 'thing'
    }],
    [{
        key1: 4,
        key2: 'stuff'
    }, {
        key1: 2,
        key2: 'stuff'
    }],
    [{
        key1: 3,
        key2: 'onlyOne'
    }]
];

array = _.filter(array, function(v) {
    return v.length >= 2;
});

array = _.map(array, function(v) {
    v.sort(function(a, b) {
        return a.key1 > b.key1;
    });
    return [v[0]];
});


console.log(array);

第二种方法，我们将映射和过滤函数合并到一个语句中。接下来还有一点困惑。

var array = [
    [{
        key1: 1,
        key2: 'thing'
    }, {
        key1: 3,
        key2: 'thing'
    }],
    [{
        key1: 4,
        key2: 'stuff'
    }, {
        key1: 2,
        key2: 'stuff'
    }],
    [{
        key1: 3,
        key2: 'onlyOne'
    }]
];

array = _.map(_.filter(array, function(v) {
    return v.length >= 2;
}), function(v) {
    v.sort(function(a, b) {
        return a.key1 > b.key1;
    });
    return [v[0]];
});


console.log(array);

输出: