Question

json对象是

var data = [{"Parent":1,"Child":[4,5,6]},{"Parent":2},{"Parent":3}]

如何使用underscore.js链/ map / pluck等...函数来获得扁平化结果

     var result = [];
for (var i = 0; i < data.length; i++) {
    result.push(data[i].Parent);
    if (data.Child != undefined) {
        for (var j = 0; j < data[i].Child.length; j++) {
            result.push(data[i].Child[j]);
        }
    }
}
console.log(result) >> //1,4,5,6,2,3

Answer 1

这是一个较短的解决方案：

flat = _.flatten(_.map(data, _.values))

Answer 2

或者，如果您想要一个可以普遍展平任何对象或数组集合的函数，

您可以使用以下内容扩展Underscore：

_.mixin({crush: function(l, s, r) {return _.isObject(l)? (r = function(l) {return _.isObject(l)? _.flatten(_.map(l, s? _.identity:r)):l;})(l):[];}});

粉碎（缺少更好的名称）可以像_.crush(list, [shallow])或_(list).crush([shallow])一样调用，其行为与Underscore的内置 Flatten的广义形式完全相同

它可以传递任何深度的嵌套对象，数组或两者的集合，并返回包含所有输入值和自身属性的单层数组。与展平一样，如果传递一个额外的参数，其值为true，则执行“浅”执行，输出仅展平一个级别。

示例1：

_.crush({ a: 1, b: [2], c: [3, { d: { e: 4 } }] }); //=> [1, 2, 3, 4]

示例2：

_.crush({ a: 1, b: [2], c: [3, { d: { e: 4 } }] }, true); //=> [1, 2, 3, { // d: { // e: 4 // } // }]

代码本身的解释如下：

_.mixin({ // This extends Underscore's native object. crush: function(list, shallow, r) { // The "r" is really just a fancy // way of declaring an extra variable // within the function without // taking up another line. return _.isObject(list)? // Arrays (being a type of object) // actually pass this test too. (r = function(list) { // It doesn't matter that "r" might have // been passed as an argument before, // as it gets rewritten here anyway. return _.isObject(list)? // While this test may seem redundant at // first, because it is enclosed in "r", // it will be useful for recursion later. _.flatten(_.map(list, shallow? // Underscore's .map is different // from plain Javascript's in // _.map will always return // that it will apply the passed // an array, which is why we // function to an object's values // can then use _.flatten. // as well as those of an array. _.identity // If "shallow" is truthy, .map uses the identity // function so "list" isn't altered any further. : r // Otherwise, the function calls itself on each value. )) : list // The input is returned unchanged if it has no children. ; })(list) // The function is both defined as "r" and executed at once. : [] // An empty array is returned if the initial input ; // was something other than an object or array. } });

希望如果有人需要它会有所帮助。：）

Answer 3

假设你想首先得到父母，然后让孩子们接受：

_.chain(data).pluck("Parent")
             .concat(_.flatten(_(data).pluck("Child")))
             .reject(_.isUndefined)
             .value()

Answer 4

如果要使用underScore.js将多个数组的数组展平为一个元素数组，那么就是这样做的。按照我的例子：

我的图表有2个系列。每个系列都有一个名称和一系列数据点{xtime，yValue}。我的目标是将所有数据点从2个系列中分解为一系列数据点，以便填写表格。

var reducedArray = // flatten an array of series of data-objects into one series of data-objects
_.flatten( _.map( AllMySeries, function ( aSeries ) {
    return ( _.map( aSeries.dataPoints, function ( aPoint ) {
                return { curveID: aSeries.legendText, xT: aPoint.x, yVal: aPoint.y };
            } ) );
} ) );

我的结果：

'Series1','2017-04-19 08:54:19',1
'Series1','2017-04-19 08:59:19',0
'Series1','2017-04-19 09:04:19',1
'Series2','2017-04-19 08:54:19',1
'Series2','2017-04-19 08:59:19',0
'Series2','2017-04-19 09:04:19',1

如何使用underscore.js生成flatten结果

4 个答案:

您可以使用以下内容扩展Underscore：