如何使用underscore.js生成flatten结果

Question

json对象是

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

如何使用underscore.js chain/map/pluck等...函数来获得展平结果

     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

这是一个更短的解决方案:

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,

您可以使用以下方式扩展Underscore:

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

Crush(缺少一个更好的名字)可以被称为_.crush(list, [shallow])或者_(list).crush([shallow])和行为完全类似于Underscore的内置Flatten的通用形式.

它可以传递嵌套对象,数组或任何深度两者的集合,并返回包含所有输入值和自身属性的单级数组.与Flatten一样,如果传递一个额外的参数,该参数的计算结果为true,则执行"浅"执行,输出仅展平一个级别.

例1:

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

//=> [1, 2, 3, 4]

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

假设你想先得到父母,然后让孩子们:

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