2016-01-20 90 views
2

我想这种结构转换:的Javascript转换对象的数组,树

var initial = [ 
{ Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" }, 
{ Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" }, 
{ Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" }, 
{ Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" }, 
{ Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" }, 
{ Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" }, 
{ Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" }, 
{ Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" } 
]; 

这一个:

var example = { 
"Phase 1": { 
    "Step 1": { 
     "Task 1": { Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" }, 
     "Task 2": { Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" } 
    } , 
    "Step 2": { 
     "Task 1": { Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" }, 
     "Task 2": { Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" } 
    } 
} , 
"Phase 2": {   
    "Step 1": { 
     "Task 1": { Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" }, 
     "Task 2": { Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" }, 
    } , 
    "Step 2": { 
     "Task 1": { Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" }, 
     "Task 2": { Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" }  
    } 
} 
}; 

,所以我可以很容易地提取这样example['Phase 2']['Step 1']['Task 1']['Value']

值我做了使用groupBy功能像这样的第一步:

function groupBy(d, arr) { 
return arr.reduce(function(acc, i) { 
    var p = i[d]; 
    var temp = acc[p] || []; 
    temp.push(i); 
    acc[p] = temp; 
    return acc; 
}, {}) 
} 

所以当我做var groupedByPhase = groupBy('Phase', initial); 我得到groupedByPhase

{ 
"Phase 1" : [ 
{ Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" }, 
{ Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" }, 
{ Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" }, 
{ Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" }, 
], 
"Phase 2": [ 
{ Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" }, 
{ Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" }, 
{ Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" }, 
{ Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" } 
] 
} 

我也通过步骤使用此功能管理到组:

function groupByInNestedObj(item, obj) { 
var x = {}; 
for (var i in obj) { 
    x[i] = groupBy(item, obj[i]); 
} 
return x; 
} 

使得到调用groupByInNestedObj('Step', groupBy('Phase', initial))

{ 
"Phase 1" : { 
    "Step 1": [ 
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" }, 
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" }], 
    "Step 2": [ 
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" }, 
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" }, 
    ]}, 
"Phase 2": { 
    "Step 1:" [ 
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" }, 
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" }], 
    "Step 2:"[ 
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" }, 
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" } 
    ]} 
} 

但是我有点st在这里,做下一个。 理想情况下,我想能够做到groupBy("Task", groupBy("Step", groupBy("Phase", initial))),让groupBy组在树的最深层次。 任何建议欢迎!

注:我曾尝试的第二步来实现此功能

function groupByInNestedObj2 (item, obj) { 
    var x = {}; 
    for(var i in obj) { 
     for (var j in obj[i]) { 
      x[i][j] = groupBy(item, obj[i][j]); 
     } 
    } 
    return x; 
} 

,但它似乎并没有工作。

注2:以前的功能工作的第二个版本,但不是纯粹的,因为它修改通过它

function groupByInNestedObj2 (item, obj) { 
    var x = {}; 
    for(var i in obj) { 
     x[i] = obj[i] 
     for (var j in x[i]) { 
      x[i][j] = groupBy('Task', x[i][j]); 
     } 
    } 
    return x; 
} 

所以当我做var groupByPhaseAndStepAndTask = groupByInNestedObj2('Task', groupByPhaseAndStep) groupByPhaseAndStep被修改过传递的对象,这是不期望副作用。仍在努力。

回答

0

一个更通用的方法:

var initial = [ 
 
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" }, 
 
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" }, 
 
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" }, 
 
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" }, 
 
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" }, 
 
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" }, 
 
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" }, 
 
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" } 
 
]; 
 

 
_.mixin({ 
 
    groupByMulti: function(obj, values, context) { 
 
    if (!values.length) return obj; 
 
    var byFirst = _.groupBy(obj, _.head(values), context); 
 
    for (var prop in byFirst) { 
 
     byFirst[prop] = _.groupByMulti(byFirst[prop], _.tail(values), context); 
 
    } 
 
    return byFirst; 
 
    } 
 
}); 
 

 
var tree = _(initial).groupByMulti(['Phase', 'Step', 'Task']); 
 
$('#pre').append(JSON.stringify(tree, null, 3));
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.0.0/lodash.min.js"></script> 
 
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 
 
<pre id='pre'></pre>

+0

它非常接近理想的结果,在任务内部是对象而不是数组,但是这样可以。虽然我用lodash做了一些事情,但我对_.mixin不太熟悉,所以答案中有太多的魔力。此外,我的目标是实现功能的组合,所以我可以做一些类似于groupBy(“Task”,groupBy(“Step”,groupBy(“Phase”,initial)))' – Bondifrench

+0

好吧我设法通过重构你的例子,使用简单的head,tail和groupBy函数,我喜欢使用递归,现在只需要修改它,所以当array.length == 1时,我得到的对象并不是一个具有单个元素的数组 – Bondifrench

1

您可以尝试使用.reduce,只是检查对象存在与否,这样

var initial = [ 
 
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" }, 
 
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" }, 
 
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" }, 
 
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" }, 
 
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" }, 
 
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" }, 
 
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" }, 
 
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" } 
 
]; 
 

 
var result = initial.reduce(function (prev, current) { 
 
    prev[current.Phase] = prev[current.Phase] || {}; 
 
    prev[current.Phase][current.Step] = prev[current.Phase][current.Step] || {}; 
 
    prev[current.Phase][current.Step][current.Task] = current; 
 
    return prev; 
 
}, {}); 
 

 
console.log(JSON.stringify(result, null, 2));

1

我们可以用一个做这种for循环:

var initial = [ 
 
{ Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" }, 
 
{ Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" }, 
 
{ Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" }, 
 
{ Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" }, 
 
{ Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" }, 
 
{ Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" }, 
 
{ Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" }, 
 
{ Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" } 
 
]; 
 
var final = {} 
 
initial.forEach(function(d){ 
 
    if (!final[d.Phase]) //phase not present so make an object 
 
    \t final[d.Phase] = {}; 
 
    if (!final[d.Phase][d.Step]) //step not present so make an object 
 
    \t final[d.Phase][d.Step] = {}; 
 
    if (!final[d.Phase][d.Step][d.Task])//task not present so make an object and store the object 
 
    \t final[d.Phase][d.Step][d.Task] = d; 
 
    
 
}) 
 
console.log(final)

使用lodash这
+0

虽然它给出正确的结果,我想实现,对结果的顶部,是组成见我的'groupBy(“Task”,groupBy(“Step”,groupBy(“Phase”,initial)))'comment – Bondifrench