D3 JSON数据转换

alt*_*lta 8 javascript json d3.js

我有这个JSON数据结构:

[
    { "dep": "d1", "name": "name1", "size": "size1" },
    { "dep": "d1", "name": "name2", "size": "size2" },
    { "dep": "d2", "name": "name1", "size": "size3" },
    { "dep": "d2", "name": "name1", "size": "size4" }
]
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我想将它转换为嵌套结构,如下所示:

{
    "name": "root",
    "children": [
        { "name": "d1",
            "children": [
                { "dep": "d1", "name": "name1", "size": "size1" },
                { "dep": "d1", "name": "name2", "size": "size2" }
            ]
        },
        { "name": "d2",
            "children": [
                { "dep": "d2", "name": "name1", "size": "size3" },
                { "dep": "d2", "name": "name2", "size": "size4" }
            ]
        }
    ]
}
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......并进一步用它来制作Reingold-Tilford树.任何人都可以指出我正确的方向,我对D3很新!

Chr*_*che 9

策略是创建一个与您想要的对应的新的空数据结构,然后通过遍历整个原始数据集来填充它.这是代码:

var data = [
    { "dep": "d1", "name": "name1", "size": "size1" },
    { "dep": "d1", "name": "name2", "size": "size2" },
    { "dep": "d2", "name": "name1", "size": "size3" },
    { "dep": "d2", "name": "name2", "size": "size4" }
]

var newData = {"name": "root", "children": {}}

data.forEach(function (d) {
    if (typeof newData.children[d.dep] !== 'undefined') {
        newData.children[d.dep].children.push(d)
    } else {
        newData.children[d.dep] = {"name": d.dep, "children": [d]}
    }
})
newData.children = Object.keys(newData.children).map(function (key) {
    return newData.children[key];
});
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最后的赋值是将对象转换为数组.

这给出了所需的结果newData:

{
    "name": "root",
    "children": [
        { "name": "d1",
            "children": [
                { "dep": "d1", "name": "name1", "size": "size1" },
                { "dep": "d1", "name": "name2", "size": "size2" }
            ]
        },
        { "name": "d2",
            "children": [
                { "dep": "d2", "name": "name1", "size": "size3" },
                { "dep": "d2", "name": "name2", "size": "size4" }
            ]
        }
    ]
}
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jsFiddle: http ://jsfiddle.net/chrisJamesC/eB4jF/

注意:此方法不适用于嵌套结构.对嵌套结构执行此操作会更加困难,但您始终可以使用递归函数.


编辑:正如@imarane在他的回答中所建议的,你可以使用d3.nest(),它比我的手工解决方案更好.你可能会接受他的回答.通过玩它,甚至很容易有多层嵌套:

var data = [
    { "dep": "d1", "name": "name1", "size": "size1" },
    { "dep": "d1", "name": "name2", "size": "size2" },
    { "dep": "d2", "name": "name1"},
    { "dep": "d2"}
]

var newData = {
    "key":"root", 
    "children": 
        d3.nest()
            .key(function(d){return d.dep})
            .key(function(d){return d.name})
            .key(function(d){return d.size})
            .entries(data)
}     
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哪个给:

{"key":"root","children":[
    {"key":"d1","values":[
        {"key":"name2","values":[
            {"dep":"d1","name":"name2","size":"size1"},
            {"dep":"d1","name":"name2","size":"size2"}
        ]}
    ]},
    {"key":"d2","values":[
        {"key":"name1","values":[
            {"dep":"d2","name":"name1"}
        ]},
        {"key":"undefined","values":[
            {"dep":"d2"}
        ]}
    ]}
]}
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其中以下数据结构(希望更好地了解整点):

var data = [
    { "dep": "d1", "name": "name2", "size": "size1" },
    { "dep": "d1", "name": "name2", "size": "size2" },
    { "dep": "d2", "name": "name1"},
    { "dep": "d2"}
]
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JsFiddle: http ://jsfiddle.net/chrisJamesC/eB4jF/2/

更多关于Nest: http ://bl.ocks.org/phoebebright/raw/3176159/