字符序列到分层 JSON - d3.sunburst

Question

我得到了许多不同的字符模式的数组，请参阅下面的data，我正在尝试将数据嵌套到分层 JSON 表单中，以便插入到旭日形可视化中。每个模式都由n 字符组成，尽管下面是 8 个字符。讨论想要的结果：

1. 在级别 0，找出有多少个唯一字符。答案：['w', 'm']。
2. 获取第一个键“w”并找到位于第 1 级的所有唯一字符，然后转到第 2 级，依此类推，直到到达模式的末尾，我们在该模式的末尾计算该唯一模式的大小。
3. 对第二个键“m”重复 #2
4. 将结果插入到root 的children 属性中。

我能够为两级案例构建代码，并且随后可以通过大量嵌套进行更深层次的嵌套，但它会非常硬编码并且超级难以理解。有谁知道我可以递归地执行此操作或使用另一种可以解决它的模式？

所需的输出样本：

var root = {"name":"evt_seq","children":[{"name":"w","children":[{"name":"w","size":8},{"name":"k","size":1}]},{"name":"m","children":[{"name":"w","size":1}]}]}

两级嵌套（寻找n级嵌套）

"use strict";
var data = [{"match": ["w", "w", "l", "w", "w", "w", "t", "w"]}, {"match": ["w", "k", "w", "A", "w", "w", "w", "w"]},
{"match": ["w", "w", "w", "w", "w", "w", "w", "w"]}, {"match": ["w", "w", "w", "w", "w", "w", "w", "w"]},
{"match": ["w", "w", "w", "w", "w", "w", "w", "w"]}, {"match": ["m", "w", "v", "v", "t", "m", "l", "m"]},
{"match": ["w", "w", "w", "l", "w", "w", "l", "l"]}, {"match": ["w", "w", "z", "w", "w", "m", "l", "w"]},
{"match": ["w", "w", "w", "w", "w", "w", "w", "w"]}, {"match": ["w", "w", "m", "w", "l", "w", "w", "w"]}
];

var root = {
    "name": "evt_seq", children: []
};
// Get initial pattern
var groupedXs = _.groupBy(data, function (d) {
    return d.match[0];
});
_.forEach(_.keys(groupedXs), function (d) {
    let _x = groupedXs[d];
    let _groupedXs = _.groupBy(_x, function (f) {
        return f.match[1];
    });
    let _children = _.map(_.keys(_groupedXs), function (f) {
        return {'name': f, 'size': _groupedXs[f].length}
    });
    root.children.push({"name": d, children: _children});
});
console.log(JSON.stringify(root));

Answer 1

这绝对是用递归算法最好的解决方案。由于语法似乎不太重要，我只给出简短的伪代码：

function subTree(inputs) {
    var children = []
    var leadingLetters = //map first letter from each array, remove duplicates
    for leadingLetter in leadingLetters {
        var matchingInputs = //filter inputs that match the first letter
        var reducedInputs = //copy of matchingInputs, but the first (matching) element is removed from each array
        children.push(subTree(reducedInputs))
    }
    return children
}
var root = subTree(data);

我没有将空子数组视为 null，因为我不完全确定当您到达字符串末尾时会发生什么。无论哪种方式，这应该让你上路。 （PS 希望这不是家庭作业！=X）

Answer 2

@Acey - 很好的指导！有关完整工作的实施，请参见下文。再次感谢，这是一个个人项目，而不是家庭作业:)

function subTree(inputs) {
    var children = [];
    var leadingLetters = _.uniq(_.map(inputs, function (d) {
        return d.match[0];
    })); //map first letter from each array, remove duplicates
    _.forEach(leadingLetters, function (leadingLetter) {
            var matchingInputs = _.filter(inputs, function (d) {
                return d.match[0] == leadingLetter;
            }); //filter inputs that match the first letter
            var reducedInputs = _.map(matchingInputs, function (d) {
                return {"match": d.match.slice(1, d.match.length)};
            }); //copy of matchingInputs, but the first (matching) element is removed from each array
            if (!reducedInputs[0].match.length) {
                children.push({name: leadingLetter, size: reducedInputs.length});
            }
            else {
                children.push({"name": leadingLetter, children: subTree(reducedInputs)});
            }
        }
    );
    return children
}

var root = subTree(data);