递归地深度压扁 JavaScript Object

Question

数据:

var data = [
    {
      "id": 1,
      "level": "1",
      "text": "Sammy",
      "type": "Item",
      "items": [
        {
          "id": 11,
          "level": "2",
          "text": "Table",
          "type": "Item",
          "items": [
            {
              "id": 111,
              "level": "3",
              "text": "Dog",
              "type": "Item",
              "items": null
            },
            {
              "id": 112,
              "level": "3",
              "text": "Cat",
              "type": "Item",
              "items": null
            }
          ]
        },
        {
          "id": 12,
          "level": "2",
          "text": "Chair",
          "type": "Item",
          "items": [
            {
              "id": 121,
              "level": "3",
              "text": "Dog",
              "type": "Item",
              "items": null
            },
            {
              "id": 122,
              "level": "3",
              "text": "Cat",
              "type": "Item",
              "items": null
            }
          ]
        }
      ]
    },
    {
      "id": 2,
      "level": "1",
      "text": "Sundy",
      "type": "Item",
      "items": [
        {
          "id": 21,
          "level": "2",
          "text": "MTable",
          "type": "Item",
          "items": [
            {
              "id": 211,
              "level": "3",
              "text": "MTDog",
              "type": "Item",
              "items": null
            },
            {
              "id": 212,
              "level": "3",
              "text": "MTCat",
              "type": "Item",
              "items": null
            }
          ]
        },
        {
          "id": 22,
          "level": "2",
          "text": "MChair",
          "type": "Item",
          "items": [
            {
              "id": 221,
              "level": "3",
              "text": "MCDog",
              "type": "Item",
              "items": null
            },
            {
              "id": 222,
              "level": "3",
              "text": "MCCat",
              "type": "Item",
              "items": null
            }
          ]
        }
      ]
    },
    {
      "id": 3,
      "level": "1",
      "text": "Bruce",
      "type": "Folder",
      "items": [
        {
          "id": 31,
          "level": "2",
          "text": "BTable",
          "type": "Item",
          "items": [
            {
              "id": 311,
              "level": "3",
              "text": "BTDog",
              "type": "Item",
              "items": null
            },
            {
              "id": 312,
              "level": "3",
              "text": "BTCat",
              "type": "Item",
              "items": null
            }
          ]
        },
        {
          "id": 32,
          "level": "2",
          "text": "Chair",
          "type": "Item",
          "items": [
            {
              "id": 321,
              "level": "3",
              "text": "BCDog",
              "type": "Item",
              "items": null
            },
            {
              "id": 322,
              "level": "3",
              "text": "BCCat",
              "type": "Item",
              "items": null
            }
          ]
        }
      ]
    }
  ];

代码:

var fdr = [];
var fd = function(n) {
  if (n.items) {
    _.forEach(n.items, function (value){
      fd(value);
    });
  }

  fdr.push(n);
};
_.forEach(data, fd);
console.log(fdr);

期望输出:

var data = [
    {
      "id": 1,
      "level": "1",
      "text": "Sammy",
      "type": "Item",
      "items": []
    },
    {
      "id": 11,
      "level": "2",
      "text": "Table",
      "type": "Item",
      "items": []
    },
    {
      "id": 111,
      "level": "3",
      "text": "Dog",
      "type": "Item",
      "items": null
    },
    {
      "id": 112,
      "level": "3",
      "text": "Cat",
      "type": "Item",
      "items": null
    },
    {
      "id": 12,
      "level": "2",
      "text": "Chair",
      "type": "Item",
      "items": []
    },
    {
      "id": 121,
      "level": "3",
      "text": "Dog",
      "type": "Item",
      "items": null
    },
    {
      "id": 122,
      "level": "3",
      "text": "Cat",
      "type": "Item",
      "items": null
    },
    {
      "id": 2,
      "level": "1",
      "text": "Sundy",
      "type": "Item",
      "items": []
    },
    {
      "id": 21,
      "level": "2",
      "text": "MTable",
      "type": "Item",
      "items": []
    },
    {
      "id": 211,
      "level": "3",
      "text": "MTDog",
      "type": "Item",
      "items": null
    },
    {
      "id": 212,
      "level": "3",
      "text": "MTCat",
      "type": "Item",
      "items": null
    },
    {
      "id": 22,
      "level": "2",
      "text": "MChair",
      "type": "Item",
      "items": []
    },
    {
      "id": 221,
      "level": "3",
      "text": "MCDog",
      "type": "Item",
      "items": null
    },
    {
      "id": 222,
      "level": "3",
      "text": "MCCat",
      "type": "Item",
      "items": null
    },
    {
      "id": 3,
      "level": "1",
      "text": "Bruce",
      "type": "Folder",
      "items": []
    },
    {
      "id": 31,
      "level": "2",
      "text": "BTable",
      "type": "Item",
      "items": []
    },
    {
      "id": 311,
      "level": "3",
      "text": "BTDog",
      "type": "Item",
      "items": null
    },
    {
      "id": 312,
      "level": "3",
      "text": "BTCat",
      "type": "Item",
      "items": null
    },
    {
      "id": 32,
      "level": "2",
      "text": "Chair",
      "type": "Item",
      "items": []
    },
    {
      "id": 321,
      "level": "3",
      "text": "BCDog",
      "type": "Item",
      "items": null
    },
    {
      "id": 322,
      "level": "3",
      "text": "BCCat",
      "type": "Item",
      "items": null
    }
  ];

条件:

Object 有未知级别。有些 child item 可能下降一级，有些可能下降 5 级。

问题

代码中的fd函数是我想出来的。我相信有一种 'cleaner' 方法可以做到这一点，只是想不出什么。另外，函数 return items object，将其渲染为循环 object.

JsBin: http://jsbin.com/debojiqove/2/edit?html,js,output

有没有办法用 lodash 或简单的 JavaScript 递归地展平 object？

Answer 1

这是使用递归函数的解决方案，我称之为 flattenNestedObjectsArray()（对于 native JavaScript）：

function flattenNestedObjectsArray(arr, part){
    var flattened = part || [], items;
    arr.forEach(function(v){
        if (Array.isArray(v.items) && v.items.length) {
            items = v.items;
            v.items = [];
            flattened.push(v);
            flattened.concat(flattened, flattenNestedObjectsArray(items, flattened));                
        } else {
            flattened.push(v);
        }        
    });
    return flattened;
}

var flattened = flattenNestedObjectsArray(data);
console.log(JSON.stringify(flattened, 0, 4));

console.log输出：

[
    {
        "id": 1,
        "level": "1",
        "text": "Sammy",
        "type": "Item",
        "items": []
    },
    {
        "id": 11,
        "level": "2",
        "text": "Table",
        "type": "Item",
        "items": []
    },
    {
        "id": 111,
        "level": "3",
        "text": "Dog",
        "type": "Item",
        "items": null
    },
    {
        "id": 112,
        "level": "3",
        "text": "Cat",
        "type": "Item",
        "items": null
    },
    {
        "id": 12,
        "level": "2",
        "text": "Chair",
        "type": "Item",
        "items": []
    },
    {
        "id": 121,
        "level": "3",
        "text": "Dog",
        "type": "Item",
        "items": null
    },
    {
        "id": 122,
        "level": "3",
        "text": "Cat",
        "type": "Item",
        "items": null
    },
    {
        "id": 2,
        "level": "1",
        "text": "Sundy",
        "type": "Item",
        "items": []
    },
    {
        "id": 21,
        "level": "2",
        "text": "MTable",
        "type": "Item",
        "items": []
    },
    {
        "id": 211,
        "level": "3",
        "text": "MTDog",
        "type": "Item",
        "items": null
    },
    {
        "id": 212,
        "level": "3",
        "text": "MTCat",
        "type": "Item",
        "items": null
    },
    {
        "id": 22,
        "level": "2",
        "text": "MChair",
        "type": "Item",
        "items": []
    },
    {
        "id": 221,
        "level": "3",
        "text": "MCDog",
        "type": "Item",
        "items": null
    },
    {
        "id": 222,
        "level": "3",
        "text": "MCCat",
        "type": "Item",
        "items": null
    },
    {
        "id": 3,
        "level": "1",
        "text": "Bruce",
        "type": "Folder",
        "items": []
    },
    {
        "id": 31,
        "level": "2",
        "text": "BTable",
        "type": "Item",
        "items": []
    },
    {
        "id": 311,
        "level": "3",
        "text": "BTDog",
        "type": "Item",
        "items": null
    },
    {
        "id": 312,
        "level": "3",
        "text": "BTCat",
        "type": "Item",
        "items": null
    },
    {
        "id": 32,
        "level": "2",
        "text": "Chair",
        "type": "Item",
        "items": []
    },
    {
        "id": 321,
        "level": "3",
        "text": "BCDog",
        "type": "Item",
        "items": null
    },
    {
        "id": 322,
        "level": "3",
        "text": "BCCat",
        "type": "Item",
        "items": null
    }
]

Answer 2

关于项目的简单 Javascript 解决方案。它不会改变源数组。

function flat(r, a) {
    var b = {};
    Object.keys(a).forEach(function (k) {
        if (k !== 'items') {
            b[k] = a[k];
        }
    });
    r.push(b);
    if (Array.isArray(a.items)) {
        b.items = a.items.map(function (a) { return a.id; });
        return a.items.reduce(flat, r);
    }
    return r;
}

var data = [{ "id": 1, "level": "1", "text": "Sammy", "type": "Item", "items": [{ "id": 11, "level": "2", "text": "Table", "type": "Item", "items": [{ "id": 111, "level": "3", "text": "Dog", "type": "Item", "items": null }, { "id": 112, "level": "3", "text": "Cat", "type": "Item", "items": null }] }, { "id": 12, "level": "2", "text": "Chair", "type": "Item", "items": [{ "id": 121, "level": "3", "text": "Dog", "type": "Item", "items": null }, { "id": 122, "level": "3", "text": "Cat", "type": "Item", "items": null }] }] }, { "id": 2, "level": "1", "text": "Sundy", "type": "Item", "items": [{ "id": 21, "level": "2", "text": "MTable", "type": "Item", "items": [{ "id": 211, "level": "3", "text": "MTDog", "type": "Item", "items": null }, { "id": 212, "level": "3", "text": "MTCat", "type": "Item", "items": null }] }, { "id": 22, "level": "2", "text": "MChair", "type": "Item", "items": [{ "id": 221, "level": "3", "text": "MCDog", "type": "Item", "items": null }, { "id": 222, "level": "3", "text": "MCCat", "type": "Item", "items": null }] }] }, { "id": 3, "level": "1", "text": "Bruce", "type": "Folder", "items": [{ "id": 31, "level": "2", "text": "BTable", "type": "Item", "items": [{ "id": 311, "level": "3", "text": "BTDog", "type": "Item", "items": null }, { "id": 312, "level": "3", "text": "BTCat", "type": "Item", "items": null }] }, { "id": 32, "level": "2", "text": "Chair", "type": "Item", "items": [{ "id": 321, "level": "3", "text": "BCDog", "type": "Item", "items": null }, { "id": 322, "level": "3", "text": "BCCat", "type": "Item", "items": null }] }] }];

document.write('<pre>' + JSON.stringify(data.reduce(flat, []), 0, 4) + '</pre>');

Answer 3

这是我的递归 flattenItems 函数版本。请注意，我已经删除了最终结果中所有级别的 items 属性。

function flattenItems(data) {
    // flat is the array that we will return by the end
    var flat = [];
    data.forEach(function(item) {
        // get child properties only
        var flatItem = {};
        Object.keys(item).forEach(function(key) {
            if(item[key] && item.hasOwnProperty(key) && !Array.isArray(item[key])) {
                flatItem[key] = item[key];
            }
            // recursive flattern on subitems
            // add recursive call results to the 
            // current stack version of "flat", by merging arrays
            else if(Array.isArray(item[key])) {
                Array.prototype.push.apply(flat, flattenItems(item[key]));
            }
        });
        flat.push(flatItem);
    });
    // sort by level before returning
    return flat.sort(function(i1, i2) {
        return parseInt(i1.level) - parseInt(i2.level);
    });
  }

这里是 fiddle 使用您的样本数据，检查控制台。

Answer 4

普通JavaScript

var data = [{ "id": 1, "level": "1", "text": "Sammy", "type": "Item", "items": [{ "id": 11, "level": "2", "text": "Table", "type": "Item", "items": [{ "id": 111, "level": "3", "text": "Dog", "type": "Item", "items": null }, { "id": 112, "level": "3", "text": "Cat", "type": "Item", "items": null }] }, { "id": 12, "level": "2", "text": "Chair", "type": "Item", "items": [{ "id": 121, "level": "3", "text": "Dog", "type": "Item", "items": null }, { "id": 122, "level": "3", "text": "Cat", "type": "Item", "items": null }] }] }, { "id": 2, "level": "1", "text": "Sundy", "type": "Item", "items": [{ "id": 21, "level": "2", "text": "MTable", "type": "Item", "items": [{ "id": 211, "level": "3", "text": "MTDog", "type": "Item", "items": null }, { "id": 212, "level": "3", "text": "MTCat", "type": "Item", "items": null }] }, { "id": 22, "level": "2", "text": "MChair", "type": "Item", "items": [{ "id": 221, "level": "3", "text": "MCDog", "type": "Item", "items": null }, { "id": 222, "level": "3", "text": "MCCat", "type": "Item", "items": null }] }] }, { "id": 3, "level": "1", "text": "Bruce", "type": "Folder", "items": [{ "id": 31, "level": "2", "text": "BTable", "type": "Item", "items": [{ "id": 311, "level": "3", "text": "BTDog", "type": "Item", "items": null }, { "id": 312, "level": "3", "text": "BTCat", "type": "Item", "items": null }] }, { "id": 32, "level": "2", "text": "Chair", "type": "Item", "items": [{ "id": 321, "level": "3", "text": "BCDog", "type": "Item", "items": null }, { "id": 322, "level": "3", "text": "BCCat", "type": "Item", "items": null }] }] }];

var r = [];

function flatten(a) {
    if (a.length == 0) return;
    var o = {};
    o.id = a[0].id;
    o.level = a[0].level;
    o.text = a[0].text;
    o.type = a[0].type
    o.items = a[0].items == null ? null : []
    r.push(o);
    if (Array.isArray(a[0].items)) {
        flatten(a[0].items);
    }
    a.shift();
    flatten(a);
}

flatten(data);

document.write('<pre>' + JSON.stringify(r, 0, 2) + '</pre>');

Answer 5

有点ES6的味道

function flatten(xs) {
  return xs.reduce((acc, x) => {
    acc = acc.concat(x);
    if (x.items) {
      acc = acc.concat(flatten(x.items));
      x.items = [];
    }
    return acc;
  }, []);
}

Answer 6

使用 reduce 和 recursion

的较短解决方案

function flatten(data){
  return data.reduce(function(result,next){
    result.push(next);
    if(next.items){
      result = result.concat(flatten(next.items));  
      next.items = [];
    }
    return result;
  },[]);
}

var data = [
    {
      "id": 1,
      "level": "1",
      "text": "Sammy",
      "type": "Item",
      "items": [
        {
          "id": 11,
          "level": "2",
          "text": "Table",
          "type": "Item",
          "items": [
            {
              "id": 111,
              "level": "3",
              "text": "Dog",
              "type": "Item",
              "items": null
            },
            {
              "id": 112,
              "level": "3",
              "text": "Cat",
              "type": "Item",
              "items": null
            }
          ]
        },
        {
          "id": 12,
          "level": "2",
          "text": "Chair",
          "type": "Item",
          "items": [
            {
              "id": 121,
              "level": "3",
              "text": "Dog",
              "type": "Item",
              "items": null
            },
            {
              "id": 122,
              "level": "3",
              "text": "Cat",
              "type": "Item",
              "items": null
            }
          ]
        }
      ]
    },
    {
      "id": 2,
      "level": "1",
      "text": "Sundy",
      "type": "Item",
      "items": [
        {
          "id": 21,
          "level": "2",
          "text": "MTable",
          "type": "Item",
          "items": [
            {
              "id": 211,
              "level": "3",
              "text": "MTDog",
              "type": "Item",
              "items": null
            },
            {
              "id": 212,
              "level": "3",
              "text": "MTCat",
              "type": "Item",
              "items": null
            }
          ]
        },
        {
          "id": 22,
          "level": "2",
          "text": "MChair",
          "type": "Item",
          "items": [
            {
              "id": 221,
              "level": "3",
              "text": "MCDog",
              "type": "Item",
              "items": null
            },
            {
              "id": 222,
              "level": "3",
              "text": "MCCat",
              "type": "Item",
              "items": null
            }
          ]
        }
      ]
    },
    {
      "id": 3,
      "level": "1",
      "text": "Bruce",
      "type": "Folder",
      "items": [
        {
          "id": 31,
          "level": "2",
          "text": "BTable",
          "type": "Item",
          "items": [
            {
              "id": 311,
              "level": "3",
              "text": "BTDog",
              "type": "Item",
              "items": null
            },
            {
              "id": 312,
              "level": "3",
              "text": "BTCat",
              "type": "Item",
              "items": null
            }
          ]
        },
        {
          "id": 32,
          "level": "2",
          "text": "Chair",
          "type": "Item",
          "items": [
            {
              "id": 321,
              "level": "3",
              "text": "BCDog",
              "type": "Item",
              "items": null
            },
            {
              "id": 322,
              "level": "3",
              "text": "BCCat",
              "type": "Item",
              "items": null
            }
          ]
        }
      ]
    }
  ];

var result = flatten(data);

document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');

Answer 7

只有一个线性函数可以完成这项工作。

var data = [{ "id": 1, "level": "1", "text": "Sammy", "type": "Item", "items": [{ "id": 11, "level": "2", "text": "Table", "type": "Item", "items": [{ "id": 111, "level": "3", "text": "Dog", "type": "Item", "items": null }, { "id": 112, "level": "3", "text": "Cat", "type": "Item", "items": null }] }, { "id": 12, "level": "2", "text": "Chair", "type": "Item", "items": [{ "id": 121, "level": "3", "text": "Dog", "type": "Item", "items": null }, { "id": 122, "level": "3", "text": "Cat", "type": "Item", "items": null }] }] }, { "id": 2, "level": "1", "text": "Sundy", "type": "Item", "items": [{ "id": 21, "level": "2", "text": "MTable", "type": "Item", "items": [{ "id": 211, "level": "3", "text": "MTDog", "type": "Item", "items": null }, { "id": 212, "level": "3", "text": "MTCat", "type": "Item", "items": null }] }, { "id": 22, "level": "2", "text": "MChair", "type": "Item", "items": [{ "id": 221, "level": "3", "text": "MCDog", "type": "Item", "items": null }, { "id": 222, "level": "3", "text": "MCCat", "type": "Item", "items": null }] }] }, { "id": 3, "level": "1", "text": "Bruce", "type": "Folder", "items": [{ "id": 31, "level": "2", "text": "BTable", "type": "Item", "items": [{ "id": 311, "level": "3", "text": "BTDog", "type": "Item", "items": null }, { "id": 312, "level": "3", "text": "BTCat", "type": "Item", "items": null }] }, { "id": 32, "level": "2", "text": "Chair", "type": "Item", "items": [{ "id": 321, "level": "3", "text": "BCDog", "type": "Item", "items": null }, { "id": 322, "level": "3", "text": "BCCat", "type": "Item", "items": null }] }] }],

flatIron = (a,b) => a.reduce((p,c) => {!!c.items ? (p.push(c), flatIron(c.items,p), c.items = []) : p.push(c); return p},b),
 flatArr = flatIron(data,[]);

document.write('<pre>' + JSON.stringify(flatArr, 0, 2) + '</pre>');

Answer 8

从 Lo-Dash 3.0.0 开始，_.flattenDeep(data) 将 return 一个你想要的深度扁平数组。还有一个 _.flatten(data) 函数可以浅层压平。

Answer 9

递归reducer函数的另一种方式

 _.reduce(data, function reducer(result, val) {
     var items = _.reduce(val.items, reducer, []);
     val.items = _.isArray(val.items) ? [] : val.items;
     return _.concat(result, val, items);
 }, []);

Answer 10

我需要做同样的事情，在解决我的问题时找到了使用 lodash 解决你的问题的方法：

function kids(node) {
    return node.items
        ? [{...node, items: []}, _.map(node.items, kids)]
        : {...node, items: null};
}

_.flatMapDeep(data, kids);

Answer 11

修改了 Роман Парадеев 答案，使其更加动态。

function flatten(xs, childSelector) {
  return xs.reduce((acc, x) => {
    acc = acc.concat(x);
    let children = childSelector(x);
    if (children) {
      acc = acc.concat(flatten(children, childSelector));
    }
    return acc;
  }, []);
}

现在 items 不是硬编码的，您可以使用它 flatten(data, x => x.items)。

Answer 12

使用 _.flatMapDeep（自 Lodash 4.7 起可用）：

var flatten = function(item) {
  return [item, _.flatMapDeep(item.items, flatten)];
}

var result = _.flatMapDeep(data, flatten);

Answer 13

此解决方案应该适用于 IE11；使用过滤器、映射和减少。

var item = function(x) {
    return {
        "id": x.id,
        "level": x.level,
        "text": x.text,
        "type": x.type,
        "items": x.items ? [] : null
    }
}

var flatten = function(a, b) {
    return a.concat(b);
};

var onlyUnique = function(acc, curr) {
    if (acc.length == 0) {
        acc.push(curr);
    } else {
        var search = acc.filter(function(x){return x.id===curr.id;})
        if (search.length == 0) {
            acc.push(curr);
        }
    }
    return acc;
}

var newData = data.map(function(x) {

    return x.items.map(function(xx) {
        return xx.items.map(function(xxx) {
            return [item(x), item(xx), item(xxx)];
        }).reduce(flatten, []);


    }).reduce(flatten, [])



}).reduce(flatten, []).reduce(onlyUnique, []);;

console.log(JSON.stringify(newData, null, 2))

Answer 14

由于这个较旧的问题已被重新提出，这里是一个现代版本。它不会改变（@Nina Scholz：或破坏！）原始数据，而是 returns 包含新对象的新数组。

const flattenItems = (xs) =>
  xs. flatMap (({items, ... node}) => [node, ... flattenItems (items || [])])

const data = [{id: 1, level: "1", text: "Sammy", type: "Item", items: [{id: 11, level: "2", text: "Table", type: "Item", items: [{id: 111, level: "3", text: "Dog", type: "Item", items: null}, {id: 112, level: "3", text: "Cat", type: "Item", items: null}]}, {id: 12, level: "2", text: "Chair", type: "Item", items: [{id: 121, level: "3", text: "Dog", type: "Item", items: null}, {id: 122, level: "3", text: "Cat", type: "Item", items: null}]}]}, {id: 2, level: "1", text: "Sundy", type: "Item", items: [{id: 21, level: "2", text: "MTable", type: "Item", items: [{id: 211, level: "3", text: "MTDog", type: "Item", items: null}, {id: 212, level: "3", text: "MTCat", type: "Item", items: null}]}, {id: 22, level: "2", text: "MChair", type: "Item", items: [{id: 221, level: "3", text: "MCDog", type: "Item", items: null}, {id: 222, level: "3", text: "MCCat", type: "Item", items: null}]}]}, {id: 3, level: "1", text: "Bruce", type: "Folder", items: [{id: 31, level: "2", text: "BTable", type: "Item", items: [{id: 311, level: "3", text: "BTDog", type: "Item", items: null}, {id: 312, level: "3", text: "BTCat", type: "Item", items: null}]}, {id: 32, level: "2", text: "Chair", type: "Item", items: [{id: 321, level: "3", text: "BCDog", type: "Item", items: null}, {id: 322, level: "3", text: "BCCat", type: "Item", items: null}]}]}]

console .log (flattenItems (data))

.as-console-wrapper {max-height: 100% !important; top: 0}

此版本不包含相当无用的 items 属性。添加它会很容易，像这样：

const flattenItems = (xs) =>
  xs. flatMap (({items, ... node}) => [
    {... node, items: items == null ? null : []}, 
    ... flattenItems (items || [])
  ])

Answer 15

简单的 map 和 reduce :

export function flatten<T>(value: T): T[] {
  const out: T[] = [];
  if (value.children) {
    const children = Object.values(value.children);
    out.push(
      value,
      ...children.map(flatten).reduce((acc, curVal) => {
        return acc.concat(curVal);
      }),
    );
  } else {
    out.push(value);
  }
  return out;
}

递归地深度压扁 JavaScript Object

Deep Flatten JavaScript Object Recursively

javascript

recursion

javascript-objects

lodash