Question

在GeoJSON文件中，某些属性由整个集合（数组）的所有“功能”（元素）共享。但是某些属性仅针对集合的子集定义。我发现了这个问题：[javascript] counting properties of the objects in an array of objects，但不能解决我的问题。

示例：

const features =
[ {"properties":{"name":"city1","zip":1234}, "geometry":{"type":"polygon","coordinates":[[1,2],[3,4] ...]}},
  {"properties":{"name":"city2","zip":1234}, "geometry":{"type":"polygon","coordinates":[[1,2],[3,4] ...]}},
  {"properties":{"name":"city3"},"geometry":{"type":"multiPolygon","coordinates":[[[1,2],[3,4] ...]]}},
// ... for instance 1000 different cities
  {"properties":{"name":"city1000","zip":1234,"updated":"May-2018"}, "geometry":{"type":"polygon","coordinates":[...]}}
];

预期结果：列出所有现有属性及其基数的信息，让我们知道数据集的完成情况。例如：

properties: 1000, properties.name: 1000, properties.zip: 890, properties.updated: 412,
geometry: 1000, geometry.type: 1000, geometry.coordinates: 1000

我有一个（相当复杂的）解决方案，但我确实怀疑有些人已经遇到了同样的问题（似乎是数据科学的经典之作），而其中的一个更好的（性能很重要）。

这是我笨拙的解决方案：

// 1: list all properties encountered in the features array, at least two levels deep
const countProps = af => af.reduce((pf,f) =>
                                            Array.from(new Set(pf.concat(Object.keys(f)))), []);
// adding all the properties of each individual feature, then removing duplicates using the array-set-array trick
const countProp2s = af => af.reduce((pf,f) =>
                                            Array.from(new Set(pf.concat(Object.keys(f.properties)))), []);
const countProp2g = af => af.reduce((pf,f) =>
                                            Array.from(new Set(pf.concat(Object.keys(f.geometry)))), []);

// 2: counting the number of defined occurrences of each property of the list 1
const countPerProp =  (ff) => pf => ` ${pf}:${ff.reduce((p,f)=> p+(!!f[pf]), 0)}`;
const countPerProp2s = (ff) => pf => ` ${pf}:${ff.reduce((p,f)=> p+(!!f.properties[pf]), 0)}`;
const countPerProp2g = (ff) => pf => ` ${pf}:${ff.reduce((p,f)=> p+(!!f.geometry[pf]), 0)}`;
const cardinalities = countProps(features).map((kk,i) => countPerProp(ff)(kk)) +
                      countProp2s(features).map(kk => countPerProp2s(ff)(kk)) +
                      countProp2g(features).map(kk => countPerProp2g(ff)(kk));

因此，存在三个问题：

-步骤1：对于一个相当简单的操作，这是很多工作（在删除大部分内容之前添加所有内容）。而且，这不是递归的，第二级是“手动强制”的。

步骤2，递归解决方案可能是更好的解决方案。

-是否可以在单个步骤中执行步骤1和2（开始计算添加新属性的时间）？

我欢迎任何想法。

Answer 1

JSON.parse reviver和JSON.stringify replacer可用于检查所有键值对：

var counts = {}, json = `[{"properties":{"name":"city1","zip":1234}, "geometry":{"type":"polygon","coordinates":[[1,2],[3,4]]}},{"properties":{"name":"city2","zip":1234}, "geometry":{"type":"polygon","coordinates":[[1,2],[3,4]]}},{"properties":{"name":"city3"},"geometry":{"type":"multiPolygon","coordinates":[[[1,2],[3,4]]]}},{"properties":{"name":"city1000","zip":1234,"updated":"May-2018"}, "geometry":{"type":"polygon","coordinates":[]}} ]`

var features = JSON.parse(json, (k, v) => (isNaN(k) && (counts[k] = counts[k] + 1 || 1), v))

console.log( counts, features )

Answer 2

考虑尝试以下方法。它只是一个减少，在forEach的内部有几个嵌套。它检查用于指示计数的键是否存在于要返回的对象中，如果不存在，则将其初始化为0。然后，无论这些键是否存在，其对应值都将递增1。

Repl在这里：https://repl.it/@dexygen/countobjpropoccur2levels，下面的代码：

const features =
[ {"properties":{"name":"city1","zip":1234}, "geometry":{"type":"polygon","coordinates":[[1,2],[3,4]]}},
  {"properties":{"name":"city2","zip":1234}, "geometry":{"type":"polygon","coordinates":[[1,2],[3,4]]}},
  {"properties":{"name":"city3"},"geometry":{"type":"multiPolygon","coordinates":[[[1,2],[3,4]]]}},
  {"properties":{"name":"city1000","zip":1234,"updated":"May-2018"}, "geometry":{"type":"polygon","coordinates":[]}}
];

const featuresCount = features.reduce((count, feature) => {
  Object.keys(feature).forEach(key => {
    count[key] = count[key] || 0;
    count[key] += 1;
    Object.keys(feature[key]).forEach(key2 => {
      let count2key = `${key}.${key2}`;
      count[count2key] = count[count2key] || 0;
      count[count2key] += 1;
    });
  });
  return count;
}, {});

console.log(featuresCount);

/*
{ properties: 4,
  'properties.name': 4,
  'properties.zip': 3,
  geometry: 4,
  'geometry.type': 4,
  'geometry.coordinates': 4,
  'properties.updated': 1 }
*/

Answer 3

使用杰克逊使用JSON的多态序列化。如下所示。您的基本界面将具有所有通用属性，并为每个变体创建子类型。依靠每种类型都会满足您的需求

@JsonTypeInfo（use = JsonTypeInfo.Id.NAME，include = JsonTypeInfo.As.PROPERTY，property =“ name”）@JsonSubTypes（{@ JsonSubTypes.Type（value = Lion.class，name =“ lion”）， @ JsonSubTypes.Type（value = Tiger.class，name =“ tiger”），}）公共接口Animal {}

给定对象数组，计算定义了多少个（可能不同）属性

3 个答案: