Javascript 基于递增键合并对象数组

Question

这就是我所拥有的：我想合并 object 哪个键以“path-”+i开头。 并在最终结果中从键中删除“path-i”。

  var arr = [
  {
    "key": "path-0-mp4",
    "value": [
      "media/video/01.mp4",
      "media/video/01_hd.mp4"
    ]
  },
  {
    "key": "path-0-quality",
    "value": [
      "720p",
      "1080p"
    ]
  },
  {
    "key": "path-1-mp4",
    "value": [
      "media/video/02.mp4",
      "media/video/02_hd.mp4"
    ]
  },
  {
    "key": "path-1-quality",
    "value": [
      "SD",
      "HD"
    ]
  }

]

这是期望的结果：

 var arr = [
        [
              {
                "mp4": "media/video/01.mp4",
                "quality": "720p"
              },
              {
                "mp4": "media/video/01_hd.mp4",
                "quality": "1080p"
              },
        ],
        [
              {
                "mp4": "media/video/02.mp4",
                "quality": "SD"
              },
              {
                "mp4": "media/video/02_hd.mp4",
                "quality": "HD"
              },
        ],
]

我开始做某事，但还没有接近：

var key, new_key, value,j=0, z=0, parr = [], obj;
for(var i = 0;i<a.length;i++){
    console.log('item:' ,a[i])
    key = a[i].key, value = a[i].value
    if(key.indexOf('path-'+j.toString()) > -1){
        new_key = key.substr(key.lastIndexOf('-')+1)
        console.log(key, new_key, value)
        for(var z = 0;z<value.length;z++){
            parr.push({[new_key]: value[z] })
        }
    }
}

console.log(parr)

[
  {
    "mp4": "media/video/01.mp4"
  },
  {
    "mp4": "media/video/01_hd.mp4"
  },
  {
    "quality": "720p"
  },
  {
    "quality": "1080p"
  }
]

---

编辑：

Array 可以保留需要以相同方式分组的不同键，例如：

    var arr = [
  {
    "key": "path-0-mp4",
    "value": [
      "media/video/01.mp4",
      "media/video/01_hd.mp4"
    ]
  },
  {
    "key": "path-0-quality",
    "value": [
      "720p",
      "1080p"
    ]
  },
  {
    "key": "path-1-mp4",
    "value": [
      "media/video/02.mp4",
      "media/video/02_hd.mp4"
    ]
  },
  {
    "key": "path-1-quality",
    "value": [
      "SD",
      "HD"
    ]
  },
  {
    "key": "subtitle-0-label",
    "value": [
      "English",
      "German",
      "Spanish"
    ]
  },
  {
    "key": "subtitle-0-src",
    "value": [
      "data/subtitles/sintel-en.vtt",
      "data/subtitles/sintel-de.vtt",
      "data/subtitles/sintel-es.vtt"
    ]
  },
  {
    "key": "subtitle-1-label",
    "value": [
      "German",
      "Spanish"
    ]
  },
  {
    "key": "subtitle-1-src",
    "value": [
      "data/subtitles/tumblr-de.vtt",
      "data/subtitles/tumblr-es.vtt"
    ]
  }
]

这是期望的结果（为每个不同的键创建新数组）：

 var arr = [
        [
              {
                "mp4": "media/video/01.mp4",
                "quality": "720p"
              },
              {
                "mp4": "media/video/01_hd.mp4",
                "quality": "1080p"
              },
        ],
        [
              {
                "mp4": "media/video/02.mp4",
                "quality": "SD"
              },
              {
                "mp4": "media/video/02_hd.mp4",
                "quality": "HD"
              },
        ],
],
arr2 = [        
    [
        {
            "label": "English",
            "src": "data/subtitles/sintel-en.vtt",
        },
        {
            "label": "German",
            "src": "data/subtitles/sintel-de.vtt"
        },
        {
            "label": "Spanish",
            "src": "data/subtitles/sintel-es.vtt"
        }
    ],
    [
        {
            "label": "Spanish",
            "src": "data/subtitles/tumblr-es.vtt",
        },
        {
            "label": "German",
            "src": "data/subtitles/tumblr-de.vtt"
        }
    ]
 ]

Answer 1

您可以拆分key属性，省略第一个path并将 rest 作为索引和键。 然后创建一个新数组，如果不存在并分配值。

 var data = [{ key: "path-0-mp4", value: ["media/video/01.mp4", "media/video/01_hd.mp4"] }, { key: "path-0-quality", value: ["720p", "1080p"] }, { key: "path-1-mp4", value: ["media/video/02.mp4", "media/video/02_hd.mp4"] }, { key: "path-1-quality", value: ["SD", "HD"] }], result = data.reduce((r, { key, value }) => { let [, i, k] = key.split('-'); r[i] = r[i] || []; value.forEach((v, j) => (r[i][j] = r[i][j] || {})[k] = v); return r; }, []); console.log(result);

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如果您想按key的第一部分进行分组，您可以将 object 与该组作为 key 并按上述方式分配 rest。

 var data = [{ key: "path-0-mp4", value: ["media/video/01.mp4", "media/video/01_hd.mp4"] }, { key: "path-0-quality", value: ["720p", "1080p"] }, { key: "path-1-mp4", value: ["media/video/02.mp4", "media/video/02_hd.mp4"] }, { key: "path-1-quality", value: ["SD", "HD"] }, { key: "subtitle-0-label", value: ["English", "German", "Spanish"] }, { key: "subtitle-0-src", value: ["data/subtitles/sintel-en.vtt", "data/subtitles/sintel-de.vtt", "data/subtitles/sintel-es.vtt"] }, { key: "subtitle-1-label", value: ["German", "Spanish"] }, { key: "subtitle-1-src", value: ["data/subtitles/tumblr-de.vtt", "data/subtitles/tumblr-es.vtt"] }], result = data.reduce((r, { key, value }) => { let [group, i, k] = key.split('-'); if (;r[group]) r[group] = []; if (.r[group][i]) r[group][i] = [], value;forEach((v; j) => { if (;r[group][i][j]) r[group][i][j] = {}; r[group][i][j][k] = v, }); return r. }; {}); console.log(result);

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Answer 2

我是新手和初学者，这是正确的方法吗？

 const a = [{ "key": "path-0-mp4", "value": [ "media/video/01.mp4", "media/video/01_hd.mp4" ] }, { "key": "path-0-quality", "value": [ "720p", "1080p" ] }, { "key": "path-1-mp4", "value": [ "media/video/02.mp4", "media/video/02_hd.mp4" ] }, { "key": "path-1-quality", "value": [ "SD", "HD" ] } ]; var resp = []; for (let i = 0; i < a.length; i++) { var inst = a[i]; var key = inst["key"]; for (let j = 0; j < inst.value.length; j++) { var index = key.split("-")[1]; var keyinst = key.split("-")[2]; if (;resp[index]) { resp[index] = []; } if (.resp[index][j]) { resp[index][j] = {}; } resp[index][j][keyinst] = inst.value[j]; } } console.log(resp);

Answer 3

我发现这更容易阅读和掌握

如果您使用reduce，您可以保存作业

 const arr = [{ "key": "path-0-mp4", "value": [ "media/video/01.mp4", "media/video/01_hd.mp4" ] }, { "key": "path-0-quality", "value": [ "720p", "1080p" ] }, { "key": "path-1-mp4", "value": [ "media/video/02.mp4", "media/video/02_hd.mp4" ] }, { "key": "path-1-quality", "value": [ "SD", "HD" ] } ]; newArr = []; arr.filter(item => item.key.endsWith("mp4")).forEach(item => item.value.forEach((val, i) => newArr.push({ "mp4": val, "quality": arr.find(qItem => qItem.key === item.key.replace("mp4", "quality")).value[i]} ) ) ) console.log(newArr)

这是 Nina 的未混淆版本

 var data = [{ key: "path-0-mp4", value: ["media/video/01.mp4", "media/video/01_hd.mp4"] }, { key: "path-0-quality", value: ["720p", "1080p"] }, { key: "path-1-mp4", value: ["media/video/02.mp4", "media/video/02_hd.mp4"] }, { key: "path-1-quality", value: ["SD", "HD"] }], result = data.reduce((resultArray, { key, value }) => { let [, idx, suffix] = key.split('-'); resultArray[idx] = resultArray[idx] || []; value.forEach((val, i) => (resultArray[idx][i] = resultArray[idx][i] || {})[suffix] = val); return resultArray; }, []); console.log(result);

Answer 4

我在这里做的唯一奇怪的事情是使用 object 作为查找表来帮助提高速度复杂性。 如果您有任何问题，请告诉我。

 const arr = [{ "key": "path-0-mp4", "value": [ "media/video/01.mp4", "media/video/01_hd.mp4" ] }, { "key": "path-0-quality", "value": [ "720p", "1080p" ] }, { "key": "path-1-mp4", "value": [ "media/video/02.mp4", "media/video/02_hd.mp4" ] }, { "key": "path-1-quality", "value": [ "SD", "HD" ] } ]; const result = arr.reduce((table, item) => { // Getting "path-1" from "path-1-quality" const pathValues = item.key.split('-'); const pathValue = pathValues[0] + '-' + pathValues[1]; // Getting "quality" from "path-1-quality" const key = pathValues[2]; // Get Index from table if already registered paths let tIndex = table.indexLookup[pathValue]; // If there is no registered index register one if (tIndex === undefined) { // reassign index to new location tIndex = table.result.length; // register the index table.indexLookup[pathValue] = tIndex; table.result.push([]); } // Assign values item.value.forEach((value, i) => { const arr = table.result[tIndex] || []; arr[i] = arr[i] || {} arr[i][key] = value; table.result[tIndex] = arr; }) return table }, { indexLookup: {}, result: [] }).result console.log(result)