根据另一个数组值对对象数组进行排序 Javascript

Question

我有两个数组： 第一个数组：

SelectedRows= 
[
   {
      "NR":"4",
      "KUNNR":"9?AMT132",
      "NAME":" AUTO TOURING HANDELS GES.M.B.H.",
      "LATLON":[
         15.4114217,
         47.0664085
      ],
      "LIN":"LIN097"
   },
   {
      "NR":"3",
      "KUNNR":"9Z?CH005",
      "NAME":" Z?CHNER ERDBAU U TRANSPORT GMBH",
      "LATLON":[
         13.4216536,
         47.3747925
      ],
      "LIN":"LIN099"
   },
   {
      "NR":"2",
      "KUNNR":"9SMTA001",
      "NAME":" SMT AUTOTEILE",
      "LATLON":[
         13.2277803,
         47.9525892
      ],
      "LIN":"LIN0102"
   },
   {
      "NR":"1",
      "KUNNR":"9REIT051",
      "NAME":" W.REITINGER GMBH",
      "LATLON":[
         14.4017044,
         48.2213305
      ],
      "LIN":"LIN0103"
   }
]

第二个数组是

LOCATIONS =
[
   [
      15.4114217,
      47.0664085
   ],
   [
      14.4017044,
      48.2213305
   ],
   [
      13.2277803,
      47.9525892
   ],
   [
      13.4216536,
      47.3747925
   ]
]

我需要根据第二个数组 (LOCATIONS)SelectedRows.LATLON 的匹配值对第一个数组 (SelectedRows) 进行排序

我试过：

var Output= LOCATIONS.filter(function(obj) { 
            return SelectedRows.LATLON.indexOf(obj) == -1; 
        });

但是没有用。因此，如果有人可以给我建议，我将不胜感激。抱歉我的英语不好，为了让我知道我发布了预期的输出：

预期输出：

Output=[
{
    "NR":"4",
    "KUNNR":"9?AMT132",
    "NAME":" AUTO TOURING HANDELS GES.M.B.H.",
    "LATLON":[
        15.4114217,
        47.0664085
    ],
    "LIN":"LIN097"
},
{
    "NR":"1",
    "KUNNR":"9REIT051",
    "NAME":" W.REITINGER GMBH",
    "LATLON":[
        14.4017044,
        48.2213305
    ],
    "LIN":"LIN0103"
},
{
    "NR":"2",
    "KUNNR":"9SMTA001",
    "NAME":" SMT AUTOTEILE",
    "LATLON":[
        13.2277803,
        47.9525892
    ],
    "LIN":"LIN0102"
},
{
    "NR":"3",
    "KUNNR":"9Z?CH005",
    "NAME":" Z?CHNER ERDBAU U TRANSPORT GMBH",
    "LATLON":[
        13.4216536,
        47.3747925
    ],
    "LIN":"LIN099"
}]

感谢您的帮助

Answer 1

如果您在 selectedRows 和 locations 中都有 n 个元素，那么在 O(n) 时间内执行此操作的最快方法（至少渐进）如下。

遍历正确顺序的位置并将位置值映射到它们（应该）所在的索引。您不能为此使用数组，因为无法使用 Map 内部使用的 Object.is() 来比较数组。所以你可以在这里使用一个连接的字符串。可能有更好/更清洁的选择，但这很简单并且可以完成工作。这需要 O(n).

地图将如下所示：

Map(4) {
  '15.4114217,47.0664085' => 0,
  '14.4017044,48.2213305' => 1,
  '13.2277803,47.9525892' => 2,
  '13.4216536,47.3747925' => 3
}

遍历 selectedRows 并为每个对象确定它的键，这又是连接的位置，并使用该键获取对象在 O(1) 中应该位于的索引。如果当前索引和目标索引匹配，一切都很好。如果不是，我们需要将对象交换到目标索引，这显然是在 O(1) 中完成的。这总共也需要 O(n).

完成。结合这些步骤将花费 O(n) 时间，因此它将是一个渐近最优算法。

const selectedRows = 
[
   {
      "NR":"4",
      "KUNNR":"9?AMT132",
      "NAME":" AUTO TOURING HANDELS GES.M.B.H.",
      "LATLON":[
         15.4114217,
         47.0664085
      ],
      "LIN":"LIN097"
   },
   {
      "NR":"3",
      "KUNNR":"9Z?CH005",
      "NAME":" Z?CHNER ERDBAU U TRANSPORT GMBH",
      "LATLON":[
         13.4216536,
         47.3747925
      ],
      "LIN":"LIN099"
   },
   {
      "NR":"2",
      "KUNNR":"9SMTA001",
      "NAME":" SMT AUTOTEILE",
      "LATLON":[
         13.2277803,
         47.9525892
      ],
      "LIN":"LIN0102"
   },
   {
      "NR":"1",
      "KUNNR":"9REIT051",
      "NAME":" W.REITINGER GMBH",
      "LATLON":[
         14.4017044,
         48.2213305
      ],
      "LIN":"LIN0103"
   }
]

const locations =
[
   [
      15.4114217,
      47.0664085
   ],
   [
      14.4017044,
      48.2213305
   ],
   [
      13.2277803,
      47.9525892
   ],
   [
      13.4216536,
      47.3747925
   ]
]

console.log("Previous", selectedRows);

const indexMap = new Map();

// Map values to indices
locations.forEach((loc, idx) => indexMap.set(`${loc[0]},${loc[1]}`, idx));

// put each object to the correct position
selectedRows.forEach((row , i) => {
    // concatenate lookup key
    const key = `${row.LATLON[0]},${row.LATLON[1]}`;
    // check if value is actually in map => should always be the case (assumption: locations and showLocation contain the same locations)
    if(indexMap.has(key)){
        const targetIndex = indexMap.get(key);
        // check if the object is at the right position
        if(!targetIndex === i){
            // position is wrong => swap positions
            const temp = selectedRows[i];
            selectedRows[i] = selectedRows[targetIndex];
            selectedRows[targetIndex] = temp;
        }
    }
})

console.log("After", selectedRows);

Note: you would need to make sure that the two variables locations and selectedRows actually contain the same locations and are of equal length.

Answer 2

你应该做的是 map LOCATIONS

在地图中 find SelectedRows 中的适当项目 ...

请注意 [1,2] !== [1,2] 因此您无法比较查找中的数组，因为您永远找不到任何东西 - 比较数组的每个元素，或者，在这种情况下，您可以检查是否array1.join() === array2.join()

const SelectedRows = [ { "NR":"4", "KUNNR":"9?AMT132", "NAME":" AUTO TOURING HANDELS GES.M.B.H.", "LATLON":[ 15.4114217, 47.0664085 ], "LIN":"LIN097" }, { "NR":"3", "KUNNR":"9Z?CH005", "NAME":" Z?CHNER ERDBAU U TRANSPORT GMBH", "LATLON":[ 13.4216536, 47.3747925 ], "LIN":"LIN099" }, { "NR":"2", "KUNNR":"9SMTA001", "NAME":" SMT AUTOTEILE", "LATLON":[ 13.2277803, 47.9525892 ], "LIN":"LIN0102" }, { "NR":"1", "KUNNR":"9REIT051", "NAME":" W.REITINGER GMBH", "LATLON":[ 14.4017044, 48.2213305 ], "LIN":"LIN0103" } ], LOCATIONS = [ [ 15.4114217, 47.0664085 ], [ 14.4017044, 48.2213305 ], [ 13.2277803, 47.9525892 ], [ 13.4216536, 47.3747925 ] ] 

const output =  LOCATIONS.map(
  a=>SelectedRows.find(({LATLON}) => LATLON.join() === a.join())
);

console.log(output);

.as-console-wrapper {max-height: 100%!important; top:0; }
.as-console-row::after { display:none !important; }

虽然，对于真正可读的代码

const SelectedRows = [ { "NR":"4", "KUNNR":"9?AMT132", "NAME":" AUTO TOURING HANDELS GES.M.B.H.", "LATLON":[ 15.4114217, 47.0664085 ], "LIN":"LIN097" }, { "NR":"3", "KUNNR":"9Z?CH005", "NAME":" Z?CHNER ERDBAU U TRANSPORT GMBH", "LATLON":[ 13.4216536, 47.3747925 ], "LIN":"LIN099" }, { "NR":"2", "KUNNR":"9SMTA001", "NAME":" SMT AUTOTEILE", "LATLON":[ 13.2277803, 47.9525892 ], "LIN":"LIN0102" }, { "NR":"1", "KUNNR":"9REIT051", "NAME":" W.REITINGER GMBH", "LATLON":[ 14.4017044, 48.2213305 ], "LIN":"LIN0103" } ], LOCATIONS = [ [ 15.4114217, 47.0664085 ], [ 14.4017044, 48.2213305 ], [ 13.2277803, 47.9525892 ], [ 13.4216536, 47.3747925 ] ] 

const output = LOCATIONS.map(([tgtLat, tgtLon]) =>
  SelectedRows.find(
    ({ LATLON: [lat, lon] }) => tgtLat === lat && tgtLon === lon
  )
);

console.log(output);

.as-console-wrapper {max-height: 100%!important; top:0; }
.as-console-row::after { display:none !important; }

Answer 3

以下代码有点脏，但它有效：

LOCATIONS_ORDER = LOCATIONS.map(e => `${e[0]};${e[1]}`);
SelectedRows.forEach(e => e.index = LOCATIONS_ORDER.indexOf(`${e.LATLON[0]};${e.LATLON[1]}`));

SelectedRows.sort((a,b) => a.index-b.index);

console.log(SelectedRows);

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