Question

我发现很难用标题来描述我遇到的问题，但是我希望举个例子会有所帮助。我有一个对象的javascript数组，看起来像这样：

const myPlayer = "jim"
const myArray = 
    [
      {"player":"bob", "points":11},
      {"player":"joe", "points":15},
      {"player":"nik", "points":18},
      {"player":"tom", "points":21},
      {"player":"jim", "points":25},
      {"player":"ron", "points":26},
      {"player":"tim", "points":32},
      {"player":"jon", "points":35},
      {"player":"len", "points":42},
      {"player":"eva", "points":51}
    ];

我的数组按键要点对对象进行排序，我想过滤该数组以保留（a） player 键与 myPlayer匹配的对象变量，以及（b）与 myPlayer 变量匹配的对象之前的2个对象和之后的2个对象。例如，在这种情况下，我想返回以下内容：

const filtArray = 
    [
      {"player":"nik", "points":18},
      {"player":"tom", "points":21},
      {"player":"jim", "points":25},
      {"player":"ron", "points":26},
      {"player":"tim", "points":32}
    ];

...因为nik和tom是jim之前的两个对象，而ron和tim是tim之后的两个对象。 player 键的每个对象都有唯一的值，因此永远不会有两个具有相同 player 键的对象。

这个问题的最后一部分-我希望过滤后的数组始终包含5个对象。因此，如果myPlayer = "bob"（其中bob是第一个对象中的 player ），我希望它简单地返回前5个对象，如下所示：

 [
  {"player":"bob", "points":11},
  {"player":"joe", "points":15},
  {"player":"nik", "points":18},
  {"player":"tom", "points":21},
  {"player":"jim", "points":25}
];

...如果myPlayer =“ joe”，或者myPlayer =“ len”或“ eva”，则输出类似。

对此表示感谢，谢谢！！

编辑：看起来我只是在代码中有一个小错误，这破坏了我的方法...具有以下内容，它使用findIndex和slice：

const index = myArray.findIndex(x => x.player === myPlayer);    
filtArray = myArray.slice(index-2, index+2)

Answer 1

您可以为第一部分做一个for loop，然后为第二部分做slice()，但是正如您所说的，如果选择eva会显示前5个，如果您想要其他内容，可以改变切片。

const myPlayer = "ron";
const myArray = [
  { player: "bob", points: 11 },
  { player: "joe", points: 15 },
  { player: "nik", points: 18 },
  { player: "tom", points: 21 },
  { player: "jim", points: 25 },
  { player: "ron", points: 26 },
  { player: "tim", points: 32 },
  { player: "jon", points: 35 },
  { player: "len", points: 42 },
  { player: "eva", points: 51 }
];

var newArray = [];

for (let i = 0; i < myArray.length; i++) {
  if (myArray[i].player == myPlayer){
      var index = i;//<-- store the index to probably be used in slice
    if ((i-2)>=0 && (i+2)<myArray.length) {
    newArray.push(myArray[i - 2]);
    newArray.push(myArray[i - 1]);
    newArray.push(myArray[i]);
    newArray.push(myArray[i + 1]);
    newArray.push(myArray[i + 2]);
  }
}
}

if(newArray.length == 0){
    newArray = myArray.slice(0, 6) 
}

console.log(newArray);

Answer 2

您可以通过找到要寻找的球员的索引，然后遍历数组并创建索引在该范围内的新球员列表来实现。

const myPlayer = "jim"
const myArray = [
    {"player":"bob", "points":11},
    {"player":"joe", "points":15},
    {"player":"nik", "points":18},
    {"player":"tom", "points":21},
    {"player":"jim", "points":25},
    {"player":"ron", "points":26},
    {"player":"tim", "points":32},
    {"player":"jon", "points":35},
    {"player":"len", "points":42},
    {"player":"eva", "points":51}
]

function indexOfPlayer(arr, player) {
    return arr.map(i => i.player).indexOf(player)
}

function getItemWithSurrounding(arr, index, around) {
    result = []
    for (i = index - around; i <= index + around; i++) {
        if (i >= 0 && i < arr.length) {
            result.push(arr[i])
        }
    }
    return result
}
playerIndex = indexOfPlayer(myArray, myPlayer)
players = getItemWithSurrounding(myArray, playerIndex, 2)
console.log(players)

输出：

[
  {
    "player": "nik",
    "points": 18
  },
  {
    "player": "tom",
    "points": 21
  },
  {
    "player": "jim",
    "points": 25
  },
  {
    "player": "ron",
    "points": 26
  },
  {
    "player": "tim",
    "points": 32
  }
]

Answer 3

您可以使用Array.findIndex方法在数组中找到正确的播放器，然后将Array.slice与两个if语句一起使用以选择数组的正确段（如果是选择第一名或第二名，如果选择了最后两名，则选择最后五名，或者选择“内部”玩家中的五名。

const myPlayer = "jim"
const myArray = 
    [
      {"player":"bob", "points":11},
      {"player":"joe", "points":15},
      {"player":"nik", "points":18},
      {"player":"tom", "points":21},
      {"player":"jim", "points":25},
      {"player":"ron", "points":26},
      {"player":"tim", "points":32},
      {"player":"jon", "points":35},
      {"player":"len", "points":42},
      {"player":"eva", "points":51}
    ];

function filterArray(array, playerName) {
  let idx = array.findIndex(o => o.player === playerName)
  if (array.length - idx <= 2) {
    return array.slice(array.length - 5)
  } else if (idx < 2) {
    return array.slice(0, 5)
  } else {
    return array.slice(idx - 2, idx + 3)
  }
}

console.log(filterArray(myArray, myPlayer)) // "jim" is an "inner" player, so the surrounding elements are returned
console.log(filterArray(myArray, "joe")) // "joe" is an "outer" player, so the first five elements are returned
console.log(filterArray(myArray, "eva")) // "eva" is an "outer" player, so the last five elements are returned

通过使对象具有匹配键以及相邻对象来过滤对象数组

3 个答案: