Javascript数组，解析每个元素将每个元素乘以'；x'；

如何通过数组进行解析，并将每个元素的值字段乘以“x”

我想通过一个包含60个元素的本机数组，如下所示：

[
     [value, key], [value, key], [value, key] 
]

在上面的示例中，键字段将是一个从60开始并倒计时到0的数字

值字段可以是任意数字

此数组从传递第一个值开始实时构建，最多可容纳60个值元素，每个值的键从60开始

每隔一秒，就会向数组传递一个新值，并取出最早的值。0被删除，60被添加，其他所有内容被1删除

每当这个更新发生而不是向数组中添加新值时，我还想解析整个数组，并将“value”字段乘以“x”例如，假设x等于1.2。

这意味着每次运行更新时，第一个值乘以1.2，第二个元素的值乘以1.2，第三个元素的值乘以1.2，依此类推

这意味着，当它到达最后一个元素0的“键”时，它将被乘以1.2 60倍

实际数据示例：

[[-30.4901691201296, 60], [-30.1833776214304, 59], [-29.7627840450973, 58], [-29.3947583209356, 57], [-28.9645892754055, 56], [-28.6354656536817, 55], [-28.2921821871286, 54], [-27.9905577131509, 53], [-27.7947946913668, 52], [-27.6543340290543, 51], [-27.6519828946371, 50], [-27.6173533427694, 49], [-27.5554196324518, 48], [-27.4347081962877, 47], [-27.3527616238956, 46], [-27.1500146810747, 45], [-26.9074687550566, 44], [-26.5520557024907, 43], [-26.3778269233317, 42], [-26.2025741177589, 41], [-25.9715337718657, 40], [-25.7909728444171, 39], [-25.6446160165696, 38], [-25.7040560541356, 37], [-25.8676731838619, 36], [-26.1857049460322, 35], [-26.5338463742982, 34], [-26.8991378853451, 33], [-27.0722352574209, 32], [-26.9933000067798, 31], [-26.5736545189266, 30], [-25.5369071865736, 29], [-24.0243166908922, 28], [-22.2063720253207, 27], [-20.4275559328569, 26], [-19.0900734751772, 25], [-17.9226541769101, 24], [-17.2615147887497, 23], [-16.8724851836878, 22], [-16.7577128443888, 21], [-17.1571347530026, 20], [-17.6471873975822, 19], [-18.6461197175468, 18], [-19.8885692353328, 17], [-21.2039894571651, 16], [-23.0079052401369, 15], [-25.1005655769037, 14], [-27.342591130044, 13], [-29.7388646710222, 12], [-32.1429579450835, 11], [-34.5906266190624, 10], [-37.0391503781189, 9], [-39.5315634976835, 8], [-41.9487262455882, 7], [-43.9289096382579, 6], [-46.00519229484, 5], [-48.6400387646249, 4], [-50.5736255352748, 3], [-52.8788720227602, 2], [-54.786535213712, 1]]

当前代码：

[[-30.4901691201296, 60], [-30.1833776214304, 59], [-29.7627840450973, 58], [-29.3947583209356, 57], [-28.9645892754055, 56], [-28.6354656536817, 55], [-28.2921821871286, 54], [-27.9905577131509, 53], [-27.7947946913668, 52], [-27.6543340290543, 51], [-27.6519828946371, 50], [-27.6173533427694, 49], [-27.5554196324518, 48], [-27.4347081962877, 47], [-27.3527616238956, 46], [-27.1500146810747, 45], [-26.9074687550566, 44], [-26.5520557024907, 43], [-26.3778269233317, 42], [-26.2025741177589, 41], [-25.9715337718657, 40], [-25.7909728444171, 39], [-25.6446160165696, 38], [-25.7040560541356, 37], [-25.8676731838619, 36], [-26.1857049460322, 35], [-26.5338463742982, 34], [-26.8991378853451, 33], [-27.0722352574209, 32], [-26.9933000067798, 31], [-26.5736545189266, 30], [-25.5369071865736, 29], [-24.0243166908922, 28], [-22.2063720253207, 27], [-20.4275559328569, 26], [-19.0900734751772, 25], [-17.9226541769101, 24], [-17.2615147887497, 23], [-16.8724851836878, 22], [-16.7577128443888, 21], [-17.1571347530026, 20], [-17.6471873975822, 19], [-18.6461197175468, 18], [-19.8885692353328, 17], [-21.2039894571651, 16], [-23.0079052401369, 15], [-25.1005655769037, 14], [-27.342591130044, 13], [-29.7388646710222, 12], [-32.1429579450835, 11], [-34.5906266190624, 10], [-37.0391503781189, 9], [-39.5315634976835, 8], [-41.9487262455882, 7], [-43.9289096382579, 6], [-46.00519229484, 5], [-48.6400387646249, 4], [-50.5736255352748, 3], [-52.8788720227602, 2], [-54.786535213712, 1]]

这是我当前用于构建阵列的用法：

setInterval(ExampleUsage, 1000);

function ExampleUsage() {
    $.getJSON(urlDefault, function (data) {
          RawDnD = data.DnD* 1;
          DnDData.unshift(RawDnD);
          DnDData = DnDData.slice(0, 60);
          DnDArray = $.map(DnDData, function (n, i) {
              return [[n, 60 - i]];
          });
          // Parse array here //
   });
}

---

我认为你不能按你的要求去做。如果正确读取代码，则每秒都在从JSON数据重建数组。之后循环遍历数组并将每个元素相乘是很简单的，但一秒钟后，您将使用新数据重建数组并丢失这些乘法

您可以通过执行以下操作来模拟效果：

var factor = 1.2;
var multiplier = factor;
for (i = 0; i < dataArray.length; i++) {
    dataArray[i] *= multiplier;
    multiplier *= factor;
}

var系数=1.2；
var乘数=因子；
对于（i=0；i

---

不过，我不确定这是否是你想要的。

你可以从这个想法开始

​$.each( arrayVariable ,function(i,v){
    console.log(v[0] + ' ---- ' + v[1] + '\n');
    // v[0] is the value
    // v[1] is the key 
});​

然后你可以做一些事情，比如

​$.each( arrayVariable ,function(i,v){
    v[1] -= 1; // decrease all keys by 1
    v[0] *= factor; // factor = 1.2;
    if (v[1] == -1) { // if key was 0 before, then 0-1 is -1
        v[1] = 60; // replace it with 60, the new one...
        v[0] = newData * factor; // new data inserted, multiplied by the factor.
    } 
});​

像这样,

setInterval(ExampleUsage, 1000);

function ExampleUsage() {
    $.getJSON(urlDefault, function (data) {
        RawDnD = data.DnD * 1;​
        // DnDData, is assumed to to have all values already
        $.each(DnDData, function (i, v) {
            v[1] -= 1; // decrease all keys by 1
            v[0] *= factor; // factor = 1.2;
            if (v[1] == -1) { // if key was 0 before, then 0-1 is -1
                v[1] = 60; // replace it with 60, the new one...
                v[0] = RawDnD * factor; // new data inserted, multiplied by the factor.
            }
        });​
    });
}

---

安排重要吗？我是说这是允许的吗<代码>[[-30.4901691201296,0]，-30.1833776214304,7]，-29.7627840450973,58]，…，只要数组包含所有60？排列就无所谓了。但默认情况下，它是从60向下排序到0的。Underline.js提供了此功能以及其他功能-请参阅：您重建数组的想法是正确的，而且效果非常好。而且非常简单。