Genereate空数据bettwen时间戳日期范围| JavaScript的 [英] Genereate empty data bettwen the range of a timestamp date | Javascript

问题描述

我得到一些时间戳统计资料。我想它绘制在图表上。
数据增加固定步骤（1秒），但他们中的一些人失踪。
我想，以填补与自我生成的时间戳这一空白。

I get some statistic data with timestamps. I want to plot it on the chart. Data are increment with fixed step (1sec) but some of them are missing. I want to fill this gaps with self generated timestamp.

// exmaple data form 18:00 to 18:47 (step 60000ms)

var data = [
    [1425056400000, 1123], //18:00
    [1425056460000, 1124], //18:01
    [1425056520000, 1125], //18:02
    [1425056580000, 1126], //18:03
    [1425056640000, 1140], //18:04
    [1425057840000, 2123], //18:24
    [1425057900000, 2133], //18:25
    [1425057960000, 2141], //18:26
    [1425059160000, 5129], //18:46
    [1425059220000, 5129]  //18:47
];


// required result
var dataParsed = [
    [1425056400000, 1123], //18:00
    [1425056460000, 1124], //18:01
    [1425056520000, 1125], //18:02
    [1425056580000, 1126], //18:03
    [1425056640000, 1140], //18:04
    [1425056700000, 0], //18:05
    [1425056760000, 0], //18:06
    [1425056820000, 0], //18:07
    //(...)
    [1425057780000, 0], //18:23
    [1425057840000, 2123], //18:24
    [1425057900000, 2133], //18:25
    [1425057960000, 2141], //18:26
    //(...)
    [1425058800000, 0], //18:40
    //(...)
    [1425059160000, 5129], //18:46
    [1425059220000, 5129]  //18:47
];

我该怎么做用JavaScript？

How can I do this with JavaScript?

推荐答案

当你遍历原始数组，检查当前的元素是否是下一个的顺序排列。如果没有，用另一个循环来生成缺少的元素：

As you loop through the original array, check whether the current element is the next one in the sequence. If not, use another loop to generate the missing elements:

var dataParsed = [];
var lastTime = data[0][0];
var timeStep = 60000;
for (var i = 0; i < data.length; i++) {
    var curTime = data[i][0];
    if (curTime > lastTime + timeStep) {
        for (var time = lastTime + timeStep; time < curTime; time += timeStep) {
            dataParse.push([time, 0]);
        }
    }
    dataParse.push(data[i]);
    lastTime = curTime;
}

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