如何在javascript数组中插入缺失的日期和数据点？

假设我有以下几点：

var sales = [5, 4, 2];
var months = ["Jan 2011", "Apr 2011", "Feb 2012"];

如果我有一个给定的范围，可以说：

var range = ["Jan 2011", "Mar 2012"];

我想在两个月之间进行“插值”，以便得到以下结果：

var sales = [5, 4, 4, 4, ..., 2, 2];
var months = ["Jan 2011", "Feb 2011", "Mar 2011", ...., "Feb 2012", "Mar 2012"];

是否有必要创建一个“按顺序排列的所有可能日期范围”数组，其中包含所有希望包含的“月+年”？或者javascript有可能不必这样做吗？ 如何做到这一点

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注意：假设月份数组中不存在日期，则该日期将填充上一个可用月份的数据。

如果不严格需要使用上一个已知的销售数字，则可以对销售数字使用一种很好的样条插值：

/* Fritsch-Carlson monotone cubic spline interpolation
   Usage example:
    var f = createInterpolant([0, 1, 2, 3], [0, 1, 4, 9]);
    var message = '';
    for (var x = 0; x <= 3; x += 0.5) {
        var xSquared = f(x);
        message += x + ' squared is about ' + xSquared + '\n';
    }
    alert(message);
*/
var createInterpolant = function(xs, ys) {
    var i, length = xs.length;

    // Deal with length issues
    if (length != ys.length) { throw 'Need an equal count of xs and ys.'; }
    if (length === 0) { return function(x) { return 0; }; }
    if (length === 1) {
        // Impl: Precomputing the result prevents problems if ys is mutated later and allows garbage collection of ys
        // Impl: Unary plus properly converts values to numbers
        var result = +ys[0];
        return function(x) { return result; };
    }

    // Rearrange xs and ys so that xs is sorted
    var indexes = [];
    for (i = 0; i < length; i++) { indexes.push(i); }
    indexes.sort(function(a, b) { return xs[a] < xs[b] ? -1 : 1; });
    var oldXs = xs, oldYs = ys;
    // Impl: Creating new arrays also prevents problems if the input arrays are mutated later
    xs = []; ys = [];
    // Impl: Unary plus properly converts values to numbers
    for (i = 0; i < length; i++) { xs.push(+oldXs[indexes[i]]); ys.push(+oldYs[indexes[i]]); }

    // Get consecutive differences and slopes
    var dys = [], dxs = [], ms = [];
    for (i = 0; i < length - 1; i++) {
        var dx = xs[i + 1] - xs[i], dy = ys[i + 1] - ys[i];
        dxs.push(dx); dys.push(dy); ms.push(dy/dx);
    }

    // Get degree-1 coefficients
    var c1s = [ms[0]];
    for (i = 0; i < dxs.length - 1; i++) {
        var m = ms[i], mNext = ms[i + 1];
        if (m*mNext <= 0) {
            c1s.push(0);
        } else {
            var dx = dxs[i], dxNext = dxs[i + 1], common = dx + dxNext;
            c1s.push(3*common/((common + dxNext)/m + (common + dx)/mNext));
        }
    }
    c1s.push(ms[ms.length - 1]);

    // Get degree-2 and degree-3 coefficients
    var c2s = [], c3s = [];
    for (i = 0; i < c1s.length - 1; i++) {
        var c1 = c1s[i], m = ms[i], invDx = 1/dxs[i], common = c1 + c1s[i + 1] - m - m;
        c2s.push((m - c1 - common)*invDx); c3s.push(common*invDx*invDx);
    }

    // Return interpolant function
    return function(x) {
        // The rightmost point in the dataset should give an exact result
        var i = xs.length - 1;
        if (x == xs[i]) { return ys[i]; }

        // Search for the interval x is in, returning the corresponding y if x is one of the original xs
        var low = 0, mid, high = c3s.length - 1;
        while (low <= high) {
            mid = Math.floor(0.5*(low + high));
            var xHere = xs[mid];
            if (xHere < x) { low = mid + 1; }
            else if (xHere > x) { high = mid - 1; }
            else { return ys[mid]; }
        }
        i = Math.max(0, high);

        // Interpolate
        var diff = x - xs[i], diffSq = diff*diff;
        return ys[i] + c1s[i]*diff + c2s[i]*diffSq + c3s[i]*diff*diffSq;
    };
};

---

如果某个日期不存在于

months

数组中，是否假设该日期的销售值与上个月相同？是的，假设某个日期不存在于months数组中，则该日期将由上个可用月份的数据填充。是否在

months

数组中使用实际日期对象？如果是这样（或者如果可以使用日期），那么您应该能够迭代月数组，跟踪索引，并将日期增加1个月，直到达到下一个索引的值，并使用上一个索引的“输入”销售值填充“结果”销售数组。请参阅在Javascript中为日期添加月份，它们可以是实际的日期对象，以生成存储在这些数组中的字符串。如果可能的话，我建议使用实际的日期对象而不是字符串，因为日期对象公开了执行日期算术的方法（我认为这基本上就是您在这里需要的）；在字符串上做同样的操作显然更难。如果最终结果必须是字符串表示，那么我建议在完成插值后（即在表示层）转换为字符串

var monthNames = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec'];
var fromMonthNumber = function(monthNumber) {
    return monthNames[monthNumber % 12] + ' ' + ((monthNumber / 12) | 0);
};

var toMonthNumber = function(monthName) {
    var date = new Date(Date.parse(monthName));
    return 12*date.getFullYear() + date.getMonth();
};

var interpolateSales = function(sales, months, range) {
    var f = createInterpolant(months.map(toMonthNumber), sales);

    var resultSales = [], resultMonths = [];
    var firstMonth = toMonthNumber(range[0]), lastMonth = toMonthNumber(range[1]);
    for (var x = firstMonth; x <= lastMonth; x++) {
        resultSales.push(Math.round(f(x)));
        resultMonths.push(fromMonthNumber(x));
    }

    return { sales: resultSales, months: resultMonths };
};

{
    sales: [5, 5, 4, 4, 4, 4, 3, 3, 3, 3, 3, 2, 2, 2, 2],
    months: ["Jan 2011", "Feb 2011", "Mar 2011", "Apr 2011", "May 2011", "Jun 2011", "Jul 2011", "Aug 2011", "Sep 2011", "Oct 2011", "Nov 2011", "Dec 2011", "Jan 2012", "Feb 2012", "Mar 2012"]
}