这是一个交叉验证问题的连续体,我在该问题中询问了该问题的合理方法。这个问题更面向编程,所以我把它贴在这里。
背景
我有一条已知日期跨越一年的曲线。该曲线的 y 值是根据每日温度和盐度记录计算的d18O 值的预测值。我还测量了由碳酸钙组成的外壳的 d18O 值。这些值是沿距离轴测量的,其中第一次和最后一次测量大约(但不完全)与曲线的开始和结束同时进行。
已知 d18O 值与曲线中的预测值在一些未知的随机误差内匹配。我想通过更改测量值的 x 轴(或至少通过将指数与曲线中的指数相匹配)来获得测量值与曲线的最佳拟合。通过这种方式,我可以获得测量值日期的估计值,并可以进一步估计壳在一年中的增长率。预计增长率是可变的,可能会出现增长中断(即增长停止)。但是,测量值之间的增长必须> 0(一个约束)。
示例数据
以下是示例数据集(curve
和meas
ured):
meas <- structure(list(index = 1:10, distance = c(0.1, 1, 3, 5, 7, 8,
13, 20, 22, 25), value = c(3.5, 4.2, 4.5, 4.4, 4.7, 4.8, 5.1,
4.9, 4.1, 3.7)), .Names = c("index", "distance", "value"), class = "data.frame",
row.names = c(NA, -10L))
curve <- structure(list(date = structure(c(15218, 15219, 15220, 15221,
15222, 15223, 15224, 15225, 15226, 15227, 15228, 15229, 15230,
15231, 15232, 15233, 15234, 15235, 15236, 15237, 15238, 15239,
15240, 15241, 15242, 15243, 15244, 15245, 15246, 15247, 15248,
15249, 15250, 15251, 15252, 15253, 15254, 15255, 15256, 15257,
15258, 15259, 15260, 15261, 15262, 15263, 15264, 15265, 15266,
15267, 15268, 15269, 15270, 15271, 15272, 15273, 15274, 15275,
15276, 15277, 15278, 15279, 15280, 15281, 15282, 15283, 15284,
15285, 15286, 15287, 15288, 15289, 15290, 15291, 15292, 15293,
15294, 15295, 15296, 15297, 15298, 15299, 15300, 15301, 15302,
15303, 15304, 15305, 15306, 15307, 15308, 15309, 15310, 15311,
15312, 15313, 15314, 15315, 15316, 15317, 15318, 15319, 15320,
15321, 15322, 15323, 15324, 15325, 15326, 15327, 15328, 15329,
15330, 15331, 15332, 15333, 15334, 15335, 15336, 15337, 15338,
15339, 15340, 15341, 15342, 15343, 15344, 15345, 15346, 15347,
15348, 15349, 15350, 15351, 15352, 15353, 15354, 15355, 15356,
15357, 15358, 15359, 15360, 15361, 15362, 15363, 15364, 15365,
15366, 15367, 15368, 15369, 15370, 15371, 15372, 15373, 15374,
15375, 15376, 15377, 15378, 15379, 15380, 15381, 15382, 15383,
15384, 15385, 15386, 15387, 15388, 15389, 15390, 15391, 15392,
15393, 15394, 15395, 15396, 15397, 15398, 15399, 15400, 15401,
15402, 15403, 15404, 15405, 15406, 15407, 15408, 15409, 15410,
15411, 15412, 15413, 15414, 15415, 15416, 15417, 15418, 15419,
15420, 15421, 15422, 15423, 15424, 15425, 15426, 15427, 15428,
15429, 15430, 15431, 15432, 15433, 15434, 15435, 15436, 15437,
15438, 15439, 15440, 15441, 15442, 15443, 15444, 15445, 15446,
15447, 15448, 15449, 15450, 15451, 15452, 15453, 15454, 15455,
15456, 15457, 15458, 15459, 15460, 15461, 15462, 15463, 15464,
15465, 15466, 15467, 15468, 15469, 15470, 15471, 15472, 15473,
15474, 15475, 15476, 15477, 15478, 15479, 15480, 15481, 15482,
15483, 15484, 15485, 15486, 15487, 15488, 15489, 15490, 15491,
15492, 15493, 15494, 15495, 15496, 15497, 15498, 15499, 15500,
15501, 15502, 15503, 15504, 15505, 15506, 15507, 15508, 15509,
15510, 15511, 15512, 15513, 15514, 15515, 15516, 15517, 15518,
15519, 15520, 15521, 15522, 15523, 15524, 15525, 15526, 15527,
15528, 15529, 15530, 15531, 15532, 15533, 15534, 15535, 15536,
15537, 15538, 15539, 15540, 15541, 15542, 15543, 15544, 15545,
15546, 15547, 15548, 15549, 15550, 15551, 15552, 15553, 15554,
15555, 15556, 15557, 15558, 15559, 15560, 15561, 15562, 15563,
15564, 15565, 15566, 15567, 15568, 15569, 15570, 15571, 15572,
15573, 15574, 15575, 15576, 15577, 15578, 15579, 15580, 15581,
15582, 15583, 15584), class = "Date"), index = 1:367, value = c(3.33,
3.35, 3.36, 3.38, 3.4, 3.42, 3.43, 3.45, 3.47, 3.48, 3.5, 3.52,
3.53, 3.55, 3.56, 3.58, 3.6, 3.61, 3.63, 3.64, 3.66, 3.67, 3.69,
3.7, 3.72, 3.73, 3.75, 3.76, 3.78, 3.79, 3.81, 3.82, 3.83, 3.85,
3.86, 3.88, 3.89, 3.9, 3.92, 3.93, 3.94, 3.96, 3.97, 3.98, 3.99,
4.01, 4.02, 4.03, 4.04, 4.06, 4.07, 4.08, 4.09, 4.1, 4.11, 4.13,
4.14, 4.15, 4.16, 4.17, 4.18, 4.19, 4.2, 4.21, 4.22, 4.23, 4.24,
4.25, 4.26, 4.27, 4.28, 4.28, 4.29, 4.3, 4.31, 4.32, 4.33, 4.33,
4.34, 4.35, 4.36, 4.36, 4.37, 4.38, 4.38, 4.39, 4.4, 4.41, 4.41,
4.42, 4.42, 4.43, 4.44, 4.44, 4.45, 4.45, 4.46, 4.46, 4.47, 4.47,
4.47, 4.48, 4.48, 4.49, 4.49, 4.49, 4.5, 4.5, 4.5, 4.51, 4.51,
4.51, 4.52, 4.52, 4.53, 4.53, 4.53, 4.54, 4.54, 4.54, 4.55, 4.55,
4.56, 4.57, 4.57, 4.58, 4.58, 4.59, 4.6, 4.61, 4.61, 4.62, 4.63,
4.64, 4.64, 4.65, 4.66, 4.67, 4.67, 4.68, 4.69, 4.7, 4.7, 4.71,
4.72, 4.72, 4.73, 4.74, 4.74, 4.75, 4.75, 4.75, 4.76, 4.76, 4.76,
4.76, 4.76, 4.76, 4.76, 4.76, 4.76, 4.75, 4.75, 4.75, 4.75, 4.74,
4.74, 4.73, 4.73, 4.73, 4.72, 4.72, 4.72, 4.71, 4.71, 4.71, 4.71,
4.7, 4.7, 4.7, 4.71, 4.71, 4.71, 4.71, 4.72, 4.72, 4.73, 4.74,
4.75, 4.75, 4.76, 4.78, 4.79, 4.8, 4.81, 4.82, 4.83, 4.84, 4.85,
4.86, 4.88, 4.89, 4.9, 4.91, 4.92, 4.92, 4.93, 4.94, 4.95, 4.95,
4.95, 4.96, 4.96, 4.96, 4.96, 4.96, 4.95, 4.95, 4.95, 4.94, 4.93,
4.92, 4.92, 4.91, 4.9, 4.89, 4.88, 4.87, 4.86, 4.85, 4.84, 4.83,
4.82, 4.8, 4.79, 4.78, 4.77, 4.76, 4.75, 4.75, 4.74, 4.73, 4.72,
4.72, 4.71, 4.71, 4.71, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7, 4.7,
4.7, 4.7, 4.7, 4.7, 4.7, 4.69, 4.69, 4.69, 4.69, 4.69, 4.69,
4.69, 4.69, 4.68, 4.68, 4.68, 4.67, 4.67, 4.67, 4.66, 4.65, 4.65,
4.64, 4.63, 4.62, 4.61, 4.6, 4.59, 4.58, 4.57, 4.56, 4.55, 4.54,
4.53, 4.51, 4.5, 4.49, 4.48, 4.47, 4.46, 4.45, 4.43, 4.42, 4.41,
4.4, 4.39, 4.38, 4.37, 4.36, 4.35, 4.34, 4.33, 4.32, 4.32, 4.31,
4.3, 4.29, 4.28, 4.28, 4.27, 4.26, 4.25, 4.24, 4.24, 4.23, 4.22,
4.21, 4.21, 4.2, 4.19, 4.18, 4.17, 4.17, 4.16, 4.15, 4.14, 4.14,
4.13, 4.12, 4.12, 4.11, 4.1, 4.09, 4.08, 4.08, 4.07, 4.06, 4.05,
4.05, 4.04, 4.03, 4.02, 4.02, 4.01, 4, 4, 3.99, 3.98, 3.97, 3.97,
3.96, 3.95, 3.94, 3.94, 3.93, 3.92, 3.92, 3.91, 3.9, 3.9, 3.89,
3.88)), .Names = c("date", "index", "value"), row.names = c(NA,
-367L), class = "data.frame")
...这是它的样子:
library(ggplot2)
library(scales)
library(gridExtra)
p.curve <- ggplot() + geom_line(data = curve, aes(x = date, y = value)) + scale_x_date(name = "Month", breaks = date_breaks("months"), labels = date_format("%b")) + labs(title = "curve")
p.meas <- ggplot(meas, aes(x = distance, y = value)) + geom_point(color = "red") + labs(title = "measured", x = "Distance (mm)")
grid.arrange(p.curve, p.meas, ncol = 1)
实践中的问题
meas
我想curve
通过更改 x 轴来找到适合 R 的数学/统计方法meas
。此外,我想获得某种拟合统计数据来比较拟合的“x轴”(以防我运行多个具有不同约束的模型)。我将“x 轴模型”称为增长模型,因为它本质上就是这样。我想通过指定meas
值之间的距离必须大于 0 来约束拟合。即Meas
,值 withindex == 2
必须出现在值 with 之后index == 1
。我还希望能够约束增长率(即两个相邻索引点之间的最大距离)。为了证明这一点,我将手动进行:
ggplot() + geom_line(data = curve, aes(x = index, y = value)) + geom_line(data = meas, aes(x = index, y = value), color = "red", linetype = 2) + scale_x_continuous(breaks = seq(0,370,10)) + scale_y_continuous(breaks = seq(3,5,0.1))
首先,meas
(红色虚线)中的一些索引必须锚定到curve
(黑线)的索引。我选择锚定第一个和最后一个点加上具有最高值的点。
anchor <- data.frame(meas.index = c(1,7,10), curve.index = c(11,215,367))
example.fit <- merge(meas, anchor, by.x = "index", by.y = "meas.index", all = T, sort = F)
example.fit <- example.fit[with(example.fit, order(distance)),]
然后,我假设这些锚定点之间存在线性增长。增长将沿着curve
指数增长。Curve
每天有一个值。因此,增长将是每天的规模。
library(zoo)
example.fit$curve.index <- round(na.approx(example.fit$curve.index),0)
在此之后,我用日期替换索引并绘制结果。
library(plyr)
example.fit$date <- as.Date(mapvalues(example.fit$curve.index, from = curve$index, to = as.character(curve$date)))
a <- ggplot() + geom_line(data = curve, aes(x = date, y = value)) + geom_point(data = example.fit, aes(x = date, y = value), color = "red") + scale_x_date(limits = range(curve$date), name = "Month", breaks = date_breaks("months"), labels = date_format("%b"))
b <- ggplot(example.fit, aes(x = date, y = distance)) + geom_line() + scale_x_date(limits = range(curve$date), name = "Month", breaks = date_breaks("months"), labels = date_format("%b"))
grid.arrange(a,b)
上图显示了基于三个锚点的拟合结果。下图显示了每日间隔随时间的建模增长。三月初增长曲线的弯曲是一些我不明白的有趣的数学工件(由于包中的na.approx
函数zoo
)。
我试过什么
从我之前的问题中,我了解到动态时间扭曲可能是一种解决方案。我还找到了一个 R 包,其中包含 dtw 函数。好的。动态时间扭曲确实适用于我在该问题中的示例数据集(设置约束除外),但我无法让它适用于这个数据集,其中curve
的数据点比meas
(称为points
在上一个问题中)。我会尽量节省一些空间,不会在这里复制代码/数字。您可以看到我在回答该问题时所做的尝试。问题似乎是,除了最简单的一种之外,没有任何一种步进模式可以处理这些类型的数据。最简单的步进模式多次将测量值与曲线匹配,这是我想避免的,因为我需要为每个测量点定义日期。同样设置测量点之间的增长率必须> 0的约束似乎很困难。
问题
我的问题有两个:首先,有没有比动态时间扭曲更好的方法来解决这个问题?其次,我如何在 R 的实践中做到这一点?.
编辑9. 2013 年 12 月我试图让问题更清楚。