确定时间序列中的开始日期

Question

我希望你过得很好。 我有一个理论上的时间序列分析问题希望你能帮我解决。

首先，您将在下面找到我的数据集的可重现示例。 日期在每天的时间范围内。 Q25是我的未转换数据的第 25 或下四分位数， Q75是我的未转换数据的第 75 或上四分位数，拟合是中位数的局部加权拟合，一阶是拟合的一阶导数，二阶是二阶导数适合。

绘制拟合图会产生两次振荡，然后数据会稳定增加。 绘制周围的四分位数会产生大量数据，这些数据会随着拟合数据的增加而缩小。 一阶导数显示拟合的变化率，这就是我的问题所在。我不确定基于一阶导数数据的拟合数据的增加从哪里开始。 从逻辑上讲，我知道信噪比开始日期必须在 3 月 7 日之后（基于四分位数）和 3 月 20 日之前（在拟合数据稳步增加之前）。 这也表示在大约相同时间间隔的一阶导数中，其中负转正拐点在 3 月 5 日发生变化，在 3 月 16 日变为正，然后产生一个平稳的时间序列。

综上所述，我的确切开始日期应该是一阶导数的拐点变化，还是 3 月 16 日的第一个正值？

感谢您花时间解决这个问题以及您可能有的任何想法！

data<-structure(list(Date = structure(c(1485950474, 1486036874, 1486123274, 
1486209674, 1486296074, 1486382474, 1486468874, 1486555274, 1486641674, 
1486728074, 1486814474, 1486900874, 1486987274, 1487073674, 1487160074, 
1487246474, 1487332874, 1487419274, 1487505674, 1487592074, 1487678474, 
1487764874, 1487851274, 1487937674, 1488024074, 1488110474, 1488196874, 
1488283274, 1488369674, 1488456074, 1488542474, 1488628874, 1488715274, 
1488801674, 1488888074, 1488974474, 1489060874, 1489147274, 1489233674, 
1489320074, 1489406474, 1489492874, 1489579274, 1489665674, 1489752074, 
1489838474, 1489924874, 1490011274, 1490097674, 1490184074, 1490270474, 
1490356874, 1490443274, 1490529674, 1490616074, 1490702474, 1490788874, 
1490875274, 1490961674, 1491048074, 1491134474, 1491220874, 1491307274, 
1491393674, 1491480074, 1491566474, 1491652874, 1491739274, 1491825674, 
1491912074, 1491998474, 1492084874, 1492171274, 1492257674, 1492344074, 
1492430474, 1492516874, 1492603274, 1492689674, 1492776074, 1492862474, 
1492948874, 1493035274, 1493121674, 1493208074, 1493294474, 1493380874, 
1493467274, 1493553674, 1493640074, 1493726474, 1493812874, 1493899274, 
1493985674, 1494072074, 1494158474, 1494244874, 1494331274, 1494417674, 
1494504074, 1494590474, 1494676874, 1494763274, 1494849674, 1494936074, 
1495022474, 1495108874, 1495195274, 1495281674, 1495368074), tzone = "UTC", class = c("POSIXct", 
"POSIXt")), Q25 = c(-1.61495132528742, -3.86616056128065, -3.92140420424278, 
-4.8011229557052, -8.64427034627082, -3.11323607034871, -4.3673083843457, 
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0.826258942687981, 3.36833418895383, -6.8991417494414, -1.15773470862185, 
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2.28752278841533, -0.275687673398348, 5.74597173218469, 6.5773422259343, 
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6.77215849921842, 9.53718510298556, 4.16562173164636, 10.4573226478082, 
7.703077796612, 7.55811710979136, 4.47194951592662, 10.2104312432178, 
11.3454383477984, 0.997649090931488, 4.84898050707927, 10.8819209584302, 
8.06296236341084, 11.3317616787558, 7.51878628894305, 7.87729934765305, 
11.9108509727303, 6.77401202490232, 5.36297357453455, 10.6362047038983, 
8.68979831512869, 4.0465996534104, 11.9579904470733, 9.41141176380086, 
10.5754750604254, 12.6944336852953, 7.61563466861022), Q75 = c(5.93775779359077, 
26.4536084846094, 7.92690107568623, 16.195405687679, 3.47567054091916, 
34.9690262666155, 15.5126126583077, 24.4425589002446, 29.7425859431597, 
23.1420118192775, 26.827758017105, 18.6306368759596, 19.759179203689, 
10.0667740183259, 30.9080218485755, 10.0628623899296, 21.1120424008512, 
12.1232187464341, 14.9571040303508, 11.4927011052638, 16.1617172813173, 
19.0606972964125, 8.39991659547325, 9.5080530252195, 10.2717546026802, 
12.018391863395, 27.2666992661895, 12.5172584337237, 19.9658806224003, 
6.90019918091751, 18.4119063276997, 23.2991253786256, 27.95161418973, 
16.9477966472485, 26.3880458021082, 19.2178725103802, 5.58699033890406, 
9.82525729279156, 6.22139350667344, 5.6625294221828, 8.18283315939774, 
4.78856479855966, 4.91215612536983, 5.35278870440784, 15.7471499356884, 
7.95473965312171, 7.58463611165082, 6.03119890210746, 9.88624343762245, 
6.66377352843609, 6.92675024060609, 7.20403099201013, 6.96877369392089, 
17.7034248870798, 6.22890341708267, 6.1624397247754, 23.3856864094132, 
7.13518162203812, 6.96344109315883, 7.69414570220079, 18.0859103957135, 
7.52300478408242, 10.1635801549871, 10.021556657451, 8.51746254314866, 
7.83000625461296, 15.4938419153615, 8.6844260972191, 8.07596479745038, 
13.1423674521087, 8.04161364299224, 10.7442773622841, 8.58410892324644, 
9.08436532340561, 8.84748510783176, 9.27529549461203, 9.01978932806698, 
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10.7107229417799, 11.452541129334, 11.9951421202043, 11.3568792509498, 
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12.8124554609003, 12.1817389611984, 13.4529860098547, 13.1808997426024, 
12.568956945967, 13.9405958892683, 14.4445923505263, 14.5816203429081, 
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15.5928587109464, 15.5111381098579, 15.7167488979248, 16.4121827249844, 
16.7700564366026), fit = c(1.3157822724014, 1.44491806546299, 
1.67963756121542, 1.96834398237369, 2.32222986513481, 2.73223146146706, 
3.16143742264514, 3.74278329406317, 4.4673163398484, 5.08529278937518, 
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7.43724432723552, 7.55279324817994, 7.67877181101032, 7.76924002146674, 
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11.6317299424136, 11.8381584436134, 12.0667779318613, 12.2724056764894, 
12.462010561811, 12.6517333832877, 12.8101492769744, 12.9055352762602, 
12.9678598772259, 13.0582354099638, 13.1489397497677, 13.2204738414797, 
13.346284619515, 13.6054940294766, 13.9436193637562, 14.2337005769519, 
14.5449448398809, 14.8895799498019, 15.0551768009747, 15.0689572800127
), firstder = c(0.0542499277820437, 0.193160412687084, 0.264645386746196, 
0.318230646770668, 0.390583391620104, 0.410606699200811, 0.484714112557398, 
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))

Answer 1

如果问题是找到拟合列开始上升的位置，则拟合一条由水平线段和斜线段（图中的红色）组成的曲线，并报告变化点（Date0 和图中的虚线）。

# calculate starting values, st
fm0 <- lm(fit ~ Date, data, subset = seq(to = nrow(data), length = 20))
st <- c(mean(data$fit[1:20]), coef(fm0))
names(st) <- c("a0", "a", "b")

fm <- nls(fit ~ pmax(a0, a + b * as.numeric(Date)), data, start = st)

# solve a0 = a + b * Date0 for Date0 using calculated a0, a and b
Date0 <- with(as.list(coef(fm)), .POSIXct((a0 - a)/b))

plot(fit ~ Date, data, ylab = "")
lines(fitted(fm) ~ Date, data, col = "red")
abline(v = Date0, lty = 2)

Date0
## [1] "2017-03-21 07:53:56 EDT"