python - 如何获得具有先验断点的分段线性回归？

Question

我需要详细解释这一点，因为我没有统计学的基础知识可以更简洁地解释。在 SO 中询问是因为我正在寻找 python 解决方案，但如果更合适的话可能会去 stats.SE。

我有井下的井数据，大概是这样的:

Rt      T
0.0000  15.0000
4.0054  15.4523
25.1858 16.0761
27.9998 16.2013
35.7259 16.5914
39.0769 16.8777
45.1805 17.3545
45.6717 17.3877
48.3419 17.5307
51.5661 17.7079
64.1578 18.4177
66.8280 18.5750
111.1613    19.8261
114.2518    19.9731
121.8681    20.4074
146.0591    21.2622
148.8134    21.4117
164.6219    22.1776
176.5220    23.4835
177.9578    23.6738
180.8773    23.9973
187.1846    24.4976
210.5131    25.7585
211.4830    26.0231
230.2598    28.5495
262.3549    30.8602
266.2318    31.3067
303.3181    37.3183
329.4067    39.2858
335.0262    39.4731
337.8323    39.6756
343.1142    39.9271
352.2322    40.6634
367.8386    42.3641
380.0900    43.9158
388.5412    44.1891
390.4162    44.3563
395.6409    44.5837

(Rt 变量可以被认为是深度的代理，T 是温度)。我也有“先验”数据给我 Rt=0 时的温度，还有一些标记，我可以用作断点、断点指南，或者至少与任何发现的断点进行比较。

这两个变量的线性关系在某些深度区间受某些过程的影响。一个简单的线性回归是

q, T0, r_value, p_value, std_err = stats.linregress(Rt, T)

看起来像这样，您可以清楚地看到偏差，以及 T0 的拟合不佳(应该是 15):

我希望能够执行一系列线性回归(在每个段的末端加入)，但我想这样做:(a) 不指定中断的数量或位置，(b) 通过指定中断的数量和位置，以及(c) 计算每段的系数

我想我可以做 (b) 和 (c)，只需将数据分开并稍微小心地分别做每一位，但我不知道 (a)，想知道是否有人有办法知道这可以更简单地完成。

我看过这个:https://stats.stackexchange.com/a/20210/9311 ，我认为 MARS 可能是处理它的好方法，但这只是因为它看起来不错；我真的不明白。我以盲目剪切'n'粘贴的方式用我的数据进行了尝试，输出如下，但我还是不明白:

Answer 1

简短的回答是，我使用 R 创建线性回归模型解决了我的问题，然后使用 segmented 包从线性模型生成分段线性回归。我能够使用 n 和 psi=NA 指定预期的断点(或结)K=n 数量，如下所示。

长答案是:

R 版本 3.0.1 (2013-05-16)
平台:x86_64-pc-linux-gnu(64 位)

# example data:
bullard <- structure(list(Rt = c(5.1861, 10.5266, 11.6688, 19.2345, 59.2882, 
68.6889, 320.6442, 340.4545, 479.3034, 482.6092, 484.048, 485.7009, 
486.4204, 488.1337, 489.5725, 491.2254, 492.3676, 493.2297, 494.3719, 
495.2339, 496.3762, 499.6819, 500.253, 501.1151, 504.5417, 505.4038, 
507.6278, 508.4899, 509.6321, 522.1321, 524.4165, 527.0027, 529.2871, 
531.8733, 533.0155, 544.6534, 547.9592, 551.4075, 553.0604, 556.9397, 
558.5926, 561.1788, 562.321, 563.1831, 563.7542, 565.0473, 566.1895, 
572.801, 573.9432, 575.6674, 576.2385, 577.1006, 586.2382, 587.5313, 
589.2446, 590.1067, 593.4125, 594.5547, 595.8478, 596.99, 598.7141, 
599.8563, 600.2873, 603.1429, 604.0049, 604.576, 605.8691, 607.0113, 
610.0286, 614.0263, 617.3321, 624.7564, 626.4805, 628.1334, 630.9889, 
631.851, 636.4198, 638.0727, 638.5038, 639.646, 644.8184, 647.1028, 
647.9649, 649.1071, 649.5381, 650.6803, 651.5424, 652.6846, 654.3375, 
656.0508, 658.2059, 659.9193, 661.2124, 662.3546, 664.0787, 664.6498, 
665.9429, 682.4782, 731.3561, 734.6619, 778.1154, 787.2919, 803.9261, 
814.335, 848.1552, 898.2568, 912.6188, 924.6932, 940.9083), Tem = c(12.7813, 
12.9341, 12.9163, 14.6367, 15.6235, 15.9454, 27.7281, 28.4951, 
34.7237, 34.8028, 34.8841, 34.9175, 34.9618, 35.087, 35.1581, 
35.204, 35.2824, 35.3751, 35.4615, 35.5567, 35.6494, 35.7464, 
35.8007, 35.8951, 36.2097, 36.3225, 36.4435, 36.5458, 36.6758, 
38.5766, 38.8014, 39.1435, 39.3543, 39.6769, 39.786, 41.0773, 
41.155, 41.4648, 41.5047, 41.8333, 41.8819, 42.111, 42.1904, 
42.2751, 42.3316, 42.4573, 42.5571, 42.7591, 42.8758, 43.0994, 
43.1605, 43.2751, 44.3113, 44.502, 44.704, 44.8372, 44.9648, 
45.104, 45.3173, 45.4562, 45.7358, 45.8809, 45.9543, 46.3093, 
46.4571, 46.5263, 46.7352, 46.8716, 47.3605, 47.8788, 48.0124, 
48.9564, 49.2635, 49.3216, 49.6884, 49.8318, 50.3981, 50.4609, 
50.5309, 50.6636, 51.4257, 51.6715, 51.7854, 51.9082, 51.9701, 
52.0924, 52.2088, 52.3334, 52.3839, 52.5518, 52.844, 53.0192, 
53.1816, 53.2734, 53.5312, 53.5609, 53.6907, 55.2449, 57.8091, 
57.8523, 59.6843, 60.0675, 60.8166, 61.3004, 63.2003, 66.456, 
67.4, 68.2014, 69.3065)), .Names = c("Rt", "Tem"), class = "data.frame", row.names = c(NA, 
-109L))


library(segmented)  # Version: segmented_0.2-9.4

# create a linear model
out.lm <- lm(Tem ~ Rt, data = bullard)

# Set X breakpoints: Set psi=NA and K=n:
o <- segmented(out.lm, seg.Z=~Rt, psi=NA, control=seg.control(display=FALSE, K=3))
slope(o)  # defaults to confidence level of 0.95 (conf.level=0.95)

# Trickery for placing text labels
r <- o$rangeZ[, 1]
est.psi <- o$psi[, 2]
v <- sort(c(r, est.psi))
xCoord <- rowMeans(cbind(v[-length(v)], v[-1]))
Z <- o$model[, o$nameUV$Z]
id <- sapply(xCoord, function(x) which.min(abs(x - Z)))
yCoord <- broken.line(o)[id]

# create the segmented plot, add linear regression for comparison, and text labels
plot(o, lwd=2, col=2:6, main="Segmented regression", res=TRUE)
abline(out.lm, col="red", lwd=1, lty=2)  # dashed line for linear regression
text(xCoord, yCoord, 
    labels=formatC(slope(o)[[1]][, 1] * 1000, digits=1, format="f"), 
    pos = 4, cex = 1.3)