我需要详细解释这一点，因为我没有统计的基础知识可以更简洁地进行解释.在SO中这样问是因为我正在寻找python解决方案，但如果更合适的话可以去stats.SE.
I need to explain this in excruciating detail because I don't have the basics of statistics to explain in a more succinct way. Asking here in SO because I am looking for a python solution, but might go to stats.SE if more appropriate.
我有井下油井数据，可能有点像这样:
I have downhole well data, it might be a bit like this:
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变量可以视为深度的代理，而T是温度).我还具有先验"数据，该数据为我提供了Rt = 0时的温度，并且未显示一些我可以用作断点，可以指导断点或至少与发现的断点进行比较的标记.
(the Rt variable can be considered a proxy for depth, and T is temperature). I also have 'a priori' data giving me the temperature at Rt=0 and, not shown, some markers that i can use as breakpoints, guides to breakpoints, or at least compare to any discovered breakpoints.
这两个变量的线性关系在某些深度范围内受某些过程的影响.一个简单的线性回归是
The linear relationship of these two variables is in some depth intervals affected by some processes. A simple linear regression is
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，并且T0的拟合度很低(应该为15):
and looks like this, where you can see the deviations clearly, and the poor fit for T0 (which should be 15):
我希望能够执行一系列线性回归(在每个段的末端连接)，但是我想要这样做: (a)没有指定休息的次数或地点， (b)通过指定休息的数量和地点，以及 (c)计算每个细分的系数
I want to be able to perform a series of linear regressions (joining at ends of each segment), but I want to do it: (a) by NOT specifying the number or locations of breaks, (b) by specifying the number and location of breaks, and (c) calculate the coefficients for each segment
我认为我可以做到(b)和(c)，只需将数据拆分并仔细处理每一部分，但是我不知道(a)，不知道是否有人知道这可以更简单地完成.
I think I can do (b) and (c) by just splitting the data up and doing each bit separately with a bit of care, but I don't know about (a), and wonder if there's a way someone knows this can be done more simply.
我已经看到了: stats.stackexchange/a/20210/9311 ，我认为MARS可能是处理它的好方法，但这仅仅是因为它看起来不错.我不太了解我以盲切粘贴的方式尝试了数据，并在下面显示了输出，但同样，我也不明白:
I have seen this: stats.stackexchange/a/20210/9311, and I think MARS might be a good way to deal with it, but that's just because it looks good; I don't really understand it. I tried it with my data in a blind cut'n'paste way and have the output below, but again, I don't understand it:
推荐答案
简短的答案是，我使用R解决了我的问题，创建了线性回归模型，然后使用了 segmented 包可从线性模型生成分段线性回归.我可以使用psi=NA和K=n指定期望的断点(或结)n数量，如下所示.
The short answer is that I solved my problem using R to create a linear regression model, and then used the segmented package to generate the piecewise linear regression from the linear model. I was able to specify the expected number of breakpoints (or knots) n as shown below using psi=NA and K=n.
长答案是:
R版本3.0.1(2013-05-16) 平台:x86_64-pc-linux-gnu(64位)
R version 3.0.1 (2013-05-16) Platform: x86_64-pc-linux-gnu (64-bit)
# 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)
更多推荐
如何获得具有先验断点的分段线性回归?
发布评论