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我正在使用 nls() 为多个鸟类种群的数据拟合一个逻辑模型(自启动;SSlogis)。我的目标是为数据拟合一个预期函数(仅使用每个数据集的一部分),并在图表上显示关于预期的方差的度量。然后,我想拟合并绘制观察到的函数(使用每个总体的整个数据集)以确定观察到的动态是否在期望的方差范围内。这是我目前为实现此目的而编写的代码:

    CE.mod = nls(CE.observed ~ SSlogis(t.CattleEgret, Asym, xmid, scal))

    with(collapse.data, plot(CE.time, CE.obs))

    CE.extrap = predict(CE.mod, data.frame(t.CattleEgret = CE.time))
    lines(CE.time, CE.extrap)

    CE.se.fit = sqrt(apply(attr(CE.extrap, "gradient"), 1, function(x) 
    sum(vcov(CE.mod)*outer(x,x))))

    matplot(CE.time, CE.extrap+outer(CE.se.fit, qnorm(c(0.5, 0.025, 0.975))),
    type = "l", lty = c(1,1,1), ylab = "Abundance (# per party hour)",
    xlab = "Time (year)", main = "Cattle Egret Collapse Analysis", 
    pch = 15, font.lab = 2, font.axis = 2, cex = 4, cex.lab = 1.5, 
    cex.axis = 2, cex.main = 2, frame.plot = FALSE, lwd = 4, 10)

    with(collapse.data, matpoints(CE.time, CE.obs, pch = 15, cex = 3))
    lines(CE.time, predict(nls(CE.obs ~ SSlogis(log(CE.time), 
    Asym, xmid, scal))), lty = 3, lwd = 4)

哪里(来自“collapse.data”文件):

    t.CattleEgret = c(1:20)
    CE.time = c(1:45)
    CE.obs = c(0.3061324, 0.0000100, 0.2361211, 0.5058240, 2.0685032, 2.1944544, 
               4.2689494, 4.9508297, 3.1334720, 3.6570752, 5.6753381, 10.9133183,
               5.4518257, 20.4166979, 15.9741054, 19.0970426, 13.7559959, 14.1358153, 
               15.9986416, 29.6762828, 10.3760667, 8.4284488, 6.1060359, 3.7099982, 
               3.3584060, 2.5981386, 2.5697082, 2.8091952, 5.5487979, 1.6505442,
               2.2696972, 2.1835692, 3.6747876, 4.8307886, 3.5019731, 2.8397137,
               1.8605288, 11.1848738, 2.6268683, 4.1215127, 2.3996210, 2.6569938, 
               2.1987387, 3.0267252, 2.4420927)
    CE.observed = c(0.3061324, 0.0000100, 0.2361211, 0.5058240, 2.0685032, 2.1944544, 
               4.2689494, 4.9508297, 3.1334720, 3.6570752, 5.6753381, 10.9133183,
               5.4518257, 20.4166979, 15.9741054, 19.0970426, 13.7559959, 14.1358153, 
               15.9986416, 29.6762828)

该代码工作正常并产生如下图:

牛背鹭塌陷分析

但是,如果我从代码的最后一行中删除“log()”,以便编写以下代码:

    lines(CE.time, predict(nls(CE.obs ~ SSlogis(CE.time, 
    Asym, xmid, scal))), lty = 3, lwd = 4),

该线不会绘制,我收到此错误:

    Error in nls(y ~ 1/(1 + exp((xmid - x)/scal)), data = xy, start = list(xmid = 
    aux[1L],  : step factor 0.000488281 reduced below 'minFactor' of 0.000976562

即使我使用 nls.controls 并更改“minFactor”值,我也无法更改。我还在为某些人群定义 mod(##.mod 部分)的初始行之后收到此错误消息。

此外,对于某些人群,我在报告此问题的最后一行代码之后收到一条错误消息:

    Error in qr.solve(QR.B, cc) : singular matrix 'a' in solve

我想不出自然对数转换数据的合理性,我只能假设我只是以允许 predict() 和 SSlogis() 的方式更改了数据(在这种情况下是任意记录的)功能正常,但我不知道为什么。对于此类问题,我无法在任何论坛中找到任何合适的答案。任何帮助将不胜感激。

*更新:我已尝试按照 Roland 的建议实现 nlsLM 功能(如下)。这确实清理了使用令人​​困惑的 log() 的代码部分:

    lines(CE.time, predict(nlsLM(CE.obs ~ Asym/(1 + exp((xmid - CE.time)/scal)), start 
    = list(Asym = max(CE.obs), xmid = popsizetime[1], scal = 1), control = 
    nls.lm.control(maxiter = 1000))

但是,对于其他人群,我在初始模型规范中遇到了与上述相同的错误消息:

    ChMa.mod = nls(ChMa.observed ~ SSlogis(t.ChestnutMannikin, Asym, xmid, scal))

    Error in nls(y ~ 1/(1 + exp((xmid - x)/scal)), data = xy, start = list(xmid = 
    aux[1L],  : step factor 0.000488281 reduced below 'minFactor' of 0.000976562

切换到:

    ChMa.mod = nlsLM(ChMa.observed ~ Asym/(1 + exp((xmid - t.ChestnutMannikin)/
scal)), start = list(Asym = max(ChMa.obs), xmid = popsizetime[2], 
scal = 1), control = nls.lm.control(maxiter = 1000))

在哪里

    ChMa.observed = c(4.02785074, 0.33847154, 0.99029776, 2.86516540, 0.59588068, 
    0.01334333, 2.07693362, 0.62485994, 3.48979515, 3.67785202, 20.84180181)
    t.ChestnutMannikin = c(1:11)
    popsizetime[2] = 11

虽然此开关确实避免了错误消息,但 nlsLM 评估函数但不评估梯度。如果没有梯度评估,我无法使用 se.fit 代码,因此无法获得用于绘图的方差估计。

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1 回答 1

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我找到了问题的答案:我需要添加我的模型的一个组件,该组件为我正在使用 nlsLM 回归的函数生成渐变。

    log.model = function(t.RedventedBulbul, Asym, xmid, scal) {
            numericDeriv(quote(Asym/(1 + exp((xmid - t.RedventedBulbul)/scal))),
            c("Asym", "xmid", "scal"), parent.frame())
    }
于 2013-01-17T02:45:43.983 回答