计算引导对象的r平方值并将其保存在r中的新数据帧中

我有一个数据帧

df

dput(df)
    structure(list(x = c(49, 50, 51, 52, 53, 54, 55, 56, 1, 2, 3, 
    4, 5, 14, 15, 16, 17, 2, 3, 4, 5, 6, 10, 11, 3, 30, 64, 66, 67, 
    68, 69, 34, 35, 37, 39, 2, 17, 18, 99, 100, 102, 103, 67, 70, 
    72), y = c(2268.14043972082, 2147.62290922552, 2269.1387550775, 
    2247.31983098201, 1903.39138268307, 2174.78291538358, 2359.51909126411, 
    2488.39004804939, 212.851575751527, 461.398994384333, 567.150629704352, 
    781.775113821961, 918.303706148872, 1107.37695799186, 1160.80594193377, 
    1412.61328924168, 1689.48879626486, 260.737164468854, 306.72700499362, 
    283.410379620422, 366.813913489692, 387.570173754128, 388.602676983443, 
    477.858510450125, 128.198042456082, 535.519377609133, 1028.8780498564, 
    1098.54431357711, 1265.26965941035, 1129.58344809909, 820.922447928053, 
    749.343583476846, 779.678206156474, 646.575242339517, 733.953282899613, 
    461.156280127354, 906.813018662913, 798.186995701282, 831.365377249207, 
    764.519073183124, 672.076289062505, 669.879217186302, 1341.47673353751, 
    1401.44881976186, 1640.27575962036)), .Names = c("x", "y"), row.names = c(NA, 
    -45L), class = "data.frame")

我基于我的数据集创建了两个非线性回归（nls1和nls2）

library(stats)
    nls1 <- nls(y~A*(x^B)*(exp(k*x)), 
                data = df, 
                start = list(A = 1000, B = 0.170, k = -0.00295))
    nls2<-nls(y~A*x^3+B*x^2+C*x+D, data=df,
        start = list(A=0.02, B=-0.6, C= 50, D=200))

然后，我们的想法是在自举的基础上，用每个非线性模型的中位数、最小值和最大值绘制一些方框图来比较它们。有人能帮我吗？提前谢谢。

从@bunk回答

stat <- function(dat, inds) { fit <- try(nls(y~A*(x^B)*(exp(k*x)), data = dat[inds,], start = list(A = 1000, B = 0.170, k = -0.00295)), silent=TRUE); f1 <- if (inherits(fit, "nls")) AIC(fit) else NA; fit2 <- try(nls(y~A*x^3+B*x^2+C*x+D, data = dat[inds,], start = list(A=0.02, B=-0.6, C= 50, D=200)), silent=TRUE); f2 <- if (inherits(fit2, "nls")) AIC(fit2) else NA; c(f1, f2) }; res <- boot(df, stat, R=200). Then, to get medians for example, apply(res$t, 2, median, na.rm=TRUE)

---

stat你在想什么r平方？对于非线性模型来说，这通常不是一个很好的衡量标准。当然，也可能是AIC或建模效率（MEF）更好。在
库（boot）
中有
boot
，您可以定义一个函数来返回感兴趣的统计数据<代码>库（引导）；对。但是，我们的想法是创建一个新的最终数据帧，它可以从n个非线性函数的自举中总结出中值、分位数等。使用统计函数，您可以在其中放入任何您想要的内容。因此，您可以在一个引导中容纳任意多个模型，返回
c（stat1、stat2等）
向量。然后，
res$t
将是所有估计值的矩阵。然后，它只是将您想要的内容（按列应用）提取到data.frame中。
structure(list(Median = c(NA, NA), Max = c(NA, NA), Min = c(NA, 
NA)), .Names = c("Median", "Max", "Min"), row.names = c("nls1", 
"nls2"), class = "data.frame")

stat <- function(dat, inds) { fit <- try(nls(y~A*(x^B)*(exp(k*x)), data = dat[inds,], start = list(A = 1000, B = 0.170, k = -0.00295)), silent=TRUE); f1 <- if (inherits(fit, "nls")) AIC(fit) else NA; fit2 <- try(nls(y~A*x^3+B*x^2+C*x+D, data = dat[inds,], start = list(A=0.02, B=-0.6, C= 50, D=200)), silent=TRUE); f2 <- if (inherits(fit2, "nls")) AIC(fit2) else NA; c(f1, f2) }; res <- boot(df, stat, R=200). Then, to get medians for example, apply(res$t, 2, median, na.rm=TRUE)