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我有一些碎岩材料的粒度质量传递累积数据，我想使用 R 拟合 Weibull 分布。我已经使用 WEIBULL.DIST() 函数在 Excel 中使用累积开关设置为 TRUE 来完成此操作。
然后我使用 excel SOLVER 使用 RMSE 推导出 alpha 和 beta 参数以获得最佳拟合。我想在 R 中重现结果。

（见附件电子表格在这里）

粒子数据和累积质量通过百分比是以下向量

d.mm <- c(20.001,6.964,4.595,2.297,1.741,1.149,
          0.871,0.574,0.287,0.082,0.062,0.020)
m.pct <- c(1.00,0.97,0.78,0.49,0.27,0.20,0.14,
         0.11,0.07,0.03,0.025,0.00)

这是我想拟合 Weibull 结果的图：

plot(log10(d.mm),m.pct)

...根据电子表格计算直径值向量的函数

   d.wei <- c(seq(0.01,0.1,0.01),seq(0.2,1,0.1),seq(2,30,1))

我使用 Solver 在 Excel 中确定的 Weibull alpha 和 beta 的最佳值分别为 1.41 和 3.31 所以我的问题是如何在 R 中重现此分析（不一定是 Solver 部分）但将 Weibull 拟合到该数据集？

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1 回答 1

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非线性最小二乘函数nls是 Execl 求解器的 R 版本。

将pweibull计算 Weibull 分布的概率分布。代码中的注释应该解释一步一步的解决方案

d.mm <- c(20.001,6.964,4.595,2.297,1.741,1.149,
          0.871,0.574,0.287,0.082,0.062,0.020)
m.pct <- c(1.00,0.97,0.78,0.49,0.27,0.20,0.14,
           0.11,0.07,0.03,0.025,0.00)

#create data frame store data
df<-data.frame(m.pct, d.mm)

#data for prediction
d.wei <- c(seq(0.01,0.1,0.01),seq(0.2,1,0.1),seq(2,30,1))

#solver (provided starting value for solution)
# alpha is used for shape and beta is used for scale
fit<-nls(m.pct~pweibull(d.mm, shape=alpha, scale=beta), data=df, start=list(alpha=1, beta=2))
print(summary(fit))

#extract out shape and scale
print(summary(fit)$parameters[,1])

#predict new values base on model
y<-predict(fit, newdata=data.frame(d.mm=d.wei))

#Plot comparison
plot(log10(d.mm),m.pct)
lines(log10(d.wei),y, col="blue")

在此处输入图像描述

于 2019-04-16T17:11:29.497 回答