r - R(或任何语言)中偏态正态分布的非线性最小二乘回归

Question

第一次海报。如果我使用了不当的礼仪或词汇，请提前道歉。

我有来自 USGS 河流调查的化学浓度 (y) 与时间 (x) 的时间序列数据。它表现出我想通过非线性最小二乘回归建模的偏斜正态分布。我能够将正态分布曲线拟合到数据，但似乎无法将“偏度”纳入模型。

我从 Whuber 在这里给出的答案得出了我的正态分布拟合...线性回归最佳多项式(或更好的使用方法)？

我的数据和代码...

y <- c(0.532431978850729, 0.609737363640599, 0.651964078008195, 0.657368066358271, 
0.741496240155044, 0.565435828629966, 0.703655525439792, 0.718855614453251, 
0.838983191559565, 0.743767469276213, 0.860155614137561, 0.81923941209205, 
1.07899884812998, 0.950877380129941, 1.01284743983765, 1.11717867112622, 
1.08452873942528, 1.14640319037414, 1.35601176845714, 1.55587090166098, 
1.81936731953165, 1.79952819117948, 2.27965075864338, 2.92158756334143, 
3.28092981974249, 1.09884083379528, 4.52126319475028, 5.50589160306292, 
6.48951979830975, 7.61196542128105, 9.56700470248019, 11.0814901164772, 
13.3072954022821, 13.8519364143597, 11.4108376964234, 8.72143939873907, 
5.12221325838613, 2.58106436004881, 1.0642701141608, 0.44945378376047, 
0.474569233285229, 0.128299654944011, 0.432876244482592, 0.445456125461339, 
0.435530646939433, 0.337503495863836, 0.456525976632425, 0.35851011819921, 
0.525854215793115, 0.381206935673774, 0.548351975353343, 0.365384673834335, 
0.418990479166088, 0.50039125911365, 0.490696977485334, 0.376809405620949, 
0.484559448760701, 0.569111550743562, 0.439671715276438, 0.353621820313257, 
0.444241243031233, 0.415197754444015, 0.474852839357701, 0.462144150397257, 
0.535339727332139, 0.480714031175711)

#creating an arbitrary vector to represent time
x <- seq(1,length(y), by=1)

#model of normal distribution 
f <- function(x, theta)  { 
  m <- theta[1]; s <- theta[2]; a <- theta[3]; b <- theta[4];
  a*exp(-0.5*((x-m)/s)^2) + b
}

# Estimate some starting values.
m.0 <- x[which.max(y)]; s.0 <- (max(x)-min(x))/4; b.0 <- min(y); a.0 <- (max(y)-min(y))

# Do the fit.  (It takes no time at all.)
fit <- nls(y ~ f(x,c(m,s,a,b)), data.frame(x,y), start=list(m=m.0, s=s.0, a=a.0, b=b.0))

# Display the estimated location of the peak and its SE.
summary(fit)$parameters["m", 1:2]

par(mfrow=c(1,1))
plot(c(x,0),c(y,f(coef(fit)["m"],coef(fit))), main="Data", type="n",
     xlab="Time", ylab="Concentration")
curve(f(x, coef(fit)), add=TRUE, col="Red", lwd=2)
points(x,y, pch=19)

那么，关于如何调整模型以适应偏度，有什么建议吗？

干杯，杰米

Answer 1

您可以使用广义加法模型 (GAM) 吗？ GAM 功能强大且灵活，但难以解释模型系数。所以决定将取决于你的目的。如果目的是评估趋势，或者目的是预测浓度(在已知时间范围内)，那么 GAM 可能是一个不错的选择。

library(mgcv)
library(ggplot2)

dat <- data.frame(x = 1:length(y), y = y)

fit_gam <- gam(y ~ s(x, k = 20), data = dat) 

ggplot(dat, aes(x = x, y = y)) +
  geom_point() +
  geom_line(data = data.frame(x = x, y = fit_gam$fitted.values),
            color = "red") +
  ggtitle("Data") +
  xlab("Cocentration") +
  ylab("Time") +
  theme_bw() +
  theme(panel.grid = element_blank())

以下是应用 stat_smooth 来拟合相同 GAM 模型的另一个选项。

ggplot(dat, aes(x = x, y = y)) +
  geom_point() +
  stat_smooth(method = "gam", formula = y ~ s(x, bs = "tp", k = 20)) +
  ggtitle("Data") +
  xlab("Cocentration") +
  ylab("Time") +
  theme_bw() +
  theme(panel.grid = element_blank())