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模拟数据的R2WinBUGS - logistic回归
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Stack Overflow用户
提问于 2011-11-25 01:30:44
回答 1查看 2.7K关注 0票数 6

我只是想知道是否有人有一些R代码,它使用R2WinBUGS软件包来运行逻辑回归-理想情况下,使用模拟数据来生成“真相”和两个连续的协变量。

谢谢。

克里斯蒂安

PS:

通过r2winbugs生成人工数据(一维情况)和运行winbug的潜在代码(目前还不起作用)。

代码语言:javascript
复制
library(MASS)
library(R2WinBUGS)

setwd("d:/BayesianLogisticRegression")

n.site <- 150

X1<- sort(runif(n = n.site, min = -1, max =1))

xb <- 0.0 + 3.0*X1 

occ.prob <- 1/(1+exp(-xb))

plot(X1, occ.prob,xlab="X1",ylab="occ.prob")

true.presence <- rbinom(n = n.site, size = 1, prob = occ.prob)

plot(X1, true.presence,xlab="X1",ylab="true.presence")

# combine data as data frame and save
data <- data.frame(X1, true.presence)
write.matrix(data, file = "data.txt", sep = "\t")

sink("model.txt")
cat("
model {

# Priors
 alpha ~ dnorm(0,0.01)
 beta ~ dnorm(0,0.01)

# Likelihood
 for (i in 1:n) {
    C[i] ~ dbin(p[i], N)        # Note p before N
    logit(p[i]) <- alpha + beta *X1[i]
 }
}
",fill=TRUE)
sink()

# Bundle data
win.data <- list(mass = X1, n = length(X1))

# Inits function
inits <- function(){ list(alpha=rlnorm(1), beta=rlnorm(1))}

# Parameters to estimate
params <- c("alpha", "beta")

# MCMC settings
nc <- 3 #Number of Chains
ni <- 1200 #Number of draws from posterior
nb <- 200 #Number of draws to discard as burn-in
nt <- 2 Thinning rate

# Start Gibbs sampling
out <- bugs(data=win.data, inits=inits, parameters.to.save=params, 
model.file="model.txt", n.thin=nt, n.chains=nc, n.burnin=nb, 
n.iter=ni, debug = TRUE)
EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2011-11-29 02:39:15

你的脚本就是这样做的。它几乎可以工作了,只需要一个简单的更改就可以使其工作:

代码语言:javascript
复制
win.data <- list(X1 = X1, n = length(X1), C = true.presence, N = 1)

它定义了哪些数据将转到WinBugs。变量C必须用true.presence填充,根据您生成的数据,N必须是1 -请注意,这是N=1的二项式分布的一个特例,称为Bernoulli -简单的“抛硬币”。

下面是输出:

代码语言:javascript
复制
> print(out, dig = 3)
Inference for Bugs model at "model.txt", fit using WinBUGS,
 3 chains, each with 1200 iterations (first 200 discarded), n.thin = 2
 n.sims = 1500 iterations saved
            mean    sd    2.5%     25%     50%     75%   97.5%  Rhat n.eff
alpha     -0.040 0.221  -0.465  -0.187  -0.037   0.114   0.390 1.001  1500
beta       3.177 0.478   2.302   2.845   3.159   3.481   4.165 1.000  1500
deviance 136.438 2.059 134.500 135.000 135.800 137.200 141.852 1.001  1500

如您所见,这些参数与用于生成数据的参数相对应。此外,如果你与频率解进行比较,你会看到它是对应的。

编辑:但典型的logistic (~二项式)回归将测量一些具有上限值Ni的计数,并且它将允许每个观察值具有不同的Ni。例如,青少年占总人口的比例(N)。这将只需要在您的模型中将索引添加到N:

代码语言:javascript
复制
C[i] ~ dbin(p[i], N[i])

数据生成将如下所示:

代码语言:javascript
复制
N = rpois(n = n.site, lambda = 50) 
juveniles <- rbinom(n = n.site, size = N, prob = occ.prob)
win.data <- list(X1 = X1, n = length(X1), C = juveniles, N = N)

(编辑结束)

有关种群生态学的更多示例,请参阅books of Marc Kéry (面向生态学家的WinBUGS入门,特别是使用WinBUGS的贝叶斯种群分析:分层透视,这是一本很棒的书)。

我使用的完整脚本-您的更正后的脚本在这里列出(与末尾的frequentist解决方案进行比较):

代码语言:javascript
复制
#library(MASS)
library(R2WinBUGS)

#setwd("d:/BayesianLogisticRegression")

n.site <- 150

X1<- sort(runif(n = n.site, min = -1, max =1))

xb <- 0.0 + 3.0*X1 

occ.prob <- 1/(1+exp(-xb)) # inverse logit

plot(X1, occ.prob,xlab="X1",ylab="occ.prob")

true.presence <- rbinom(n = n.site, size = 1, prob = occ.prob)

plot(X1, true.presence,xlab="X1",ylab="true.presence")

# combine data as data frame and save
data <- data.frame(X1, true.presence)
write.matrix(data, file = "data.txt", sep = "\t")

sink("tmp_bugs/model.txt")
cat("
model {

# Priors
 alpha ~ dnorm(0,0.01)
 beta ~ dnorm(0,0.01)

# Likelihood
 for (i in 1:n) {
    C[i] ~ dbin(p[i], N)        # Note p before N
    logit(p[i]) <- alpha + beta *X1[i]
 }
}
",fill=TRUE)
sink()

# Bundle data
win.data <- list(X1 = X1, n = length(X1), C = true.presence, N = 1)

# Inits function
inits <- function(){ list(alpha=rlnorm(1), beta=rlnorm(1))}

# Parameters to estimate
params <- c("alpha", "beta")

# MCMC settings
nc <- 3 #Number of Chains
ni <- 1200 #Number of draws from posterior
nb <- 200 #Number of draws to discard as burn-in
nt <- 2 #Thinning rate

# Start Gibbs sampling
out <- bugs(data=win.data, inits=inits, parameters.to.save=params, 
model.file="model.txt", n.thin=nt, n.chains=nc, n.burnin=nb, 
n.iter=ni, 
working.directory = paste(getwd(), "/tmp_bugs/", sep = ""),
debug = TRUE)

print(out, dig = 3)

# Frequentist approach for comparison
m1 = glm(true.presence ~ X1, family = binomial)
summary(m1)

# normally, you should do it this way, but the above also works:
#m2 = glm(cbind(true.presence, 1 - true.presence) ~ X1, family = binomial)
票数 5
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/8260771

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