2015-04-05 35 views
1

我正在嘗試使用Rstan來擬合來自Christensen,Johnson,Branscum和Hanson的貝葉斯理念和數據分析的示例模型:科學家和統計人員的介紹。作者使用WinBUGS,所以一些適應是必要的。數據是here和WinBUGS代碼複製在這篇文章的底部。這是一個非常簡單的模型,但我是一個完整的初學者,我無法弄清楚如何解決我得到的錯誤。我斯坦代碼如下:簡單多元線性模型的Rstan代碼

data { 
    int N_subjects; 
    int N_items; 
    matrix[N_subjects,N_items] y; 
} 

parameters { 
    vector[N_items] mu; 
    real<lower=0> sigma; 
    real<lower=-1,upper=1> rho; 
} 

transformed parameters { 
    cov_matrix[N_items] Sigma; 
    for (j in 1:N_items) 
    for (k in 1:N_items) 
     Sigma[j,k] <- pow(sigma,2)*pow(rho,step(abs(j-k)-0.5)); 
} 

model { 
    sigma ~ uniform(0,100); 
    rho ~ uniform(0,1); 
    mu ~ multi_normal(0,100); 
    for (i in 1:N_subjects) 
    y[i] ~ multi_normal(mu,Sigma); 
} 

分析器引發以下錯誤:

Error in stanc(file = file, model_code = model_code, model_name = model_name, : 
    failed to parse Stan model 'model' with error message: 
SYNTAX ERROR, MESSAGE(S) FROM PARSER: 
no matches for function name="multi_normal_log" 
    arg 0 type=vector 
    arg 1 type=int 
    arg 2 type=int 
available function signatures for multi_normal_log: 
0. multi_normal_log(vector, vector, matrix) : real 
1. multi_normal_log(vector, row vector, matrix) : real 
2. multi_normal_log(row vector, vector, matrix) : real 
3. multi_normal_log(row vector, row vector, matrix) : real 
4. multi_normal_log(vector, vector[1], matrix) : real 
5. multi_normal_log(vector, row vector[1], matrix) : real 
6. multi_normal_log(row vector, vector[1], matrix) : real 
7. multi_normal_log(row vector, row vector[1], matrix) : real 
8. multi_normal_log(vector[1], vector, matrix) : real 
9. multi_normal_log(vector[1], row vector, matrix) : real 
10. multi_normal_log(row vector[1], vector, matrix) : real 
11. multi_normal_log(row vector[1], row vector, matrix) : real 
12. multi_normal_log(vector[1], vector 

(我認爲)我明白,解析器告訴我,我試圖通過不恰當的數據類型模型塊中的multi_normal函數,但我無法弄清楚它是從哪裏來的。我懷疑我做錯了在定義協方差矩陣,但現在看來,不止一個參數有一個不正確的數據類型...

WinBUGS軟件代碼,我在我的造型斯坦代碼:

model{ 
for(i in 1:30){ 
for(j in 1:6){ 
logy[i,j] <- log(y[i,j]) 
} 
} 
for(i in 1:30){logy[i,1:6]~dmnorm(m[1:6],precision[1:6,1:6])} 
for(j in 1:6){ 
for(k in 1:6){ 
covariance[j,k] <- sigma2*pow(rho, step(abs(j-k)-0.5)) 
} 
} 
for(i in 1:6){ m[i] <- mu } 
precision[1:6,1:6] <- inverse(covariance[1:6,1:6]) 
sigma ~ dunif(0,100) 
mu ~ dnorm(0,0.001) 
L <- -1/(6-1) 
rho ~ dunif(L,1) 
sigma2 <- sigma*sigma 
tau <- 1/sigma2 
} 
作爲要傳遞一個矢量畝,整數0和整數100。我想

回答

2

的錯誤來自

mu ~ multi_normal(0,100); 

mu ~ normal(0,100); 

它將mu的元素視爲獨立且相同的正態分佈,均值爲0,標準差爲100.

+0

D'oh!謝謝!我知道這件事很簡單,我只是看着它過去... – psychometriko 2015-04-06 12:34:33