Hacktoberfest 2026:维护者为十月标记出来的 issue,仍然开放、适合新手。 浏览 Hacktoberfest issue

Relabelling helper function for working with factors in rstan

未关闭
#69 1 条评论 0 个 reaction 已指派 0 人 在 GitHub 查看

还没有人认领这个 Issue。

评估

难度
5/5
预计耗时
一周以上
新手友好度
30/100
Issue 类型
功能
描述清晰度
基本清楚
活跃度
停滞
技术栈
r
领域
tooling

调研方向

未提及任何仓库文件或测试。首先检查 rstantools 如何处理 rstan 拟合对象和参数名称,然后将提议的 delabel 和 relabel 示例与现有接口进行比较。要视为完成,需要就通用 helper 的设计达成一致并进行相应的验证,但该 issue 未定义验收标准。

由索引模型根据 Issue 内容生成。

描述

Summary:

When using factor variables (which are have attached meaningful level names), would it be useful/possible to have a function that relabels variables in the Stan fit object with the relevant factor levels?

Description:

When using categorical variables in stan you have to drop the level labels. This is fine and sensible, but with variables that have a number of levels, the output can be difficult to read. Would it be possible to write a helper function that relabels the numeric index with the relevant label for easier use? @jgabry thought it might be, but wasn't sure if it would be too complex to create something that was useful in the more general.

Reproducible Steps:

For example. With the 8 schools model the schools are coded as numbers. However, in real world data, they would often be recorded with meaningful names (here we use states).

schools_df <- data.frame(schools = state.name[1:8],
y = c(28,  8, -3,  7, -1,  1, 18, 12),
sigma = c(15, 10, 16, 11,  9, 11, 10, 18))
schools_df$schools = as.factor(schools_df$schools)

I am proposing a pair of functions that force the factors to numeric for use in rstan, and then relabels the relevant variables in the stan fit object. e.g. for the simple 8 schools case.

delabel <- function(x){
  return(as.numeric(x))
}

relabel <- function(x,stan_fit,variables){
   current_names <- names(stan_fit)
   match <- cbind(levels(x),levels(as.factor(as.numeric(x))))
   for (i in 1:length(variables)){
     loc_var <- grepl(paste0("^",variables[i]),current_names)
     for(j in 1:nrow(match)){
       current_names[loc_var] <- gsub(pattern = match[j,2], replacement = match[j,1],current_names[loc_var])
     }
   }
   names(stan_fit) <- current_names
   return(stan_fit)
}

The idea is that is would work something like this:

schools_dat <- list(J = 8, 
                    school = delabel(schools_df$schools),
                    y = schools_df$y,
                    sigma =schools_df$sigma)

library(rstan)

fit <- stan(file = '8schools.stan', data = schools_dat)

fit2 <- relabel(schools_df$schools,fit,c('theta','eta'))

colnames(as.matrix(fit2))
library(bayesplot)
mcmc_areas_ridges(as.matrix(fit2, pars = "theta"))
主要语言
R
星标
49
派生
24
平均合并
23 小时 46 分钟
30 天内合并 PR
1

环境准备

  • 没有 Dockerfile 或 Docker Compose 文件
  • 没有 Pull Request 模板
  • 阅读贡献指南

从这里开始

  1. 先读完整个 Issue,再读项目的贡献指南。
  2. 在 Issue 下留言说明你要接手 —— 这能避免两个人做同样的事。
  3. Fork 仓库,在一个分支上完成修改。
  4. 提交 Pull Request,并在描述里引用这个 Issue 编号。

stan-dev/rstantools 的其他 Issue

查看 stan-dev/rstantools 的全部 Issue

相似的 Issue

更多 R Issue

把新 issue 发到你的邮箱

精选适合新手参与的 GitHub issue 摘要。