R Markdown生成Qualtrics调研单用户PDF报告格式问题咨询

先修复循环代码的小问题 #

你现有循环里存在两个可优化点：

1. for循环自带迭代逻辑，不需要手动写i <- i + 1
2. 文件名拼接重复使用了name2，应该是name1+name2的组合
   修改后的循环代码如下：

# 提前提取列映射表，避免循环内重复计算
col_map <- attr(df, "column_map") %>% select(qname, description)

for(i in 1:nrow(df)){
  user_data <- slice(df, i)
  rmarkdown::render(
    input = 'test.Rmd',
    output_file = paste0(user_data$name2, "_", user_data$name1, "_", user_data$yob, "_", user_data$StartDate, '.pdf'),
    # 通过params传递参数，避免全局变量冲突
    params = list(
      user_data = user_data,
      col_map = col_map
    )
  )
}

再修改test.Rmd文件解决两个核心问题 #

修改后的完整test.Rmd内容如下，核心逻辑是：

• 将单行用户数据从宽表转成长表，实现纵向展示
• 关联列映射表替换列名为实际问题标签

---
header-includes:
- \usepackage{booktabs}
- \usepackage{longtable}
- \usepackage{array}
- \usepackage{multirow}
- \usepackage{wrapfig}
- \usepackage{float}
- \usepackage{colortbl}
- \usepackage{pdflscape}
- \usepackage{tabu}
- \usepackage{threeparttable}
- \usepackage{threeparttablex}
- \usepackage[normalem]{ulem}
- \usepackage{makecell}
- \usepackage{xcolor}
# 新增参数接收入口
params:
  user_data: NULL
  col_map: NULL
title:  "Pacjent: `r params$user_data$name1` `r params$user_data$name2`" 
author: "YOB: `r params$user_data$yob`; e-mail: `r params$user_data$mail`" 
date: "date: `r params$user_data$StartDate`"
output:
  pdf_document:
    toc: false
editor_options: 
  chunk_output_type: console
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo =FALSE, warning = FALSE, message = FALSE)

library(kableExtra)
library(tidyverse)

# 宽表转长表，实现纵向展示
data_long <- params$user_data %>% 
  pivot_longer(everything(), names_to = "qname", values_to = "回答") %>% 
  # 关联问题标签替换列名
  left_join(params$col_map, by = "qname") %>% 
  select(问题 = description, 回答) %>% 
  # 可选：过滤空回答，让报告更整洁
  filter(!is.na(回答))

kbl(data_long, longtable = T, booktabs = T) %>% 
  kable_styling(latex_options = c("repeat_header", "striped"), full_width = T) %>%
  column_spec(1, width = "15em") %>%
  column_spec(2, width = "25em")

## 可选优化
如果需要隐藏问卷系统自带的元字段（如IP地址、ResponseId等），可以在`data_long`处理步骤中添加过滤逻辑，比如：
```r
data_long <- params$user_data %>% 
  pivot_longer(everything(), names_to = "qname", values_to = "回答") %>% 
  left_join(params$col_map, by = "qname") %>% 
  select(qname, 问题 = description, 回答) %>% 
  # 过滤不需要展示的字段
  filter(!qname %in% c("IPAddress", "ResponseId", "RecipientLastName", "Status")) %>% 
  select(-qname) %>% 
  filter(!is.na(回答))

内容的提问来源于stack exchange，提问作者blazej