繁体   English   中英

rmarkdown:递归块评估?

[英]rmarkdown: recursive chunk evaluation?

我想重新评估rmarkdown文档中R块产生的字符串-可以吗? 为了演示起见,请考虑以下文档:

---
params:
  B: 'test'
---

```{r simple, results='asis', echo=FALSE}
write(params[['B']], file = "")
```

```{r recursive-evaluation-questionmark, results='asis', echo=FALSE}
write(
  "How to get \"params[['B']]\" evaluated here? This \"`r params[['B']]`\" is
    expected to be \"test\" ...",
  file = "")
```

当前,编织时产生以下内容:

test

How to get “params[[‘B’]]” evaluated here? This “r params[['B']]” is expected to be “test” …

但我想要:

test

How to get “params[[‘B’]]” evaluated here? This “test” is expected to be “test” …

您不能让knitr将R代码视为文本,但是可以使用纯R代码解决该问题。 我还建议使用cat()而不是write() 例如,

write(
  paste0("How to get \"params[['B']]\" evaluated here? This \"", params[['B']], "\" is
  expected to be \"test\" ..."),
 file = "")

或更简单

cat("How to get \"params[['B']]\" evaluated here? This \"", params[['B']], "\" is
  expected to be \"test\" ...", sep = "")

对于更复杂的类似宏的替换,您可能需要使用sub()gsub()函数,例如

msg <- "How to get \"params[['B']]\" evaluated here? This \"%param%\" is
  expected to be \"test\" ..."
cat(sub("%param%", params[['B']], msg))

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM