First: can you take this quick survey please?

Dynamic Documents For Computational Reproducibility

• Based on principles of literate programming aims at combining code and paper in one single document
• Best framework to achieve the holy grail of one-click reproducible workflow
• Best two current implementations: RMarkdown (R) & Jupyter (Python). Stata is catching up (more at the end)

Currently code and narrative components live in separate universes

Dynamic Documents: integrate the two universes!

Dynamic Documents: A Reciepe

• 1 simple language that can combine text and code: Markdown
• 1 statistical package to do the analysis (R, Python, 3S's?)
• 1 machinery to combine analysis and text to create a single output: Pandoc
• [Optional-but-not-really] 1 program to bring all the elements together: RStudio/RMarkdown, Jupyter

For our excercise: R Markdown

• R: open source programming language design for statistical analysis.
  
• RStudio: free software that provides and Integrated Development Environment (IDE)
  
• RStudio combines all together: R + Markdown + Pandoc to produce multiple outputs

R Markdown

Basic Structure

• A header
• Text
• Code: inline and chunks

Basic Structure: Header

---
title: "Sample Paper"
author: "Fernando Hoces de la Guardia"
output: html_document
---

Basic Structure: Body of Text

---
header
---

This is where you write your paper. Nothing much to add. You can check Markdown syntax here. And it can use can type equations using LaTex syntax!

Basic Structure: Code Chunks and Inline

---
header
---

Body of text.

To begin a piece of code (“code chunk”). Enclose them in the following expression (Ctrl/Cmd + shift/optn + i)

```{r, eval=TRUE}
here goes the code
```

To write inline use only one Backtick to open followed by an “r”" and one to close `r 1+1` in the output.

Hands-on excercise: the birthday problem!

As an illustration lets write a report using the participants in this workshop to illustrate the famous birthday problem.

What is the probability that at least two people this room share the same birthday?

Is it something like \(\frac{1}{365} \times N =\) 0.11?

Create a new RMarkdown File

1 - In RStudio: File-> New File -> RMarkdown...
2 - Name it, and save it.
3 - Review/edit the header, and delete all the default body of text except for one code chunk.
4 - Define a seed (set.seed = 1234 and number of people in the room (n.pers = ?)

The birthday problem: the math

Code for the math (https://goo.gl/ZFQvba)

Don’t look at this: just copy and paste into your report

\begin{align} 
 1 - \bar p(n) &= 1 \times \left(1-\frac{1}{365}\right) 
 \times \left(1-\frac{2}{365}\right) \times \cdots \times 
 \left(1-\frac{n-1}{365}\right) \nonumber  \\  
 &= \frac{ 365 \times 364 \times \cdots \times 
   (365-n+1) }{ 365^n } \nonumber \\ 
 &= \frac{ 365! }{ 365^n (365-n)!} = 
   \frac{n!\cdot\binom{365}{n}}{365^n}\\
p(n= `r n.pers`) &= `r  
 round(1 - factorial(n.pers) * 
         choose(365,n.pers)/ 365^n.pers, 3)`\nonumber
\end{align}

Don’t like math? Let’s run a simple simulation!

1 - Simulate 10,000 rooms with \(n = 40\) random birthdays, and store the results in matrix where each row represents a room.
2 - For each room (row) compute the number of unique birthdays.
3 - Compute the average number of times a room has 40 unique birthdays, across 10,000 simulations, and report the complement.

Code for the simulation (https://goo.gl/ZFQvba)

birthday.prob = function(n.pers, n.sims) {
  # simulate birthdays
  birthdays = matrix(round(runif(n.pers * n.sims, 1, 365)), 
                      nrow = n.sims, ncol = n.pers)
  # for each room (row) get unique birthdays
  unique.birthdays = apply(birthdays, 1, unique)
  # Indicator with 1 if all are unique birthdays
  all.different = (lapply(unique.birthdays, length) == n.pers)
  # Compute average time all have different birthdays 
  result = 1 - mean(all.different)
return(result)
}
n.pers.param = 43
n.sims.param = 1e4
birthday.prob(n.pers.param,n.sims.param)

## [1] 0.9274

Results

• Many people originally think of a prob ~ \(\frac{1}{365} \times N =\) 0.118
• However the true probability is of \(p(n= 43) = 0.924\)
• And the simulated probability is of 0.9296

Final Remarks & More Resources

• With DD with can achieve a one-click reproducible workflow.
• This is particularly helpful to understand/present results that are hard to digest.
• Stata just develop an internal version of DD for v15. Review Here
• More great examples here
• Want to learn more: great free books (can you guess how they were written?)