Using a DAG to simulate data with the dagR library

Directed acyclic graphs (DAGs), and causal graphs in general, provide a framework for making assumptions explicit and identifying confounders or mediators of the relationship between the exposure of interest and outcome that need to be adjusted for in analysis. Recently, I ran into the need to generate data from a DAG for a paper I am writing with my peers Kevin McIntyre and Joshua Wiener. After a quick Google search, I was pleasantly surprised to see there were several options to do so. In particular, the dagR library provides “functions to draw, manipulate, [and] evaluate directed acyclic graphs and simulate corresponding data”.

Besides dagR‘s reference manual, a short letter published in Epidemiology, and a limited collection of examples, I couldn’t find too many resources regarding how to use the functionality provided by dagR. The goal of this blog post is to provide an expository example of how to create a DAG and generate data from it using the dagR library.

To simulate data from a DAG with dagR, we need to:

  1. Create the DAG of interest using the dag.init function by specifying its nodes (exposure, outcome, and covariates) and their directed arcs (directed arrows to/from nodes).
  2. Pass the DAG from (1) to the dag.sim function and specify the number of observations to be generated, arc coefficients, node types (binary or continuous), and parameters of the node distributions (Normal or Bernoulli).

For this tutorial, we are going to try to replicate the simple confounding/common cause DAG presented in Figure 1b as well as the more complex DAG in Figure 2a of Shier and Platt’s (2008) paper, Reducing bias through directed acyclic graphs.

library(dagR)
set.seed(12345)

Continue reading Using a DAG to simulate data with the dagR library

Advent of Code 2017 in R: Day 2

Day 2 of the Advent of Code provides us with a tab delimited input consisting of numbers 2-4 digits long and asks us to calculate its “checksum”. checksum is defined as the sum of the difference between each row’s largest and smallest values. Awesome! This is a problem that is well-suited for base R.

I started by reading the file in using read.delim, specifying header = F in order to ensure that numbers within the first row of the data are not treated as variable names.

When working with short problems like this where I know I won’t be rerunning my code or reloading my data often, I will use file.choose() in my read.whatever functions for speed. file.choose() opens Windows Explorer, allowing you to navigate to your file path.

input <- read.delim(file.choose(), header = F)

# Check the dimensions of input to ensure the data read in correctly.
dim(input)

After checking the dimensions of our input, everything looks good. As suspected, this is a perfect opportunity to use some vectorization via the apply function.

row_diff <- apply(input, 1, function(x) max(x) - min(x))
checksum <- sum(row_diff)
checksum

Et voilà, the answer is 45,972! Continue reading Advent of Code 2017 in R: Day 2

Advent of Code 2017 in R: Day 1

My boyfriend recently introduced me to Advent of Code while I was in one of my “learn ALL of the things!” phases. Every year starting December 1st, new programming challenges are posted daily leading up to Christmas. They’re meant to be quick 5-10 minute challenges, so, wanting to test my programming skills, I figured why not try to do all of them in base R!

I went with base R because I know I can dplyr and stringr my way to victory with some of these challenges. I really want to force myself to really go back to basics and confirm that I have the knowledge to do these things on my own without Hadley Wickham‘s (very much appreciated in any other situation) assistance.

Since I’ve started, I’ve also seen a couple of other bloggers attempt to do these challenges in R so I’m really curious how my solutions will compare to theirs.

The first day of the challenge provides you with a string of numbers and asks you to sum all of the digits that match the next digit in a circular list, i.e. the digit after the last digit is the first digit.

My string was…

8231753674683997878179259195565332579493378483264978184143341284379682788518559178822225126625428318115396632681141871952894291898364781898929292614792884883249356728741993224889167928232261325123447569829932951268292953928766755779761837993812528527484487298117739869189415599461746944992651752768158611996715467871381527675219481185217357632445748912726487669881876129192932995282777848496561259839781188719233951619188388532698519298142112853776942545211859134185231768952888462471642851588368445761489225786919778983848113833773768236969923939838755997989537648222217996381757542964844337285428654375499359997792679256881378967852376848812795761118139288152799921176874256377615952758268844139579622754965461884862647423491918913628848748756595463191585555385849335742224855473769411212376446591654846168189278959857681336724221434846946124915271196433144335482787432683848594487648477532498952572515118864475621828118274911298396748213136426357769991314661642612786847135485969889237193822718111269561741563479116832364485724716242176288642371849569664594194674763319687735723517614962575592111286177553435651952853878775431234327919595595658641534765455489561934548474291254387229751472883423413196845162752716925199866591883313638846474321161569892518574346226751366315311145777448781862222126923449311838564685882695889397531413937666673233451216968414288135984394249684886554812761191289485457945866524228415191549168557957633386991931186773843869999284468773866221976873998168818944399661463963658784821796272987155278195355579386768156718813624559264574836134419725187881514665834441359644955768658663278765363789664721736533517774292478192143934318399418188298753351815388561359528533778996296279366394386455544446922653976725113889842749182361253582433319351193862788433113852782596161148992233558144692913791714859516653421917841295749163469751479835492713392861519993791967927773114713888458982796514977717987598165486967786989991998142488631168697963816156374216224386193941566358543266646516247854435356941566492841213424915682394928959116411457967897614457497279472661229548612777155998358618945222326558176486944695689777438164612198225816646583996426313832539918

My first thought was that I would need to separate this string such that each character was the element of an object, either a vector or a list. I kept things simple and started by just copy-pasting the string into R. I could import it as a .txt file or otherwise but I figured that was unnecessary for such a quick problem. I stored the string as a variable named input.

# Split string after each character.

input_split <- strsplit(input, "")

# As a result, input_split is a list with 1 element:
# a vector containing each character of input as an 
# element. Annoying. Let's unlist() it to extract
# *just* the vector.

char_vector <- unlist(input_split)

# The problem now is that if we are going to sum
# the elements of our string, we need them to be
# numeric and not characters. Easy enough...

num_vector <- as.numeric(char_vector)

# Now lets just initialize our sum...

num_sum = 0

# And use a loop...

for(i in 1:length(num_vector)){

  # If we have the last element of the input string, 
  # set the next number equal to the first element
  # of the string, else select element i + 1.
  next_num <- ifelse(i = length(num_vector), 
  				num_vector[1],
  				num_vector[i + 1])


  # If our current element is equal to the next element,
  # update the sum.
  if(num_vector[i] == next_num){

  	num_sum = num_sum + num_vector[i]

  }
}

num_sum

Our sum is 1390 which is correct, huzzah.

Continue reading Advent of Code 2017 in R: Day 1