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From n00b to Pro

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From n00b to Pro

Jack Davis

Andrew Henrey

Create a simulator from scratch that:

- Generates data from a variety of distributions
- Makes a response variable from a known function of the data (plus an error term)
- Constructs a linear model that estimates the coefficients of the function
- Repeats generation and modeling many times to compare the average estimates of the linear model to the known parameters.
- Package the whole thing nicely into a function that we can call in a single line in later work.
- If you’re experienced, the commands themselves may seem trivial

- 1) Learning how to learn
- 2) Randomly Generating Data
- 3) Data Frames and Manipulation
- 4) Linear Models
- BREAK – Quality of presenter improves
- 5) Running loops
- 6) Function Definition
- 7) More advanced function topics
- 8) Using functions
- 9) A short simulation study

- Google CRAN Packages to get the package list
- From here you can get a description of every command in a package.

- ??<term> searches for commands related to <term>
- ??plot will find commands related to plot

- ?<command> calls up the help file for that command
- ?ablinegives the help file for the abline() command.

- Exercises:
- Name one function in the darts game package.
- What is the e-mail of the author of the Texas Holdem simulation package?
- (Bonus) Tell the author about your day via e-mail; s/he likes hearing from fans.
- Find a function to make a histogram
- Find some example code on the heatmap() command.

- The r<dist> commands randomly generate data from a distribution
- rnorm( n , mean, sd) Generates from normal distribution (default N(0,1))
- rexp( n, rate)
- rbinom( n, size, prob)
- rt( n, df) From Student’s T. (Mean is zero, so setting a mean is up to you)
- set.seed() Allows you to generate the same data every time, so you or others can verify work.

- Set a random seed
- Generate a vector of 50 values from the Normal (mean=10,sd=4) distribution, name the vector x1.
- Do the same with
- Poisson ( lambda = 5), named x2,
- Exponential (rate = 1/7) named x3,
- Student’s t distribution (df =5), with a mean of 5, named x4,
- Normal (mean=0, sd=20), named err

- Make a new variable y, let it be 3 + 20x1 + 15x2 – 12x3 – 10x4 + err

- data.frame() makes a dataframe object of the vectors listed in the ()
- The advantage of having a data frame is that it can be treated as a single object..
- Data frames, models, and even matrix decompositions can be objects in R.
- You can call parts of objects by name using $
- model$coefor model$coefficient will bring up the estimated coefficients
- If no such aspect exists, then you’ll get a null response.
- Example:Andrew$height

- Exercises:
- Make a data.frame() of x1,x2,x3,x4, and y Name it dat
- (if you’re stuck from the last part, run “Q3-dataframethis.txt” first)
- Use index indicators like dat[4,3], dat [2:7,3], dat [4,], and dat [4,-1] to get
- The 3rd row, 5th entry of dat
- The 2nd – 7th values of the 5th column
- The entire 3rd row
- The 3rd row without the 1st entry

- The results of the lm() function are an object.
- Example: mod = lm(y ~ x1 + I(x2^2) + x1:x2, data=dat)
- Useful aspects
- mod$fitted
- mod$residuals

- Useful functions
- summary(mod)
- predict(mod, newdata)

- Use the lm command to create a linear model of y as a function of x1,x2,x3, and x4 additively using dat data, name it mod. (No interactions or transformations)
- Get the summary of mod
- Display the estimated coefficients with no other values.

- This slide unintentionally left 98% blank

- 1) Learning how to learn
- 2) Randomly Generating Data
- 3) Data Frames and Manipulation
- 4) Linear Models
- BREAK – Quality of presenter improves deteriorates
- 5) Running loops
- 6) Function Definition
- 7) More advanced function topics
- 8) Using functions
- 9) A short simulation study

- Similar to other programming languages, loops in R allow you to repeat the same block of code several times
- Unlike other programming languages, large loops in R are exceedingly slow
- Any loop of less than about 100,000 total iterations is not going to give you much trouble in terms of time

- An R loop that executes a million commands takes about a second. Conditions vary wildly
- Generating 100,000 data sets of size 50,000 and looping through the dataset to calculate a mean for each one would take longer to run than Jack Davis heading up Burnaby Mountain (ouch)

Sup d00dz late for tutorial

Jack

- Loop syntax:
for (i in 1:n)

{

#TellVicEverything

}

- No need to run from 1:K
- Can use an arbitrary vector instead
- Runs for length(vect) iterations
- Takes on the ith value of the vector each iteration
- e.g.
- V = c(1,5,3,-6)
- for (count in V) {print(count);}
- ## 1, 5 , 3 , -6

- Exercises:
- A) Define a variable runs to be the number 10,000
- B) Define a matrix() called mat with 5 columns and runs rows (10,000 rows)
- C) Put a for() loop around the code found in q5-loopthis.txt . Loop from 1:runs.
Use index indicators like a[k,] to save the estimated coefficients of the model in a new row of mat.

OR, if you think you are a total coding BOSS, then put the loop around your code in parts 2-4 that generates data and finds the linear model estimates of the betas.

- Functions are a slightly abstract concept
- Mathematics: f(x) = x2+4x-16
- Computing: mean(x) = sum(x)/length(x) – All 3 are functions!
- Functions map INPUTS to OUTPUT
- Possibly no inputs
- In one way or another, always some form of output
- Example functions:
- SORT, MEDIAN, OPTIM / NLM, LM/GLM

- Function syntax
- Simple function:
F = function()

{

return (5)

}

>> F()

5

- Exercises:
- Make a function out of the code you wrote in part 5. The syntax should be similar to the previous slide. The function:
- Should be called simulate.lm
- Should include everything needed to generate the data several times, find a linear model, and extract the coefficients
- Does NOT take any inputs
- Should return the matrix of 10,000 runs of coefficients

- Use the function and save the results to a matrix called test
- If nothing is working (), you can use the example code in q6 – function this.txt

- A more complicated example:
MSE = function(X=c(0,3,11),Y)

{

return (mean((X-Y)^2))

}

- Observe that X has default values
>>MSE(Y=c(4,5,6))

15

- If an input argument to a function has default values, you don’t have to specify them when calling the function
- If an input argument has no default values, running the function without specifying them gives you an error

- Exercises:
- Modify simulate.lm() by adding input parameters. Include:
- nruns, the number of runs in the simulation, with no default
- seed, the random initial seed of the simulation, defaulting to 1337
- verbose, a Boolean true/false to report progress, defaulting to FALSE (caps matter)
- Set runs to nruns at the beginning of your function
- Use set.seed(seed) in your function
- Add code that prints out how far along you are in the loop, but only when verbose is true
- Run this new function to overwrite the old one

- Exercises:
- A) Run your simulate.lm() with 25000 runs. Store these results as a variable called betas
- B) Use hist() on the first column of betas to see the sampling distribution of the intercept
- C) Use summary(), mean() , and sd() on this column as well
- D) Use par(mfrow=c(2,2)) and then some hist() commands to display histograms of the other four sampling distributions in a 2x2 grid
- E) Compare your results to the known values (The means of the sampling distributions to the true values, and the standard deviations to the estimated values in Q4)

- Idea: You have a binomial experiment with 9 successes and 3 failures.
- You would like to construct a 95% CI for the true proportion of successes
- You DON’T know whether the normal approximation is appropriate
- How can we find out whether or not it’s OK?

- Overall procedure:
- Construct a LOT of samples from a population with 12 trials and p=0.75
- For each sample, calculate the 95% CI using the normal approximation
- For each sample, see whether the CI overlaps with 0.75
- Count the number of samples for which the CI overlaps with 0.75
- The proportion of the samples that have a CI that overlaps is called the “true coverage probability”
- If the true coverage probability is close to 95% , the normal approximation to the sampling distribution of p is a good one.

- Steps:
- Generate 100,000 samples of binomial(12,0.75) data using rbinom()
- For each sample, calculate the usual estimate of p
- For each sample, calculate SE = sqrt(p*(1-p)/12)
- For each sample, calculate the lower and upper bounds of the 95% CI
- Find out how many intervals actually contain 0.75
- Optional: Look at hist(x) to gain intuition of why the normal approximation isn’t perfect

- Leave plztyvm