Instructions

Complete all exercises below by filling in your answers in the designated spaces. When finished:

  1. Use this .Rmd file to complete the homework
  2. Complete the homework creating code chunks where needed
  3. Knit the document to HTML (click the “Knit” button or press Ctrl+Shift+K / Cmd+Shift+K)
  4. Submit both your .Rmd file and the knitted Word/PDF (HTMLs are not ‘uploadable’, but can be converted to PDF) to Blackboard

Due: Wednesday, March 18 by 11:59 PM

Part 1: Building Your First Functions

Question 1 (1 point)

Foresters use diameter at breast height (DBH, in cm) to estimate a tree’s basal area — the cross-sectional area of the trunk. The formula is:

\[BA = \frac{\pi}{4} \times DBH^2\]

where \(BA\) is in cm\(^2\).

  1. Write a function called basal_area that takes a single argument dbh (in cm) and returns the basal area. Give dbh a default value of 25.

  2. Test your function on these inputs and show the output:

    • basal_area() (using the default)
    • basal_area(10)
    • basal_area(c(10, 25, 40, 55))

  3. Notice that your function works on a single number and on a vector. In one sentence, explain why this happens without you having to write any special loop or extra code.

Delete this text and write your explanation here.

Question 2 (1 point)

Wildlife biologists sometimes need to convert animal body mass between units depending on the journal, the country, or the dataset they inherited. Writing a converter saves time and prevents arithmetic mistakes.

  1. Write a function called mass_convert that takes three arguments:

    • mass — a numeric vector (no default; the user must supply this)
    • from — a character string, with a default of "kg"
    • to — a character string, with a default of "lb"

    The function should handle conversions between "kg", "lb", and "g" (grams). Use the conversion factors: 1 kg = 2.20462 lb = 1000 g. If from and to are the same, just return the original mass.

    Hint: one clean approach is to first convert everything to kg (regardless of from), then convert from kg to the target unit.

  2. Show the output of each of the following:

mass_convert(100)                          # 100 kg to lb
mass_convert(100, from = "lb", to = "kg")  # 100 lb to kg
mass_convert(5000, from = "g", to = "lb")  # 5000 g to lb
mass_convert(80, from = "kg", to = "kg")   # no conversion needed
  1. The average mass of an adult male black bear in the Adirondacks is roughly 125 kg. Use your function to express that in pounds and in grams.

Delete this text and show your code here.

Part 2: Functions That Summarize Data

Question 3 (1.5 points)

Aquatic ecologists assess water quality partly by looking at dissolved oxygen (DO) readings from a stream over time. A quick health check might report the mean, the minimum (stressful for fish), and how many readings fall below a critical threshold.

  1. Write a function called do_report that takes two arguments:

    • do_vec — a numeric vector of dissolved oxygen readings (mg/L)
    • threshold — a numeric value representing the stress threshold (default = 5 mg/L)

    The function should return a named list with four elements:

    • mean_do — the mean of do_vec (handle NAs)
    • min_do — the minimum value (handle NAs)
    • n_below — the count of readings below the threshold (ignore NAs)
    • pct_below — the percentage of readings below the threshold, rounded to one decimal place (ignore NAs)

  2. Create a test vector of dissolved oxygen readings to simulate a warm summer month on a sluggish creek:

creek_do <- c(6.2, 5.8, 4.1, 3.9, 7.0, 5.5, 4.8, 6.1, NA, 3.2, 
              5.0, 4.5, 6.8, 5.3, 4.0, 7.2, 5.9, 3.5, 4.7, 6.0)

Run do_report(creek_do) and show the output.

  1. Now run do_report(creek_do, threshold = 4.5). How does raising the bar change the picture?

  2. Your function returns a list. Show how you would extract just the pct_below element from the result using $ notation.

Delete this text and write your explanation here.

Question 4 (1 point)

The trees dataset built into R contains measurements of 31 black cherry trees: Girth (diameter in inches), Height (in feet), and Volume (in cubic feet). Foresters estimate timber volume from these measurements using various allometric equations. A simple one is:

\[V = a \times G^b \times H^c\]

where \(G\) is girth, \(H\) is height, and \(a\), \(b\), \(c\) are fitted constants.

  1. Write a function called est_volume that takes five arguments: girth, height, a, b, and c. Give the constants these defaults: a = 0.003, b = 2.0, c = 1.0. The function should return the estimated volume.

  2. Load the trees dataset. Use your function to estimate volume for every tree using the defaults:

data(trees)
v_est <- est_volume(trees$Girth, trees$Height)
  1. Now compare your estimates to the actual Volume column. Compute the mean absolute error:

\[MAE = \frac{1}{n}\sum_{i=1}^{n}|V_{estimated,i} - V_{actual,i}|\]

You can do this in one line using mean() and abs().

  1. Try tweaking the constants to reduce the MAE. Report the values of a, b, and c you settled on and the resulting MAE. You don’t need to find the best possible fit — just show that changing the arguments changes the output.

Delete this text and write your explanation here.

Part 3: Scope and Debugging

Question 5 (0.75 points)

  1. Run the following code exactly as written and explain the output of each exists() call:
wind_chill <- function(temp_f, wind_mph) {
  wc <- 35.74 + 0.6215 * temp_f - 35.75 * wind_mph^0.16 + 
        0.4275 * temp_f * wind_mph^0.16
  return(wc)
}

wind_chill(20, 15)
exists("wc")
exists("wind_chill")
exists("temp_f")

  1. In your own words, explain why wc and temp_f do not exist in your global environment after you call the function, even though both were clearly created or used inside it.

  2. Suppose you wanted to store the wind chill for a 20-degree day with 15 mph winds so you could use it later in your script. Write the one line of code that accomplishes this.

Delete this text and write your explanation here.

Question 6 (0.75 points)

A colleague in the wildlife department wrote a function to classify black bear body condition from mass (in kg), but it’s not working. Here is their code:

bear_condition <- function(mass_kg) {
  if (mass_kg < 60) {
    condition <- "underweight"
  } else if (mass_kg >= 60 & mass_kg <= 100) {
    condition <- "normal"
  } else if (mass_kg > 100) {
    condtion <- "heavy"
  }
  return(condition)
}

  1. Test the function with bear_condition(50) and bear_condition(80). These work fine. Now test it with bear_condition(130). What happens?

  2. Identify the bug. Describe the problem in one sentence.

  3. Fix the function and show that it works.

  4. Add browser() as the first line inside the function. Run bear_condition(130) again. In the browser console, step through the code line by line (type n at the prompt). At each step, check whether condition exists yet by typing condition. At what point does the problem become visible? Type Q to quit the browser. Remove browser() when finished.

Explain in your own words how browser() helped you find the issue:

Delete this text and write your explanation here.

Question 7 (1 pt.)

The following function was presented in the lecture on loops:

computeFib <- function(n){
  fib <- rep(NA, n)
  fib[1] <- 1
  fib[2] <- 1
  for(i in 3:length(fib)){
    fib[i] <- fib[i-1] + fib[i-2]
  }
  fib
}

  1. Modify this function so that (1) returns just the final (nth) value of the Fibonnaci sequence, rather than the whole sequence, and (2) takes as an argument a vector of two initial values, that can be arbitrary.

  2. Use this function to test whether the ratio of consecutive elements of the Fibonacci sequence converges to the golden ratio \(1 + \sqrt{5} \over 2\) regardless of the initial values.

Submission Checklist

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This homework is worth 6 points total.