Basic Practice

Please try to complete the following problems and email your solutions to Chris at christopher_magnano@hms.harvard.edu. It is okay if you are not able to complete the exercises, you may also email Chris whatever you are able to complete within an hour. You can email your code as an R script or report, or inside a document.

Set 1: Complete by September 8

1. What will the class and values in vec1 to vec4 be after the following lines of code? Try to figure it out without running the code, then check your work:

vec1 <- c("4", T, F, F)
vec2 <- c(9, "banana", 16, "apple")
vec3 <- c(TRUE + TRUE, TRUE, FALSE, TRUE)
vec4 <- c(3**2, 19%%2)

Solution

vec1 will be a character vector with "4" "TRUE" "FALSE" "FALSE".

vec2 will be a character vector with "9" "banana" "16" "apple".

vec3 will be an integer vector with 2 1 0 1. Integer is a type similar to numeric but only for whole numbers.

vec4 will be a numeric vector with 9 1.

2. Write an R command or set of commands to make the vec1 vector twice as large and contain all of the values of the original vector twice.

For example, using this command [1,2,3] should become [1,2,3,1,2,3], ["a"] should become ["a", "a"], and [2,3,1,2,1] should become [2,3,1,2,1,2,3,1,2,1].

Solution

There many ways this can be done in R, but lets look at two of the most common ways. The first (and one we learning during the bootcamp) is to use c():

vec1 <- c(vec1, vec1)

We could also use the rep function:

rep(vec1, 2)

Using rep is nice as we can easily change the number of times we want to repeat the vector, as opposed to having to manually type out each vec1.

3. For this problem you will need to load in the nhanes example data and, if desired, load the tidyverse package.

library(tidyverse)

nhanes <- read.csv("https://raw.githubusercontent.com/ccb-hms/hsdm-r-course/main/session-materials/session0/session0Data.csv")

1. Calculate the mean plasma glucose in the nhanes dataset, ignoring missing values.

Solution

Here we can just use mean with the na.rm argument set to TRUE. We might also want to round the mean to the nearest whole number, so it is simlilar to the rest of the plasma glucose values.

library(tidyverse)

── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
✔ dplyr     1.1.3     ✔ readr     2.1.4
✔ forcats   1.0.0     ✔ stringr   1.5.0
✔ ggplot2   3.4.4     ✔ tibble    3.2.1
✔ lubridate 1.9.3     ✔ tidyr     1.3.0
✔ purrr     1.0.2     
── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
✖ dplyr::filter() masks stats::filter()
✖ dplyr::lag()    masks stats::lag()
ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors

nhanes <- read.csv("https://raw.githubusercontent.com/ccb-hms/hsdm-r-course/main/session-materials/session0/session0Data.csv")

plasma_mean <- round(mean(nhanes$plasma.glucose, na.rm = TRUE))

2. Create a dataframe or tibble, nhanes_mean_filled, where all missing plasma glucose values in nhanes have been replaced by the mean value calculated in part a. Hint: Check out the replace_na function, you can find its documentation here.

Solution

Now that we have the mean plasma glucose, we can use it to fill in the NA values in the dataset. Note that in R we can set multiple indices of a vector to a single value. To do this in Tidyverse we would need to find the replace_na function, but it base R we can use simple indexing.

# Solution from part a
plasma_mean <- round(mean(nhanes$plasma.glucose, na.rm = TRUE))

# The base R way:
nhanes_mean_filled <- nhanes
nhanes_mean_filled$plasma.glucose[is.na(nhanes_mean_filled$plasma.glucose)] <- plasma_mean

# The Tidyverse way
nhanes_mean_filled <- nhanes |>
  mutate(plasma.glucose = replace_na(plasma.glucose, plasma_mean))

#double check NA count
sum(is.na(nhanes_mean_filled$plasma.glucose))

[1] 0

Set 2: Due September 22

Start by loading tidyverse and reading in the nhanes example data from the bootcamp:

library(tidyverse)

nhanes <- read_csv("https://raw.githubusercontent.com/ccb-hms/hsdm-r-course/main/session-materials/session0/session0Data.csv")

1. The dyplr package, a part of the tidyverse set of packages, contains an arrange function.

1. What does the arrange function do?
2. What is the arrange function’s default behavior towards NA values?

Hint: Recall that you can look up information on a function with the command ?function.

Solution

1. arrange orders the rows of a data frame by the values of selected columns. In other words, arrange is sorting our dataframe by the chosen column.

2. arrange will always put missing data at the end, no matter which way you sorted the data. The documentation also mentions an exception for remote data, but we don’t need to worry about that.

2. Sort nhanes by BMI from lowest to highest and store the result in a new variable, nhanes_sorted.

Solution

nhanes_sorted <- nhanes |>
  arrange(bmi)

3. Add an additional column to nhanes_sorted called bmi_rank which contains each participant’s BMI rank, meaning that the value for the lowest BMI would be 1, the second lowest BMI would be 2, etc. all the way up to the highest BMI. Don’t worry about ties or preserving the original order of the data.

Hint: You might want to use the seq function or colon notation (1:num) to generate a sequence of numbers for the new column.

Solution

Here, we can use seq with mutate to create a new column which contains the ranks. Since the data is already sorted, if we just increase the value by 1 each row we will naturally get the correct ranking value.

# We can lay out the length
nhanes_sorted <- nhanes_sorted |>
  mutate(bmi_rank = seq(1:nrow(nhanes_sorted)))

# Or just give the column directly as an argument to seq. 
# Seq will create a sequence of the same length as an input vector. 
nhanes_sorted <- nhanes_sorted |>
  mutate(bmi_rank = seq(bmi))

Set 3: Due October 13

Start by loading tidyverse and reading in the nhanes example data from the bootcamp:

library(tidyverse)

nhanes <- read_csv("https://raw.githubusercontent.com/ccb-hms/hsdm-r-course/main/session-materials/session0/session0Data.csv")

1. Write R statements using your choice of base R or Tidyverse to compute the following counts:

1. The number of participants 14 or older.
2. The number of participants 14 or older, or whose BMI is greater than 25 (or both).
3. The number of participants who meet the criteria for part b and have dental decay present.

2. Right now, in the nhanes example data the birthplace variable is a bit unwieldy. We can see all of the values with unique:

unique(nhanes$birthplace)

[1] "Born in 50 US States or Washi"  "Born in Mexico"                
[3] "Born Elsewhere"                 "Born in Other Non-Spanish Spea"
[5] "Born in Other Spanish Speaking" "Refused"                       
[7] "Don't Know"                    

These category names are generally cut-off, and many of the categories contain very few participants, which we can see with count:

nhanes |>
  count(birthplace)

                      birthplace    n
1                 Born Elsewhere   58
2  Born in 50 US States or Washi 2938
3                 Born in Mexico  230
4 Born in Other Non-Spanish Spea   62
5 Born in Other Spanish Speaking   56
6                     Don't Know    1
7                        Refused    1

Write R code to convert the birthplace variable into a factor with only two levels: "In USA" and "Outside USA". Remove the rows with Don't Know or Refused responses.

Hint: There are multiple ways to go about the steps needed here. Note that birthplace is currently a character column. You can remove the Don't Know or Refused rows using negative selection with indexing or filter. You can then construct a large OR (|) statement inside mutate, use indexing to set values to new text value-by-value, try using factor all at once, or you can take a sneak peek at future weeks and try one of the specialized functions in the reading on recoding values.

3. Write R statements using your choice of base R or Tidyverse to compute the difference between the mean family poverty income ratio (PIR) of participants born in the United States versus outside the United States.

Bonus: Try also calculating the difference in the percent of participants with dental caries between those born in or outside the United States.