differences for PR #620

swcarpentry · Apr 19, 2024 · e7d9bc2 · e7d9bc2
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diff --git a/01-starting-with-data.md b/01-starting-with-data.md
@@ -121,17 +121,13 @@ read.csv(file = "data/inflammation-01.csv")
 
 :::::::::::::::  solution
 
-## Solution
-
 R will construct column headers from values in your first row of data,
 resulting in `X0 X0.1 X1 X3 X1.1 X2 ...`.
 
 Note that the character `X` is prepended: a standalone number would not be a valid variable
 name. Because column headers are variables, the same naming rules apply.
 Appending `.1`, `.2` etc. is necessary to avoid duplicate column headers.
 
-
-
 :::::::::::::::::::::::::
 
 ::::::::::::::::::::::::::::::::::::::::::::::::::
@@ -146,8 +142,6 @@ Take a look at `?read.csv` and write the code to load a file called `commadec.tx
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 read.csv(file = "data/commadec.txt", sep = ";", dec = ",")
 ```
@@ -366,8 +360,6 @@ age <- age - 20
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 mass <- 47.5
 age <- 122
@@ -827,8 +819,6 @@ animal[4:6]
 
 :::::::::::::::  solution
 
-## Solutions
-
 1. `animal[4:1]`
 
 2. `"o" "n" "k" "e" "y"` and `"m" "o" "n" "e" "y"`, which means that a
@@ -857,8 +847,6 @@ Which of the following lines of R code gives the correct answer?
 
 :::::::::::::::  solution
 
-## Solution
-
 Answer: 3
 
 Explanation: You want to extract the part of the dataframe representing data for patient 5 from days three to seven. In this dataframe, patient data is organised in rows and the days are represented by the columns. Subscripting in R follows the `[i, j]` principle, where `i = rows` and `j = columns`. Thus, answer 3 is correct since the patient is represented by the value for i (5) and the days are represented by the values in j, which is a subset spanning day 3 to 7.
@@ -880,8 +868,6 @@ Let's pretend there was something wrong with the instrument on the first five da
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 whichPatients <- seq(2, 60, 2) # i.e., which rows
 whichDays <- seq(1, 5)         # i.e., which columns
@@ -916,8 +902,6 @@ apply call and check your intuition by applying the mean function.
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 # 1.
 apply(dat[1:5, ], 1, mean)
@@ -987,8 +971,6 @@ Create a plot showing the standard deviation of the inflammation data for each d
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 sd_day_inflammation <- apply(dat, 2, sd)
 plot(sd_day_inflammation)
@@ -998,8 +980,6 @@ plot(sd_day_inflammation)
 
 ::::::::::::::::::::::::::::::::::::::::::::::::::
 
-
-
 :::::::::::::::::::::::::::::::::::::::: keypoints
 
 - Use `variable <- value` to assign a value to a variable in order to record it in memory.
@@ -1014,5 +994,3 @@ plot(sd_day_inflammation)
 - Use `plot` to create simple visualizations.
 
 ::::::::::::::::::::::::::::::::::::::::::::::::::
-
-
diff --git a/02-func-R.md b/02-func-R.md
@@ -188,8 +188,6 @@ highlight(best_practice, asterisk)
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 highlight <- function(content, wrapper) {
   answer <- c(wrapper, content, wrapper)
@@ -214,8 +212,6 @@ edges(dry_principle)
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 edges <- function(v) {
    first <- v[1]
@@ -268,8 +264,6 @@ mySum <- function(input_1, input_2 = 10) {
 
 :::::::::::::::  solution
 
-## Solution
-
 1. The solution is `a.`.
 
 2. Read the error message: `argument "input_1" is missing, with no default`
@@ -535,8 +529,6 @@ Be sure to document your function with comments.
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 analyze <- function(filename) {
   # Plots the average, min, and max inflammation over time.
@@ -567,8 +559,6 @@ Test that your `rescale` function is working properly using `min`, `max`, and `p
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 rescale <- function(v) {
   # Rescales a vector, v, to lie in the range 0 to 1.
@@ -775,8 +765,6 @@ both are given the same input vector and parameters?
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 rescale <- function(v, lower = 0, upper = 1) {
   # Rescales a vector, v, to lie in the range lower to upper.

diff --git a/03-loops-R.md b/03-loops-R.md
@@ -271,8 +271,6 @@ print_N(3)
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 print_N <- function(N) {
   nseq <- seq(N)
@@ -306,8 +304,6 @@ total(ex_vec)
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 total <- function(vec) {
   # calculates the sum of the values in a vector
@@ -351,8 +347,6 @@ expo(2, 4)
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 expo <- function(base, power) {
   result <- 1
@@ -525,8 +519,6 @@ and runs `analyze` for each file whose name matches the pattern.
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 analyze_all <- function(folder = "data", pattern) {
   # Runs the function analyze for each file in the given folder

diff --git a/04-cond.md b/04-cond.md
@@ -331,8 +331,6 @@ plot_dist(dat[1:5, 10], threshold = 10)  # samples (rows) 1-5 on day (column) 10
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 plot_dist <- function(x, threshold) {
   if (length(x) > threshold) {
@@ -385,8 +383,6 @@ plot_dist(dat[1:5, 10], threshold = 10)                    # samples (rows) 1-5
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 plot_dist <- function(x, threshold, use_boxplot = TRUE) {
    if (length(x) > threshold && use_boxplot) {
@@ -438,8 +434,6 @@ print(average_inf_max)
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 # Add your code here ...
 if (patient_average_inf > average_inf_max) {
@@ -605,8 +599,6 @@ update `analyze`, and then recreate all the figures with `analyze_all`.
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 analyze <- function(filename, output = NULL) {
   # Plots the average, min, and max inflammation over time.

diff --git a/05-cmdline.md b/05-cmdline.md
@@ -344,8 +344,6 @@ Rscript arith.R 3 - 4
 
 :::::::::::::::  solution
 
-## Solution
-
 
 ```bash
 cat arith.R
@@ -387,8 +385,6 @@ main()
 
 :::::::::::::::  solution
 
-## Solution
-
 An error message is returned due to "invalid input."
 This is likely because '\*' has a special meaning in the shell, as a wildcard.
 
@@ -412,8 +408,6 @@ print-args.R
 
 :::::::::::::::  solution
 
-## Solution
-
 
 ```bash
 cat find-pattern.R
@@ -531,8 +525,6 @@ What is the best way to test your program?
 
 :::::::::::::::  solution
 
-## Solution
-
 
 ```bash
 cat check.R
@@ -674,8 +666,6 @@ Separately, modify the program so that if no action is specified (or an incorrec
 
 :::::::::::::::  solution
 
-## Solution
-
 
 ```bash
 cat readings-short.R
@@ -867,8 +857,6 @@ Write a program called `line-count.R` that works like the Unix `wc` command:
 
 :::::::::::::::  solution
 
-## Solution
-
 
 ```bash
 cat line-count.R

diff --git a/06-best-practices-R.md b/06-best-practices-R.md
@@ -164,8 +164,6 @@ rm(list = ls()) # If you want to delete all the objects in the workspace and sta
 
 :::::::::::::::  solution
 
-## Solution
-
 
 ```bash
 cat inflammation.R

diff --git a/10-supp-addressing-data.md b/10-supp-addressing-data.md
@@ -116,8 +116,6 @@ Think about the number of rows and columns you would expect as the result.
 
 :::::::::::::::  solution
 
-## Solution
-
 
 ```r
 dat[1, 1]
@@ -141,8 +139,6 @@ What will be returned by `dat[, 2]`?
 
 :::::::::::::::  solution
 
-## Solution
-
 
 ```r
 dat[, 2]
@@ -184,8 +180,6 @@ Use the colon operator to index just the aneurism count data (columns 6 to 9).
 
 :::::::::::::::  solution
 
-## Solution
-
 
 ```r
 dat[, 6:9]
@@ -591,8 +585,6 @@ plot(dat[dat$Group == 'Control', ]$BloodPressure)
 
 :::::::::::::::  solution
 
-## Solution
-
 1. The code for such a plot:
 
   ```r
@@ -603,8 +595,6 @@ plot(dat[dat$Group == 'Control', ]$BloodPressure)
 2. In addition to
   `dat$Group != 'Control'`, one could use
   `dat$Group %in% c("Treatment1", "Treatment2")`.
-
-
 
 :::::::::::::::::::::::::
 
@@ -634,8 +624,6 @@ Combine the addressing and assignment operations to convert all values to lowerc
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 dat[dat$Gender == 'M', ]$Gender <- 'm'
 dat[dat$Gender == 'F', ]$Gender <- 'f'

diff --git a/11-supp-read-write-csv.md b/11-supp-read-write-csv.md
@@ -263,8 +263,6 @@ without importing the data with `stringsAsFactors=FALSE`.
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 carSpeeds <- read.csv(file = 'data/car-speeds.csv')
 # Replace 'Blue' with 'Green' in cars$Color without using the stringsAsFactors
@@ -335,8 +333,6 @@ Is there a way to specify more than one 'string', such as "Black" and "Blue", to
 
 :::::::::::::::  solution
 
-## Solution
-
 ```r
 read.csv(file = "data/inflammation-01.csv", na.strings = "0")
 ```