Ch. 19: Functions

Key questions:

  • 19.3.1. #1
  • 19.4.4. #2, 3
  • 19.5.5. #2 (actually make a function that fixes this)
Functions and notes:
  • function_name <- function(input1, input2) {}
  • if () {}
  • else if () {}
  • else {}
  • || (or) , && (and) – used to combine multiple logical expressions
  • | and & are vectorized operations not to be used with if statements
  • any checks if any values in vector are TRUE
  • all checks if all values in vector are TRUE
  • identical strict check for if values are the same
  • dplyr::near for more lenient comparison and typically better than identical as unaffected by floating point numbers and fine with different types
  • switch evaluate selected code based on position or name (good for replacing long chain of if statements). e.g. 
Operation_Times2 <- function(x, y, op){
   first_op <- switch(op,
     plus = x + y,
     minus = x - y,
     times = x * y,
     divide = x / y,
     stop("Unknown op!")
   )
   
   first_op * 2
  }

Operation_Times2(5, 7, "plus")
## [1] 24
  • stop stops expression and executes error action, note that .call default is FALSE
  • warning generates a warning that corresponds with arguments, suppressWarnings may also be useful to no
  • stopifnot test multiple args and will produce error message if any are not true – compromise that prevents you from tedious work of putting multiple if(){} else stop() statements
  • ... useful catch-all if your function primarily wraps another function (note that misspelled arguments will not raise an error), e.g. 
commas <- function(...) stringr::str_c(..., collapse = ", ")
commas(letters[1:10])
## [1] "a, b, c, d, e, f, g, h, i, j"
  • cut can be used to discretize continuous variables (also saves long if statements)
  • return allows you to return the function early, typically reserve use for when the function can return early with a simpler function
  • cat function to print label in output
  • invisible input does not get printed out

19.2: When should you write a function?

19.2.1

  1. Why is TRUE not a parameter to rescale01()? What would happen if x contained a single missing value, and na.rm was FALSE?

    • TRUE doesn’t change between uses.
    • The output would be NA

  2. In the second variant of rescale01(), infinite values are left unchanged. Rewrite rescale01() so that -Inf is mapped to 0, and Inf is mapped to 1.

    rescale01_inf <- function(x){

  rng <- range(x, na.rm = TRUE, finite = TRUE)
  x_scaled <- (x - rng[1]) / (rng[2] - rng[1])
  is_inf <- is.infinite(x)
  is_less0 <- x < 0
  x_scaled[is_inf & is_less0] <- 0
  x_scaled[is_inf & (!is_less0)] <- 1
  x_scaled
}

x <-  c(Inf, -Inf, 0, 3, -5)
rescale01_inf(x)
    ## [1] 1.000 0.000 0.625 1.000 0.000

  3. Practice turning the following code snippets into functions. Think about what each function does. What would you call it? How many arguments does it need? Can you rewrite it to be more expressive or less duplicative?

    mean(is.na(x))

x / sum(x, na.rm = TRUE)

sd(x, na.rm = TRUE) / mean(x, na.rm = TRUE)
    • See solutions below:
    x <- c(1, 4, 2, 0, NA, 3, NA)

#mean(is.na(x))
perc_na <- function(x) {
  is.na(x) %>% mean()
}

perc_na(x)
    ## [1] 0.2857143
    #x / sum(x, na.rm = TRUE)
prop_weighted <- function(x) {
  x / sum(x, na.rm = TRUE)
}
prop_weighted(x)
    ## [1] 0.1 0.4 0.2 0.0  NA 0.3  NA
    #sd(x, na.rm = TRUE) / mean(x, na.rm = TRUE)
CoefficientOfVariation <- function(x) {
  sd(x, na.rm = TRUE) / mean(x, na.rm = TRUE)
}

CoefficientOfVariation(x)
    ## [1] 0.7905694

  4. Follow http://nicercode.github.io/intro/writing-functions.html to write your own functions to compute the variance and skew of a numeric vector.

    • Re-do below to write measures for skew and variance (e.g. kurtosis, etc.)
    var_bry <- function(x){
  sum((x - mean(x)) ^ 2) / (length(x) - 1)
}

skewness_bry <- function(x) {
  mean((x - mean(x)) ^ 3) / var_bry(x) ^ (3 / 2)
}

    Let’s create some samples of distributions – normal, t (with 7 degrees of freedom), unifrom, poisson (with lambda of 2).

    Note that an example with a cauchy distribution and looking at difference in kurtosis between that and a normal distribution has been moved to the Appendix section 19.2.1.4.

    nd <- rnorm(10000)
td_df7 <- rt(10000, df = 7)
ud <- runif(10000)
pd_l2 <- rpois(10000, 2)

    Verify that these functions match with established functions

    dplyr::near(skewness_bry(pd_l2), e1071::skewness(pd_l2, type = 3))
    ## [1] TRUE
    dplyr::near(var_bry(pd_l2), var(pd_l2))
    ## [1] TRUE

    Let’s look at the distributions as well as their variance an skewness

    distributions_df <- tibble(normal_dist = nd,
       t_7df_dist = td_df7,
       uniform_dist = ud,
       poisson_dist = pd_l2)

distributions_df %>% 
  gather(normal_dist:poisson_dist, value = "sample", key = "dist_type") %>% 
  mutate(dist_type = factor(forcats::fct_inorder(dist_type))) %>% 
  ggplot(aes(x = sample))+
  geom_histogram()+
  facet_wrap(~ dist_type, scales = "free")
    ## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

    tibble(dist_type = names(distributions_df),
skewness = purrr::map_dbl(distributions_df, skewness_bry), 
variance = purrr::map_dbl(distributions_df, var_bry))
    ## # A tibble: 4 x 3
##   dist_type    skewness variance
##   <chr>           <dbl>    <dbl>
## 1 normal_dist   0.0561    0.988 
## 2 t_7df_dist   -0.0263    1.40  
## 3 uniform_dist -0.00678   0.0833
## 4 poisson_dist  0.680     1.98

  5. Write both_na(), a function that takes two vectors of the same length and returns the number of positions that have an NA in both vectors.

    both_na <- function(x, y) {
  if (length(x) == length(y)) {
  sum(is.na(x) & is.na(y))
  } else
  stop("Vectors are not equal length")
}

x <- c(4, NA, 7, NA, 3)
y <- c(NA, NA, 5, NA, 0)
z <- c(NA, 4)

both_na(x, y)
    ## [1] 2
    both_na(x, z)
    ## Error in both_na(x, z): Vectors are not equal length

  6. What do the following functions do? Why are they useful even though they are so short?

    is_directory <- function(x) file.info(x)$isdir
is_readable <- function(x) file.access(x, 4) == 0
    • first checks if what is being referred to is actually a directory
    • second checks if a specific file is readable

  7. Read the complete lyrics to “Little Bunny Foo Foo”. There’s a lot of duplication in this song. Extend the initial piping example to recreate the complete song, and use functions to reduce the duplication.

19.3: Functions are for humans and computers

  • Recommends snake_case over camelCase, but just choose one and be consistent
  • When functions have a link, common prefix over suffix (i.e. input_select, input_text over, select_input, text_input)
  • ctrl + shift + r creates section breaks in R scripts like below

# test label --------------------------------------------------------------

  • (though these cannot be made in markdown documents)

19.3.1

  1. Read the source code for each of the following three functions, puzzle out what they do, and then brainstorm better names.

    f1 <- function(string, prefix) {
  substr(string, 1, nchar(prefix)) == prefix
}
f2 <- function(x) {
  if (length(x) <= 1) return(NULL)
  x[-length(x)]
}
f3 <- function(x, y) {
  rep(y, length.out = length(x))
}
    • f1: check_prefix
    • f2: return_not_last
    • f3: repeat_for_length

  2. Take a function that you’ve written recently and spend 5 minutes brainstorming a better name for it and its arguments.

    • done seperately

  3. Compare and contrast rnorm() and MASS::mvrnorm(). How could you make them more consistent?

    • uses mu = and Sigma = instead of mean = and sd = , and has extra parameters like tol, empirical, EISPACK
    • Similar in that both are pulling samples from gaussian distribution
    • mvrnorm is multivariate though, could change name to rnorm_mv

  4. Make a case for why norm_r(), norm_d() etc would be better than rnorm(), dnorm(). Make a case for the opposite.

    • norm_* would show the commonality of them being from the same distribution. One could argue the important commonality though may be more related to it being either a random sample or a density distribution, in which case the r* or d* coming first may make more sense. To me, the fact that the help pages has all of the ‘normal distribution’ functions on the same page suggests the former may make more sense. However, I actually like having it be set-up the way it is, because I am more likely to forget the name of the distribution type I want over the fact that I want a random sample, so it’s easier to type r and then do ctrl + space and have autocomplete help me find the specific distribution I want, e.g. rnorm, runif, rpois, rbinom…

19.4: Conditional execution

Function example that uses if statement:

has_name <- function(x) {
  nms <- names(x) 
  if (is.null(nms)) { 
    rep(FALSE, length(x))
  } else {
    !is.na(nms) & nms != ""
  }
}
  • note that if all names are blank, it returns the one-unit vector value NULL, hence the need for the if statement here…33

19.4.4.

  1. What’s the difference between if and ifelse()? Carefully read the help and construct three examples that illustrate the key differences.

    • ifelse is vectorized, if is not
      • Typically use if in functions when giving conditional options for how to evaluate
      • Typically use ifelse when changing specific values in a vector
    • If you supply if with a vector of length > 1, it will use the first value
    x <- c(3, 4, 6)
y <- c("5", "c", "9")

# Use `ifelse` simple transformations of values
ifelse(x < 5, 0, x)
    ## [1] 0 0 6
    # Use `if` for single condition tests
cutoff_make0 <- function(x, cutoff = 0){
  if(is.numeric(x)){
    ifelse(x < cutoff, 0, x)
  } else stop("The input provided is not a numeric vector")
}

cutoff_make0(x, cutoff = 4)
    ## [1] 0 4 6
    cutoff_make0(y, cutoff = 4)
    ## Error in cutoff_make0(y, cutoff = 4): The input provided is not a numeric vector

  2. Write a greeting function that says “good morning”, “good afternoon”, or “good evening”, depending on the time of day. (Hint: use a time argument that defaults to lubridate::now(). That will make it easier to test your function.)

    greeting <- function(when) {

  time <- hour(when)

  if (time < 12 && time > 4) {
    greating <- "good morning"
  } else if (time < 17 && time >= 12) {
    greeting <- "good afternoon"
  } else greeting <- "good evening"

  when_char <- as.character(when)
  mid <- ", it is: "
  cat(greeting, mid, when_char, sep = "")
}

greeting(now())
    ## good evening, it is: 2019-06-05 19:28:02

  3. Implement a fizzbuzz function. It takes a single number as input. If the number is divisible by three, it returns “fizz”. If it’s divisible by five it returns “buzz”. If it’s divisible by three and five, it returns “fizzbuzz”. Otherwise, it returns the number. Make sure you first write working code before you create the function.

    fizzbuzz <- function(x){
  if(is.numeric(x) && length(x) == 1){
    y <- ""
    if (x %% 5 == 0) y <- str_c(y, "fizz")
    if (x %% 3 == 0) y <- str_c(y, "buzz")
    if (str_length(y) == 0) {
      print(x)
    } else print(y)
  } else stop("Input is not a numeric vector with length 1")
}

fizzbuzz(4)
    ## [1] 4
    fizzbuzz(10)
    ## [1] "fizz"
    fizzbuzz(6)
    ## [1] "buzz"
    fizzbuzz(30)
    ## [1] "fizzbuzz"
    fizzbuzz(c(34, 21))
    ## Error in fizzbuzz(c(34, 21)): Input is not a numeric vector with length 1

  4. How could you use cut() to simplify this set of nested if-else statements?

    if (temp <= 0) {
  "freezing"
} else if (temp <= 10) {
  "cold"
} else if (temp <= 20) {
  "cool"
} else if (temp <= 30) {
  "warm"
} else {
  "hot"
}
    • Below is example of fix
    temp <- seq(-10, 50, 5)
cut(temp, 
    breaks = c(-Inf, 0, 10, 20, 30, Inf),  #need to include negative and positive infiniity
    labels = c("freezing", "cold", "cool", "warm", "hot"),
    right = TRUE,
    oredered_result = TRUE)
    ##  [1] freezing freezing freezing cold     cold     cool     cool    
##  [8] warm     warm     hot      hot      hot      hot     
## Levels: freezing cold cool warm hot

    How would you change the call to cut() if I’d used < instead of <=? What is the other chief advantage of cut() for this problem? (Hint: what happens if you have many values in temp?)

    • See below change to right argument
    cut(temp, 
    breaks = c(-Inf, 0, 10, 20, 30, Inf),  #need to include negative and positive infiniity
    labels = c("freezing", "cold", "cool", "warm", "hot"),
    right = FALSE,
    oredered_result = TRUE)
    ##  [1] freezing freezing cold     cold     cool     cool     warm    
##  [8] warm     hot      hot      hot      hot      hot     
## Levels: freezing cold cool warm hot

  5. What happens if you use switch() with numeric values?

    • It will return the index of the argument.
      • In example below, I input ‘3’ into switch value so it does the times argument
    math_operation <- function(x, y, op){
   switch(op,
     plus = x + y,
     minus = x - y,
     times = x * y,
     divide = x / y,
     stop("Unknown op!")
   )
  }

math_operation(5, 4, 3)
    ## [1] 20

  6. What does this switch() call do? What happens if x is “e”?

    x <- "e"

switch(x, 
  a = ,
  b = "ab",
  c = ,
  d = "cd"
)

    Experiment, then carefully read the documentation.

    • If x is ‘e’ nothing will be outputted. If x is ‘c’ or ‘d’ then ‘cd’ is outputted. If ‘a’ or ‘b’ then ‘ab’ is outputeed. If blank it will continue down list until reaching an argument to output.

19.5: Function arguments

Common non-descriptive short argument names:

  • x, y, z: vectors.
  • w: a vector of weights.
  • df: a data frame.
  • i, j: numeric indices (typically rows and columns).
  • n: length, or number of rows.
  • p: number of columns.

19.5.5.

  1. What does commas(letters, collapse = "-") do? Why?

    • commas function is below
    commas <- function(...) stringr::str_c(..., collapse = ", ")

commas(letters[1:10])
    ## [1] "a, b, c, d, e, f, g, h, i, j"
    commas(letters[1:10], collapse = "-")
    ## Error in stringr::str_c(..., collapse = ", "): formal argument "collapse" matched by multiple actual arguments
    • The above fails because are essentially specifying two different values for the collapse argument
    • Takes in vector of mulitple strings and outputs one-unit character string with items concatenated together and seperated by columns
    • Is able to do this via use of ... that turns this into a wrapper on stringr::str_c with the collapse value specified

  2. It’d be nice if you could supply multiple characters to the pad argument, e.g. rule("Title", pad = "-+"). Why doesn’t this currently work? How could you fix it?

    • current rule function is below
    rule <- function(..., pad = "-") {
  title <- paste0(...)
  width <- getOption("width") - nchar(title) - 5
  cat(title, " ", stringr::str_dup(pad, width), "\n", sep = "")
}

# Note that `cat` is used instead of `paste` because paste would output it as a character vector, whereas `cat` is focused on just ouptut, could also have used `print`, though print does more conversion than cat does (apparently)

rule("Tis the season"," to be jolly")
    ## Tis the season to be jolly --------------------------------------------
    • doesn’t work because pad ends-up being too many characters in this situation
    rule("Tis the season"," to be jolly", pad="+-")
    ## Tis the season to be jolly +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
    • instead would need to make the number of times pad is duplicated dependent on its length, see below for fix
    rule_pad_fix <- function(..., pad = "-") {
  title <- paste0(...)
  width <- getOption("width") - nchar(title) - 5
  width_fix <- width %/% stringr::str_length(pad)
  cat(title, " ", stringr::str_dup(pad, width_fix), "\n", sep = "")
}

rule_pad_fix("Tis the season"," to be jolly", pad="+-")
    ## Tis the season to be jolly +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-

  3. What does the trim argument to mean() do? When might you use it?

    • trim specifies proportion of data to take off from both ends, good with outliers
    mean(c(-1000, 1:100, 100000), trim = .025) 
    ## [1] 50.5

  4. The default value for the method argument to cor() is c("pearson", "kendall", "spearman"). What does that mean? What value is used by default?

    • is showing that you can choose from any of these, will default to use pearson (value in first position)

19.6: Return values

show_missings <- function(df) {
  n <- sum(is.na(df))
  cat("Missing values: ", n, "\n", sep = "")
  invisible(df)
}

x <- show_missings(mtcars)
## Missing values: 0
str(x)
## 'data.frame':    32 obs. of  11 variables:
##  $ mpg : num  21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
##  $ cyl : num  6 6 4 6 8 6 8 4 4 6 ...
##  $ disp: num  160 160 108 258 360 ...
##  $ hp  : num  110 110 93 110 175 105 245 62 95 123 ...
##  $ drat: num  3.9 3.9 3.85 3.08 3.15 2.76 3.21 3.69 3.92 3.92 ...
##  $ wt  : num  2.62 2.88 2.32 3.21 3.44 ...
##  $ qsec: num  16.5 17 18.6 19.4 17 ...
##  $ vs  : num  0 0 1 1 0 1 0 1 1 1 ...
##  $ am  : num  1 1 1 0 0 0 0 0 0 0 ...
##  $ gear: num  4 4 4 3 3 3 3 4 4 4 ...
##  $ carb: num  4 4 1 1 2 1 4 2 2 4 ...
  • can still use in pipes
mtcars %>% 
  show_missings() %>% 
  mutate(mpg = ifelse(mpg < 20, NA, mpg)) %>% 
  show_missings() 
## Missing values: 0
## Missing values: 18

Appendix

19.2.1.4

Function for Standard Error:

x <- c(5, -2, 8, 6, 9)
sd(x, na.rm = TRUE) / sqrt(sum(!is.na(x)))
## [1] 1.933908
sample_se <- function(x) {
  sd(x, na.rm = TRUE) / sqrt(sum(!is.na(x)) - 1)
}    
#sqrt(var(x)/sum(!is.na(x)))

sample_se(x)
## [1] 2.162175

Function for kurtosis:

kurtosis_type3 <- function(x){
  sum((x - mean(x)) ^ 4) / length(x) / sd(x) ^ 4 - 3
}

Notice differences between cauchy and normal distribution

set.seed(1235)
norm_exp <- rnorm(10000)

set.seed(1235)
cauchy_exp <- rcauchy(10000)
hist(norm_exp)
hist(cauchy_exp)

kurtosis

kurtosis_type3(norm_exp)
## [1] 0.06382172
kurtosis_type3(cauchy_exp)
## [1] 1197.052

19.2.3.5

position_both_na <- function(x, y) {
  if (length(x) == length(y)) {
  (c(1:length(x)))[(is.na(x) & is.na(y))]
  } else
  stop("Vectors are not equal length")
}

x <- c(4, NA, 7, NA, 3)
y <- c(NA, NA, 5, NA, 0)
z <- c(NA, 4)
both_na(x, y)
## [1] 2
both_na(x, z)
## Error in both_na(x, z): Vectors are not equal length
  • specifies position where both are NA
  • second example shows returning of ‘stop’ argument

  1. NULL types are not vectors but only single length elements of a different type. is.null is not vectorized in the way that is.na. E.g. it’s job is to return TRUE if given a NULL. For example, a list of NULLs is a list type (not a NULL type) therefore the following: is.null(list(NULL, NULL)) would be FALSE – to return a list of TRUE values you would need to run map(list(NULL, NULL), is.null).↩