10 Base R

Settling In

Sit with whomever you want.

Introduce yourself:

Name, preferred pronouns
Macalester connections (e.g., majors/minors/concentrations, clubs, teams, events regularly attended)

You can download a template Quarto file to start from here. Put this file in a folder called programming within a folder for this course.

Data Storytelling Moment

Go to https://www.nytimes.com/interactive/2017/08/29/opinion/climate-change-carbon-budget.html

What is the data story?
What is effective?
What could be improved?

Learning goals

After this lesson, you should be able to:

Identify and define the properties of common structures in R
Subset vectors and lists with [ by index, name, logical vector, and indirectly with objects
Subset data frames and lists with $ and [[
Use the str() function to examine the structure of an unfamiliar object and extract components from the object
Apply printing strategies to streamline the debugging and development process

Common R Object Structures

Vector: A vector is a collection of elements of the same type (e.g., numeric, integer, character, logical).

. . .

num_vec <- vector("numeric", length = 2) #empty vector: Zeros
num_vec

[1] 0 0

class(num_vec)

[1] "numeric"

length(num_vec)

[1] 2

. . .

log_vec <- vector("logical", length = 3) #empty vector: FALSE
log_vec

[1] FALSE FALSE FALSE

class(log_vec)

[1] "logical"

length(log_vec)

[1] 3

. . .

chr_vec <- vector("character", length = 4) #empty vector: empty strings
chr_vec

[1] "" "" "" ""

class(chr_vec)

[1] "character"

length(chr_vec)

[1] 4

. . .

Fun Fact: A vector can have names for each of its elements.

named_vec <- c('name1' = 1, 'name2' = 2) # Named numeric vector
named_vec

name1 name2 
    1     2

class(named_vec)

[1] "numeric"

length(named_vec)

[1] 2

. . .

List: A list is a collection of elements (e.g., vectors, matrices, data frames, other lists).

A list can have different types of elements.
A list can have names for its elements.

. . .

ex_list <- list(a = 1:3, b = c("a", "b", "c"), c = matrix(1:6, nrow = 2))
ex_list

$a
[1] 1 2 3

$b
[1] "a" "b" "c"

$c
     [,1] [,2] [,3]
[1,]    1    3    5
[2,]    2    4    6

class(ex_list)

[1] "list"

length(ex_list) # number of elements in a list

[1] 3

Other Common R Object Structures

Array: An array is a vector with a dimension attribute.

Like a vector, an array can only have one type of data (e.g., numeric, character).

ary <- array(NA, dim = c(2,3,4))
ary

, , 1

     [,1] [,2] [,3]
[1,]   NA   NA   NA
[2,]   NA   NA   NA

, , 2

     [,1] [,2] [,3]
[1,]   NA   NA   NA
[2,]   NA   NA   NA

, , 3

     [,1] [,2] [,3]
[1,]   NA   NA   NA
[2,]   NA   NA   NA

, , 4

     [,1] [,2] [,3]
[1,]   NA   NA   NA
[2,]   NA   NA   NA

class(ary)

[1] "array"

length(ary) # The number of elements in a array is the product of its dimensions

[1] 24

dim(ary) # Get the dimensions of the array

[1] 2 3 4

. . .

Matrix: A matrix is an array with only 2 dimensions (rows, columns).

Like a vector, a matrix can only have one type of data (e.g., numeric, character).

m <- matrix(NA, nrow = 2, ncol = 3)
m

     [,1] [,2] [,3]
[1,]   NA   NA   NA
[2,]   NA   NA   NA

class(m)

[1] "matrix" "array"

length(m) # The number of elements in a matrix is the product of its dimensions

[1] 6

dim(m) # Get the dimensions of the matrix

[1] 2 3

. . .

Data Frame/tibble: A data frame is a named list with elements of equal length.

Each element is a “column” in the data frame.
The columns can be of different types (e.g., character, numeric, logical, lists, etc.).
Data frames are the most common way to store data in R.
Tibbles do less and complain more than base data.frames

mod_df <- tibble(x = 1:10, y = 1:10 + rnorm(10))

df <- tibble(a = 1:3, b = c("constant", "x", "x squared"), d = list(lm(y ~ 1, data = mod_df), lm(y ~ x, data = mod_df), lm(y ~ x + I(x^2), data = mod_df)))
df

# A tibble: 3 × 3
      a b         d     
  <int> <chr>     <list>
1     1 constant  <lm>  
2     2 x         <lm>  
3     3 x squared <lm>

length(df) # number of "elements" in a data frame is the number of "columns"

[1] 3

Base R Subsetting

The content here comes from Chapter 27 of R4DS, with some small additions.

Selecting elements with `[`

We can subset common R structures and maintain the class structure with [ ].

There are four main types of things that you can subset with, i.e., that can be the i in x[i]:

A vector of positive integers. Subsetting with positive integers keeps the elements at those positions:

# Vectors
x <- c("one", "two", "three", "four", "five")
x[c(3, 2, 5)]

[1] "three" "two"   "five"

x[2:4]

[1] "two"   "three" "four"

class(x[2:4]) # result is a character vector

[1] "character"

. . .

# Lists
y <- list(a = 1:3, b = c("a", "b", "c"), c = matrix(1:6, nrow = 2))
y[c(1, 2)]

$a
[1] 1 2 3

$b
[1] "a" "b" "c"

class(y[c(1)]) # result is a list

[1] "list"

. . .

By repeating a position, you can actually make a longer output than input, making the term “subsetting” a bit of a misnomer.

# Vector
x[c(1, 1, 2)]

[1] "one" "one" "two"

# List
y[c(1, 1, 2)]

$a
[1] 1 2 3

$a
[1] 1 2 3

$b
[1] "a" "b" "c"

. . .

A vector of negative integers. Negative values drop the elements at the specified positions:

# Vector
x[c(-1, -3, -5)]

[1] "two"  "four"

# List
y[c(-1)]

$b
[1] "a" "b" "c"

$c
     [,1] [,2] [,3]
[1,]    1    3    5
[2,]    2    4    6

. . .

A logical vector. Subsetting with a logical vector only keeps values corresponding to TRUE. This is generally used with comparison functions and operators.

# Vector
x <- c(10, 3, NA, 5, 8, 1, NA)

# All non-missing values of x
x[!is.na(x)]

[1] 10  3  5  8  1

# All values greater than 5, with NAs
x[x > 5]

[1] 10 NA  8 NA

# All non-missing values greater than 5
x[x > 5 & !is.na(x)]

[1] 10  8

. . .

Unlike filter(), NA indices will be included in the output as NAs unless you explicitly remove them (filter() removes instances of missing values by default.

# Compare with filter 
filter(tibble(x = x), x > 5)

# A tibble: 2 × 1
      x
  <dbl>
1    10
2     8

. . .

# List
y[c(TRUE, FALSE, TRUE)]

$a
[1] 1 2 3

$c
     [,1] [,2] [,3]
[1,]    1    3    5
[2,]    2    4    6

y[y %>% map_lgl(~ is.numeric(.x))] # example of a map function!

$a
[1] 1 2 3

$c
     [,1] [,2] [,3]
[1,]    1    3    5
[2,]    2    4    6

y[y %>% map_lgl(~ is.character(.x))]

$b
[1] "a" "b" "c"

. . .

A character vector. If you have a named vector or list, you can subset it with a character vector:

# Named Vector
x <- c(abc = 1, def = 2, xyz = 5)
x[c("xyz", "def")]

xyz def 
  5   2

x[c("xyz","xyz","xyz", "def")]

xyz xyz xyz def 
  5   5   5   2

#Named List
y[c('a','a','c')]

$a
[1] 1 2 3

$a
[1] 1 2 3

$c
     [,1] [,2] [,3]
[1,]    1    3    5
[2,]    2    4    6

As with subsetting with positive integers, you can use a character vector to duplicate individual entries.

. . .

Be very wary of vector recycling when doing this! The number of things that you’re inserting should either be 1 or the size of the x[i] subset.

x <- c(first = "one", second = "two", third = "three", fourth = "four")
x

  first  second   third  fourth 
  "one"   "two" "three"  "four"

. . .

x[c(1, 3)] <- "new" # Replacement length is 1
x

 first second  third fourth 
 "new"  "two"  "new" "four"

. . .

x <- c(first = "one", second = "two", third = "three", fourth = "four")
x[c(1, 3)] <- c("new1", "new2") # Replacement length is 2, and length of subset is 2
x

 first second  third fourth 
"new1"  "two" "new2" "four"

. . .

x <- c(first = "one", second = "two", third = "three", fourth = "four")
x[c(1, 3, 4)] <- c("new1", "new2") # BAD! Replacement length is 2, and length of subset is 3
x

 first second  third fourth 
"new1"  "two" "new2" "new1"

. . .

x <- c(first = "one", second = "two", third = "three", fourth = "four")
x[c(1, 3)] <- c("new1", "new2", "new3") # BAD! Replacement length is 3, and length of subset is 2
x

 first second  third fourth 
"new1"  "two" "new2" "four"

Subsetting Matricies and Data Frames with `[`

All of the above subsetting options can be used for subsetting matrices and data frames (named list of elements of equal length).

m <- matrix(1:12, nrow = 3, ncol = 4)
m

     [,1] [,2] [,3] [,4]
[1,]    1    4    7   10
[2,]    2    5    8   11
[3,]    3    6    9   12

m[1:5] # Matrix = vector (down the columns) with dimensions

[1] 1 2 3 4 5

You can use a comma to subset by rows and columns separately.

m[1,] # Get 1st row

[1]  1  4  7 10

m[,1] # Get 1st column

[1] 1 2 3

. . .

m[1,3] # Get 1st row and 3rd column

[1] 7

m[c(1,3),] # Get 1st and 3rd rows

     [,1] [,2] [,3] [,4]
[1,]    1    4    7   10
[2,]    3    6    9   12

m[,c(1,3)] # Get 1st and 3rd columns

     [,1] [,2]
[1,]    1    7
[2,]    2    8
[3,]    3    9

m[c(1,3),c(1,3)] # Get 1st and 3rd rows and 1st and 3rd columns

     [,1] [,2]
[1,]    1    7
[2,]    3    9

m[-1,] # Get all rows except 1st

     [,1] [,2] [,3] [,4]
[1,]    2    5    8   11
[2,]    3    6    9   12

m[c(TRUE, FALSE, FALSE),] # Get the 1st row via a logical

[1]  1  4  7 10

# Add row and column names to the matrix
colnames(m) <- str_c("col", 1:4)
rownames(m) <- str_c("row", 1:3)
m["row1",]

col1 col2 col3 col4 
   1    4    7   10

Selecting a single element with `$` and `[[`

We can use $ and [[ to extract a single column of a data frame or an element within a list. This breaks out of the original class structure.

mtcars

                     mpg cyl  disp  hp drat    wt  qsec vs am gear carb
Mazda RX4           21.0   6 160.0 110 3.90 2.620 16.46  0  1    4    4
Mazda RX4 Wag       21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4
Datsun 710          22.8   4 108.0  93 3.85 2.320 18.61  1  1    4    1
Hornet 4 Drive      21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1
Hornet Sportabout   18.7   8 360.0 175 3.15 3.440 17.02  0  0    3    2
Valiant             18.1   6 225.0 105 2.76 3.460 20.22  1  0    3    1
Duster 360          14.3   8 360.0 245 3.21 3.570 15.84  0  0    3    4
Merc 240D           24.4   4 146.7  62 3.69 3.190 20.00  1  0    4    2
Merc 230            22.8   4 140.8  95 3.92 3.150 22.90  1  0    4    2
Merc 280            19.2   6 167.6 123 3.92 3.440 18.30  1  0    4    4
Merc 280C           17.8   6 167.6 123 3.92 3.440 18.90  1  0    4    4
Merc 450SE          16.4   8 275.8 180 3.07 4.070 17.40  0  0    3    3
Merc 450SL          17.3   8 275.8 180 3.07 3.730 17.60  0  0    3    3
Merc 450SLC         15.2   8 275.8 180 3.07 3.780 18.00  0  0    3    3
Cadillac Fleetwood  10.4   8 472.0 205 2.93 5.250 17.98  0  0    3    4
Lincoln Continental 10.4   8 460.0 215 3.00 5.424 17.82  0  0    3    4
Chrysler Imperial   14.7   8 440.0 230 3.23 5.345 17.42  0  0    3    4
Fiat 128            32.4   4  78.7  66 4.08 2.200 19.47  1  1    4    1
Honda Civic         30.4   4  75.7  52 4.93 1.615 18.52  1  1    4    2
Toyota Corolla      33.9   4  71.1  65 4.22 1.835 19.90  1  1    4    1
Toyota Corona       21.5   4 120.1  97 3.70 2.465 20.01  1  0    3    1
Dodge Challenger    15.5   8 318.0 150 2.76 3.520 16.87  0  0    3    2
AMC Javelin         15.2   8 304.0 150 3.15 3.435 17.30  0  0    3    2
Camaro Z28          13.3   8 350.0 245 3.73 3.840 15.41  0  0    3    4
Pontiac Firebird    19.2   8 400.0 175 3.08 3.845 17.05  0  0    3    2
Fiat X1-9           27.3   4  79.0  66 4.08 1.935 18.90  1  1    4    1
Porsche 914-2       26.0   4 120.3  91 4.43 2.140 16.70  0  1    5    2
Lotus Europa        30.4   4  95.1 113 3.77 1.513 16.90  1  1    5    2
Ford Pantera L      15.8   8 351.0 264 4.22 3.170 14.50  0  1    5    4
Ferrari Dino        19.7   6 145.0 175 3.62 2.770 15.50  0  1    5    6
Maserati Bora       15.0   8 301.0 335 3.54 3.570 14.60  0  1    5    8
Volvo 142E          21.4   4 121.0 109 4.11 2.780 18.60  1  1    4    2

mtcars$mpg

 [1] 21.0 21.0 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 17.8 16.4 17.3 15.2 10.4
[16] 10.4 14.7 32.4 30.4 33.9 21.5 15.5 15.2 13.3 19.2 27.3 26.0 30.4 15.8 19.7
[31] 15.0 21.4

mtcars[["mpg"]]

 [1] 21.0 21.0 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 17.8 16.4 17.3 15.2 10.4
[16] 10.4 14.7 32.4 30.4 33.9 21.5 15.5 15.2 13.3 19.2 27.3 26.0 30.4 15.8 19.7
[31] 15.0 21.4

mtcars %>% pull(mpg)

 [1] 21.0 21.0 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 17.8 16.4 17.3 15.2 10.4
[16] 10.4 14.7 32.4 30.4 33.9 21.5 15.5 15.2 13.3 19.2 27.3 26.0 30.4 15.8 19.7
[31] 15.0 21.4

Exercises

Subsetting Functions For each of the tasks below, write a function that take a vector as input returns the desired output:

The elements at even-numbered positions. (Hint: use the seq() function.)
Every element except the last value.
Only even values (and no missing values).

Exploring the structure of an object with `str()`

The str() function shows you the structure of an object and is useful for exploring model objects and objects created from packages that are new to you.

In the output of str() dollar signs indicate named components of a list that can be extracted via $ or [[.

. . .

We see that both mod and mod_summ are lists, so we can also interactively view these objects with View(mod) and View(mod_summ) in the Console.

mod <- lm(mpg ~ hp + wt, data = mtcars)
mod_summ <- summary(mod)

str(mod)

List of 12
 $ coefficients : Named num [1:3] 37.2273 -0.0318 -3.8778
  ..- attr(*, "names")= chr [1:3] "(Intercept)" "hp" "wt"
 $ residuals    : Named num [1:32] -2.572 -1.583 -2.476 0.135 0.373 ...
  ..- attr(*, "names")= chr [1:32] "Mazda RX4" "Mazda RX4 Wag" "Datsun 710" "Hornet 4 Drive" ...
 $ effects      : Named num [1:32] -113.65 -26.046 -15.894 0.447 0.662 ...
  ..- attr(*, "names")= chr [1:32] "(Intercept)" "hp" "wt" "" ...
 $ rank         : int 3
 $ fitted.values: Named num [1:32] 23.6 22.6 25.3 21.3 18.3 ...
  ..- attr(*, "names")= chr [1:32] "Mazda RX4" "Mazda RX4 Wag" "Datsun 710" "Hornet 4 Drive" ...
 $ assign       : int [1:3] 0 1 2
 $ qr           :List of 5
  ..$ qr   : num [1:32, 1:3] -5.657 0.177 0.177 0.177 0.177 ...
  .. ..- attr(*, "dimnames")=List of 2
  .. .. ..$ : chr [1:32] "Mazda RX4" "Mazda RX4 Wag" "Datsun 710" "Hornet 4 Drive" ...
  .. .. ..$ : chr [1:3] "(Intercept)" "hp" "wt"
  .. ..- attr(*, "assign")= int [1:3] 0 1 2
  ..$ qraux: num [1:3] 1.18 1.08 1.09
  ..$ pivot: int [1:3] 1 2 3
  ..$ tol  : num 1e-07
  ..$ rank : int 3
  ..- attr(*, "class")= chr "qr"
 $ df.residual  : int 29
 $ xlevels      : Named list()
 $ call         : language lm(formula = mpg ~ hp + wt, data = mtcars)
 $ terms        :Classes 'terms', 'formula'  language mpg ~ hp + wt
  .. ..- attr(*, "variables")= language list(mpg, hp, wt)
  .. ..- attr(*, "factors")= int [1:3, 1:2] 0 1 0 0 0 1
  .. .. ..- attr(*, "dimnames")=List of 2
  .. .. .. ..$ : chr [1:3] "mpg" "hp" "wt"
  .. .. .. ..$ : chr [1:2] "hp" "wt"
  .. ..- attr(*, "term.labels")= chr [1:2] "hp" "wt"
  .. ..- attr(*, "order")= int [1:2] 1 1
  .. ..- attr(*, "intercept")= int 1
  .. ..- attr(*, "response")= int 1
  .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv> 
  .. ..- attr(*, "predvars")= language list(mpg, hp, wt)
  .. ..- attr(*, "dataClasses")= Named chr [1:3] "numeric" "numeric" "numeric"
  .. .. ..- attr(*, "names")= chr [1:3] "mpg" "hp" "wt"
 $ model        :'data.frame':  32 obs. of  3 variables:
  ..$ mpg: num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
  ..$ hp : num [1:32] 110 110 93 110 175 105 245 62 95 123 ...
  ..$ wt : num [1:32] 2.62 2.88 2.32 3.21 3.44 ...
  ..- attr(*, "terms")=Classes 'terms', 'formula'  language mpg ~ hp + wt
  .. .. ..- attr(*, "variables")= language list(mpg, hp, wt)
  .. .. ..- attr(*, "factors")= int [1:3, 1:2] 0 1 0 0 0 1
  .. .. .. ..- attr(*, "dimnames")=List of 2
  .. .. .. .. ..$ : chr [1:3] "mpg" "hp" "wt"
  .. .. .. .. ..$ : chr [1:2] "hp" "wt"
  .. .. ..- attr(*, "term.labels")= chr [1:2] "hp" "wt"
  .. .. ..- attr(*, "order")= int [1:2] 1 1
  .. .. ..- attr(*, "intercept")= int 1
  .. .. ..- attr(*, "response")= int 1
  .. .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv> 
  .. .. ..- attr(*, "predvars")= language list(mpg, hp, wt)
  .. .. ..- attr(*, "dataClasses")= Named chr [1:3] "numeric" "numeric" "numeric"
  .. .. .. ..- attr(*, "names")= chr [1:3] "mpg" "hp" "wt"
 - attr(*, "class")= chr "lm"

str(mod_summ)

List of 11
 $ call         : language lm(formula = mpg ~ hp + wt, data = mtcars)
 $ terms        :Classes 'terms', 'formula'  language mpg ~ hp + wt
  .. ..- attr(*, "variables")= language list(mpg, hp, wt)
  .. ..- attr(*, "factors")= int [1:3, 1:2] 0 1 0 0 0 1
  .. .. ..- attr(*, "dimnames")=List of 2
  .. .. .. ..$ : chr [1:3] "mpg" "hp" "wt"
  .. .. .. ..$ : chr [1:2] "hp" "wt"
  .. ..- attr(*, "term.labels")= chr [1:2] "hp" "wt"
  .. ..- attr(*, "order")= int [1:2] 1 1
  .. ..- attr(*, "intercept")= int 1
  .. ..- attr(*, "response")= int 1
  .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv> 
  .. ..- attr(*, "predvars")= language list(mpg, hp, wt)
  .. ..- attr(*, "dataClasses")= Named chr [1:3] "numeric" "numeric" "numeric"
  .. .. ..- attr(*, "names")= chr [1:3] "mpg" "hp" "wt"
 $ residuals    : Named num [1:32] -2.572 -1.583 -2.476 0.135 0.373 ...
  ..- attr(*, "names")= chr [1:32] "Mazda RX4" "Mazda RX4 Wag" "Datsun 710" "Hornet 4 Drive" ...
 $ coefficients : num [1:3, 1:4] 37.22727 -0.03177 -3.87783 1.59879 0.00903 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : chr [1:3] "(Intercept)" "hp" "wt"
  .. ..$ : chr [1:4] "Estimate" "Std. Error" "t value" "Pr(>|t|)"
 $ aliased      : Named logi [1:3] FALSE FALSE FALSE
  ..- attr(*, "names")= chr [1:3] "(Intercept)" "hp" "wt"
 $ sigma        : num 2.59
 $ df           : int [1:3] 3 29 3
 $ r.squared    : num 0.827
 $ adj.r.squared: num 0.815
 $ fstatistic   : Named num [1:3] 69.2 2 29
  ..- attr(*, "names")= chr [1:3] "value" "numdf" "dendf"
 $ cov.unscaled : num [1:3, 1:3] 3.80e-01 2.21e-05 -1.09e-01 2.21e-05 1.21e-05 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : chr [1:3] "(Intercept)" "hp" "wt"
  .. ..$ : chr [1:3] "(Intercept)" "hp" "wt"
 - attr(*, "class")= chr "summary.lm"

Exercise

CI Function Write a function that fits a linear model on the dataset using the given outcome and predictor variables and return a data frame (tibble) with the coefficient estimate and CI for the predictor of interest. It should take the following inputs:

data: A dataset
yvar: Outcome variable to be used in a linear model (a length-1 character vector)
preds: Predictor variables to be used in a linear model (a character vector)
pred_of_interest: The variable whose coefficient estimate and confidence interval are of interest (a length-1 character vector and should be one of preds)

Development tip: As you develop, it will help to create objects for the arguments so that you can see what output looks like interactively:

Test your function on the mtcars dataset.

data <- mtcars
yvar <- "mpg"
preds <- c("hp", "wt")
pred_of_interest <- "hp"

When you’re done developing your function, remove these objects to declutter your environment by entering rm(data, yvar, preds, pred_of_interest) in the Console.

fit_mod_and_extract <- function(___) {
    # Use str_c to create a string (formula_str) that looks like "yvar ~ pred1 + pred2"
    # Look at the documentation for a helpful argument
    mod_formula_str <- 
    mod_form <- as.formula(mod_formula_str)
    
    # Fit a linear model using the constructed formula and given data
    mod <- lm(mod_form, data = data)
    
    # Obtain 95% confidence interval
    ci <- confint(mod, level = 0.95)
    
    # Return the coefficient estimate and CI for the predictor of interest
    tibble(
        which_pred = pred_of_interest,
        estimate = ___,
        ci_lower = ___,
        ci_upper = ___
    )
}

Debugging Strategies

When writing functions and working with functions that you wrote, you may encounter errors that are hard to figure out.

Here are some strategies to help you debug the issues you encounter:

Use print() and cat() to print out intermediate results and messages within a function.
- Examples: print(x), cat("The value of x is", x, "\n")

My_own_sum <- function(x){
  print(x)
  return(sum(x))
}
My_own_sum(c(1,2,3))

My_own_sum <- function(x){
  cat("The value of x is", x, "\n")
  cat("The class of x is", class(x), "\n")
  return(sum(x))
}

My_own_sum(c(1,2,3))

. . .

Use browser() to pause the function at a certain point and interactively explore the environment. Press “Next” or type n to run the next line of code. Type the name of an object in the Console to see its value at this point in the function. You can type Q to quit the browser.
- Example below:

fit_mod_and_extract <- function(data, yvar, preds, pred_of_interest) {
    # Use str_c to create a string (formula_str) that looks like "yvar ~ pred1 + pred2"
    # Look at the documentation for a helpful argument
    mod_formula_str <- str_c(yvar, "~", str_c(preds, collapse = "+"))
    mod_form <- as.formula(mod_formula_str)
    
    # Add browser() to where in the function you'd like to pause and interact in the function environment using the Console
    browser()
    
    # Fit a linear model using the constructed formula and given data
    mod <- lm(mod_form, data = data)
    
    # Obtain 95% confidence interval
    ci <- confint(mod, level = 0.95)
    
    # Return the coefficient estimate and CI for the predictor of interest
    tibble(
        which_pred = pred_of_interest,
        estimate = mod$coefficients[pred_of_interest],
        ci_lower = ci[pred_of_interest, "2.5 %"],
        ci_upper = ci[pred_of_interest, "97.5 %"]
    )
}


fit_mod_and_extract(data = mtcars, yvar = "mpg", preds = c("hp", "wt"), pred_of_interest = "hp")

. . .

Use try() to catch errors and print out a message when an error occurs.
- Example below:

My_own_sum <- function(x){
  return(sum(x))
}

results <- My_own_sum(c("a","b","c"))

Error in sum(x): invalid 'type' (character) of argument

class(results)

Error: object 'results' not found

results <- try(My_own_sum(c("a","b","c")), silent = TRUE)
class(results)

[1] "try-error"

. . .

Include if else statements within a function to ensure that you are passing the right type of input to a function. You can create you own custom error message with stop().
- Example below:

My_own_sum <- function(x){
  if(!is.numeric(x)){
    stop("Input must be numeric")
  }
  return(sum(x))
}

results <- My_own_sum(c("a","b","c"))

Error in My_own_sum(c("a", "b", "c")): Input must be numeric

class(results)

[1] "try-error"

results <- try(My_own_sum(c("a","b","c")), silent = TRUE)
class(results)

[1] "try-error"

Solutions

Subsetting Functions

Solutions

get_even_pos <- function(x) {
    if (length(x) <= 1) {
        print("No even positions")
    } else {
        idx <- seq(2, length(x), by = 2)
        x[idx]
    }
}
get_even_pos(1:10)

[1]  2  4  6  8 10

get_even_pos(1:9)

[1] 2 4 6 8

get_even_pos(1)

[1] "No even positions"

get_all_but_last <- function(x) {
    head(x, -1)
    # x[1:(length(x)-1)]
}
get_all_but_last(1:10)

[1] 1 2 3 4 5 6 7 8 9

get_evens <- function(x) {
    x[x %% 2 == 0 & !is.na(x)]
}

get_evens(c(1, 2, 7, NA))

[1] 2

get_evens(c(1, 2, 7, 8, NA))

[1] 2 8

CI Function

Solutions

fit_mod_and_extract <- function(data, yvar, preds, pred_of_interest) {
    # Use str_c to create a string (formula_str) that looks like "yvar ~ pred1 + pred2"
    # Look at the documentation for a helpful argument
    mod_formula_str <- str_c(yvar, "~", str_c(preds, collapse = "+"))
    mod_form <- as.formula(mod_formula_str)
    
    # Fit a linear model using the constructed formula and given data
    mod <- lm(mod_form, data = data)
    
    # Obtain 95% confidence interval
    ci <- confint(mod, level = 0.95)
    
    # Return the coefficient estimate and CI for the predictor of interest
    tibble(
        which_pred = pred_of_interest,
        estimate = mod$coefficients[pred_of_interest],
        ci_lower = ci[pred_of_interest, "2.5 %"],
        ci_upper = ci[pred_of_interest, "97.5 %"]
    )
}


fit_mod_and_extract(data = mtcars, yvar = "mpg", preds = c("hp", "wt"), pred_of_interest = "hp")

# A tibble: 1 × 4
  which_pred estimate ci_lower ci_upper
  <chr>         <dbl>    <dbl>    <dbl>
1 hp          -0.0318  -0.0502  -0.0133

10.1 After Class

Go back to the Missing Data activity.
- Look for all base R subsetting examples.
- You’ll see iteration with for loops (next time).
Take a look at the Schedule page to see how to prepare for the next class.
Finish Homework 5.
- Complete Reflection in Google Doc shared with you
Project Work
- Look at my feedback/questions for Milestone 1.
- Next Milestone 2 due next week.
I’ll posted HW6 on Wednesday.

Settling In

Data Storytelling Moment

Learning goals

Common R Object Structures

Other Common R Object Structures

Base R Subsetting

Selecting elements with [

Subsetting Matricies and Data Frames with [

Selecting a single element with $ and [[

Exercises

Exploring the structure of an object with str()

Exercise

Debugging Strategies

Solutions

10.1 After Class

Selecting elements with `[`

Subsetting Matricies and Data Frames with `[`

Selecting a single element with `$` and `[[`

Exploring the structure of an object with `str()`