tidyverse

Question 1

count

Answer 1

rawmorg05 %>% count(month, sort = TRUE)
# A tibble: 12 × 2
month n
1 1 [January] 27346
2 8 [August] 27217

add_count adds a column with the counts by group

Question 2

summarize

Answer 2

With grouping:
mtcars %>%
group_by(cyl) %>%
summarise(mean = mean(disp), n = n())
#> # A tibble: 3 × 3
#> cyl mean n
#> <dbl> <dbl> <int>
#> 1 4 105. 11
#> 2 6 183. 7
#> 3 8 353. 14</int></dbl></dbl>

Question 3

summarize with group_by

Answer 3

group_by() tells R that we want to look at the dataset in terms of different groups,
instead of just a single block. If we use group_by()beforehand, we are subtly splitting
up the data into different groups. Then, summarize() will work on each group of data. Below, we use group_by() to tell R that we want to calculate the average for each student’s tests.

avg_score_by_student <- student_scores %>%
group_by(names) %>%
summarize(
avg_writing = mean(new_writing_score)
)

Question 4

across

Answer 4

graphtrain %>%
summarize(across(where(is.numeric), mean))
remember to give the function without parentheses

 We can find that out by supplying two functions to across(): one to compute the median and the other to count the missing values. You supply multiple functions by using a named list to .fns:

df_miss |>
summarize(
across(a:d, list(
median = function(x) median(x, na.rm = TRUE),
n_miss = function(x) sum(is.na(x))
)),
n = n()
)
#> # A tibble: 1 × 9
#> a_median a_n_miss b_median b_n_miss c_median c_n_miss d_median d_n_miss
#> <dbl> <int> <dbl> <int> <dbl> <int> <dbl> <int>
#> 1 0.139 1 -1.11 1 -0.387 2 1.15 0
#> # ℹ 1 more variable: n <int></int></int></dbl></int></dbl></int></dbl></int></dbl>

Question 5

c_across

Answer 5

c_across rowwise operations, e.g. creating a column that is a sum of a few columns
df <- tibble(id = 1:4, w = runif(4), x = runif(4), y = runif(4), z = runif(4))
df %>%
rowwise() %>%
mutate(
sum = sum(c_across(w:z)),
sd = sd(c_across(w:z))
)
#> # A tibble: 4 × 7
#> # Rowwise:
#> id w x y z sum sd
#> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 1 0.790 0.588 0.142 0.225 1.74 0.305
#> 2 2 0.892 0.514 0.781 0.207 2.39 0.305
#> 3 3 0.327 0.317 0.456 0.659 1.76 0.159
#> 4 4 0.351 0.408 0.234 0.715 1.71 0.205</dbl></dbl></dbl></dbl></dbl></dbl></int>

Question 6

case_when

Answer 6

case_when(
x %% 35 == 0 ~ “fizz buzz”,
x %% 5 == 0 ~ “fizz”,
x %% 7 == 0 ~ “buzz”,
.default = as.character(x)
)

Question 7

contains()

Answer 7

sat_results <- sat_results %>%
mutate(across(contains(“SAT”),as.numeric))

contains with multiple choices, acts like an OR

mtcars %>%
select(contains(c(“m”, “ar”))

This does the same thing as the following:

mtcars %>%
select(matches(‘m|ar’)) %>%
head(2)
# mpg am gear carb
#Mazda RX4 21 1 4 4
#Mazda RX4 Wag 21 1 4 4

Question 8

distinct

Answer 8

distinct set .keep_all to TRUE to keep all the columns, default is FALSE (surprisingly to me)
mtcars |> distinct(gear, .keep_all = TRUE)

Question 9

filter something containing a certain character string

Answer 9

mtcars$type <- rownames(mtcars)
mtcars %>%
filter(str_detect(type, ‘Toyota|Mazda’))

Question 10

joins

Answer 10

inner_join(), right_join(), full_join() have the same interface as left_join(). The
difference is which rows they keep: left join keeps all the rows in x, the right join keeps all rows in y, the full join keeps all rows in either x or y, and the inner join only keeps rows that occur in both x and y.

Question 11

matches

Answer 11

You want to return every column in your data whose name contains a specific string or regular expression.
Solution

table1 %>%
select(matches(“o.*u”))

Question 12

Variable name in character vector, but no function call issue

Answer 12

for (var in names(mtcars)) {
mtcars %>% count(.data[[var]]) %>% print()
}

Question 13

Passing name of variable in var, and then putting in tidyverse function

Answer 13

var_summary <- function(data, var) { data %>% summarise(n = n(), min = min({{ var }}), max = max({{ var }})) }

mtcars %>% group_by(cyl) %>% var_summary(mpg)

Question 14

Generating a variable name programmatically

Answer 14

name <- “susan”
tibble(“{name}” := 2)

Question 15

Select variables from names in a character vector

Answer 15

vars <- c(“mpg”, “vs”)
mtcars %>% select(all_of(vars))
mtcars %>% select(!all_of(vars))

select(any_of(vars)). would select all of them if there, but not give error of they don’t exist

Question 16

magine you have this simple tibble and you want to count the number of observations, and compute the median of every column

df <- tibble(
a = rnorm(10),
b = rnorm(10),
c = rnorm(10),
d = rnorm(10)
)

Answer 16

df |> summarize(
n = n(),
across(a:d, median),
)
#> # A tibble: 1 × 5
#> n a b c d
#> <int> <dbl> <dbl> <dbl> <dbl>
#> 1 10 -0.246 -0.287 -0.0567 0.144</dbl></dbl></dbl></dbl></int>

Question 17

Use two functions with across

Answer 17

df_miss |>
summarize(
across(a:d, list(
median = function(x) {median(x, na.rm = TRUE)},
n_miss = function(x) {sum(is.na(x))}
)),
n = n()
)
#> # A tibble: 1 × 9
#> a_median a_n_miss b_median b_n_miss c_median c_n_miss d_median d_n_miss
#> <dbl> <int> <dbl> <int> <dbl> <int> <dbl> <int>
#> 1 0.139 1 -1.11 1 -0.387 2 1.15 0
#> # ℹ 1 more variable: n <int></int></int></dbl></int></dbl></int></dbl></int></dbl>

Question 18

filter to just rows with at least one of columns a to d being NA

Answer 18

same as df_miss |> filter(is.na(a) | is.na(b) | is.na(c) | is.na(d))

df_miss |> filter(if_any(a:d, is.na))
#> # A tibble: 4 × 4
#> a b c d
#> <dbl> <dbl> <dbl> <dbl>
#> 1 0.434 -1.25 NA 1.60
#> 2 NA -1.43 -0.297 0.776
#> 3 -0.156 -0.980 NA 1.15
#> 4 1.11 NA -0.387 0.704</dbl></dbl></dbl></dbl>

df_miss |> filter(if_all(a:d, is.na))
#> # A tibble: 0 × 4
#> # ℹ 4 variables: a <dbl>, b <dbl>, c <dbl>, d <dbl></dbl></dbl></dbl></dbl>

Question 19

Function to summarize over all of a set of variables

Answer 19

summarize_means <- function(df, summary_vars = where(is.numeric)) {
df |>
summarize(
across({{ summary_vars }}, function(x) {mean(x, na.rm = TRUE)}),
n = n(),
.groups = “drop”
)
}
diamonds |>
group_by(cut) |>
summarize_means()
#> # A tibble: 5 × 9
#> cut carat depth table price x y z n
#> <ord> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <int>
#> 1 Fair 1.05 64.0 59.1 4359. 6.25 6.18 3.98 1610
#> 2 Good 0.849 62.4 58.7 3929. 5.84 5.85 3.64 4906
#> 3 Very Good 0.806 61.8 58.0 3982. 5.74 5.77 3.56 12082
#> 4 Premium 0.892 61.3 58.7 4584. 5.97 5.94 3.65 13791
#> 5 Ideal 0.703 61.7 56.0 3458. 5.51 5.52 3.40 21551</int></dbl></dbl></dbl></dbl></dbl></dbl></dbl></ord>

diamonds |>
group_by(cut) |>
summarize_means(c(carat, x:z))
#> # A tibble: 5 × 6
#> cut carat x y z n
#> <ord> <dbl> <dbl> <dbl> <dbl> <int>
#> 1 Fair 1.05 6.25 6.18 3.98 1610
#> 2 Good 0.849 5.84 5.85 3.64 4906
#> 3 Very Good 0.806 5.74 5.77 3.56 12082
#> 4 Premium 0.892 5.97 5.94 3.65 13791
#> 5 Ideal 0.703 5.51 5.52 3.40 21551</int></dbl></dbl></dbl></dbl></ord>

Question 20

Compute length of each column using map (just as an example of using map, obviously it’s the same for all of them)

Answer 20

df %>%
map(length)

Question 21

get type of a vector

Answer 21

typeof(vec)

Question 22

pivot_longer

Answer 22

songs

track wk1 wk2 wk3
Song A 1 5 9
Song B 4 3 3

songs |>
pivot_longer(
cols = starts_with(“wk”),
names_to = “week”,
values_to = “rank”
)
track week rank
Song A 1 1
Song A 2 5
Song A 3 9
Song B 1 4

Question 23

pivot_wider

Answer 23

Obs CaseType Accepted
A Single Yes
A Family No
B Single No
B Family Yes

pivot_wider(names_from=”CaseType”, values_from=”Accepted”)

Obs Single Family
A Yes No
B No Yes

Question 24

S4 class in R

Answer 24

create a class “Student_Info” with three member variables

setClass(“Student_Info”, slots=list(name=”character”, age=”numeric”, GPA=”numeric”))

student1 <- new(“Student_Info”, name = “John”, age = 21, GPA = 3.5)

student1

Question 25

give col sums of a data frame as a vector, as example of map, of course there is an easier way to do this

Answer 25

df |> map_dbl(sum)

Question 26

pass variables in function to group_by, for example

Answer 26

group_by() https://dplyr.tidyverse.org/reference/group_by.html uses
data-masking, not tidy-selection. We can work around that problem by using
the handy pick() https://dplyr.tidyverse.org/reference/pick.html
function, which allows you to use tidy-selection inside data-masking
functions:

df %>% count(pick(starts_with(“z”))) #> # A tibble: 3 × 3

count_missing <- function(df, group_vars, x_var) {
df |>
group_by(pick({{ group_vars }})) |>
summarize(
n_miss = sum(is.na({{ x_var }})),
.groups = “drop”
)
}

Question 27

search for function you can’t remember

Answer 27

apropos(“replace”)

Question 28

apply rows vs. columns

Answer 28

You want apply (see the docs for it). apply(var,1,fun) will apply to rows, apply(var,2,fun) will apply to columns.

a
c.1..2..3. c.10..0..6.
1 1 10
2 2 0
3 3 6

> apply(a,1,min)
[1] 1 0 3

Question 29

assign with a list of things

Answer 29

paramlist=list(test.this = .20, vec.test=c(“hi”,”bye”))

for (ii.param in 1:length(paramlist)){
assign(x=names(paramlist)[[ii.param]],value = paramlist[[ii.param]])
}

assigns test.this to value .20
and vec.test to (“hi”,”bye”)

Question 30

time a few lines

Answer 30

start.time <- proc.time()
cat(“It took “, as.numeric(getElement(proc.time(),”elapsed”) - getElement(start.time,”elapsed”)), “seconds to download the data.\n”)

Question 31

lubridate

Answer 31

ymd(“2017-01-31”), ymd_hms(“2017-01-31 20:11:59”), year(), wday() weekday label=TRUE to get Tuesday;

expand_dates <- function(df) {
df |>
mutate(
across(where(is.Date), list(year = year, month = month, day = mday))
)
}

Question 32

lubridate time differences

Answer 32

durations (exact # of seconds), periods (weeks and months), intervals (start and end point)

Question 33

get number distinct, with summarize

Answer 33

mtcars |> summarize(across(everything(), n_distinct))

Question 34

break

Answer 34

use break, not break()

Question 35

formulas

Answer 35

as.formula,

formula.gbm.without.msa <-
reformulate(response=”weekpay”, termlabels= predictors.including.msas)

Question 36

apply list of functions to something

Answer 36

x2 <- list(c(5,6,7))
x2
[[1]]
[1] 5 6 7
for (f2 in c(“mean”,”min”)){
+ print(do.call(f2,x2))
+ }

[1] 6
[1] 5

Question 37

ggplot2 aes basic example

Answer 37

mtcars %>% ggplot(mapping=aes(x=wt, y=mpg)) +
geom_point()

with colors for different values of cyl:
mtcars %>% ggplot(mapping=aes(x=wt, y=mpg, color=cyl)) + geom_point()

Question 38

labels and title

Answer 38

labs(x=”mpg”, y=”hwy”, title=”A Nice title”)

Question 39

bar plot

Answer 39

ggplot(mpg, aes(x = drv, fill = drv)) +
geom_bar()

Note that fill is here set to be equal to the same variable as x.

Question 40

bar plot, side by side

Answer 40

diamonds %>%
ggplot(aes(x=color, fill = cut)) +
geom_bar(position=”dodge”)

Question 41

frequency polygon

Answer 41

geom_freqpoly

Question 42

facet_wrap

Answer 42

facet wrap draws 5 plots, one for each level of marstat:
graphtrain %>%
ggplot(aes(x=age, y=weekpay)) +
geom_point() +
facet_wrap(~ marstat)

Question 43

facet_grid

Answer 43

facet grid, draws grid for levels of 2 variables

Question 44

flipping axes ggplot2

Answer 44

ggplot(data = diamonds, mapping = aes(x = cut)) +
geom_bar() +
coord_flip()

Question 45

counts over 2 continuous variables

Answer 45

ggplot(smaller, aes(x = carat, y = price)) +
geom_bin2d()

Question 46

counts over 2 discrete variables

Answer 46

ggplot(diamonds, aes(x = cut, y = color)) +
geom_count()

Question 47

limit y axis

Answer 47

ggplot(diamonds, aes(x = y)) +
geom_histogram(binwidth = 0.5) +
coord_cartesian(ylim = c(0, 50))

Question 48

save ggplot

Answer 48

defaults to the last plot
ggsave(filename, device=”pdf”)

also can do pdf() with print(ggplot() …)

Question 49

grep

Answer 49

x <- c(“hid”, “hai”, “hirsute”, “bla”)
grep(pattern=”hi”,x)
[1] 1 3
> grep(pattern=”hi”,x, value=TRUE)
[1] “hid” “hirsute”

grep(pattern,
x, value = TRUE) to return the values

Question 50

flatten lists

Answer 50

list_flatten, in purrr

Question 51

get first element of each element of a list, or element labeled x of each element

Answer 51

sapply(templist[1:3], FUN=”[“,1)

sapply(tt.cases, FUN= function(x){ x$id})

Question 52

eval(parse())

Answer 52

x <- list(CA = c(1,2), NV =c(3,4), AZ=c(5,6))
state = “CA”
eval(parse(text= paste(“y<- x$”, state, sep=””)))

This actually does (and evaluates):

y <- x$CA

REMEMBER to include the ‘text=’ part!!

Question 53

pick

Answer 53

use pick() within something like mutate, so you can use the things you can do with
select()
(mutate uses data-masking, select use tidy selection

pick() provides a way to easily select a subset of columns from your data using
select() semantics while inside a “data-masking” function like mutate() or
summarise(). pick() returns a data frame containing the selected columns for the
current group.

my_group_by <- function(data, cols) {
group_by(data, pick({{ cols }}))
}

df %>% my_group_by(c(x, starts_with(“z”)))

Question 54

modular arithmetic

Answer 54

n %% 20

Question 55

math expressions in plots

Answer 55

or maybe use quote with ggplot2?
plot(x,y,
xlab = expression(paste(“Text here “, hat(x), “ here “, z^rho, “ and here”)),
ylab = expression(paste(“Here is some text of “, phi^{rho})),
main = “Expressions with Text”)

Question 56

pivot longer more complicated

Answer 56

household
#> # A tibble: 5 × 5
#> family dob_child1 dob_child2 name_child1 name_child2
#> <int> <date> <date> <chr> <chr>
#> 1 1 1998-11-26 2000-01-29 Susan Jose
#> 2 2 1996-06-22 NA Mark NA
#> 3 3 2002-07-11 2004-04-05 Sam Seth
#> 4 4 2004-10-10 2009-08-27 Craig Khai
#> 5 5 2000-12-05 2005-02-28 Parker Gracie</chr></chr></date></date></int>

Note that we have two pieces of information (or values) for each child: their name and
their dob (date of birth). These need to go into separate columns in the result. Again
we supply multiple variables to names_to, using names_sep to split up each variable name. Note the special name .value: this tells pivot_longer() that that part of the column name specifies the “value” being measured (which will become a variable in the output).

household %>%
pivot_longer(
cols = !family,
names_to = c(“.value”, “child”),
names_sep = “_”,
values_drop_na = TRUE
)

Question 57

pivot wider

Answer 57

pivot_wider() is the opposite of pivot_longer(): it makes a dataset wider by increasing the number of columns and decreasing the number of rows. It’s relatively rare to need pivot_wider() to make tidy data, but it’s often useful for creating summary tables for presentation, or data in a format needed by other tools.

fish_encounters
#> # A tibble: 114 × 3
#> fish station seen
#> <fct> <fct> <int>
#> 1 4842 Release 1
#> 2 4842 I80_1 1
#> 3 4842 Lisbon 1
#> 4 4842 Rstr 1</int></fct></fct>

gets converted into

fish_encounters %>%
pivot_wider(
names_from = station,
values_from = seen
)

Question 58

Type 1 error , type 2

Answer 58

False positive, i.e mistaken rejection of null hypothesis

False negative or mistaken lack of rejection

Question 59

Sensitivity

Specificity

Answer 59

Sensitivity: Rate of getting a positive, given that it’s a positive

Rate of getting a negative, given that it’s actually a negative

Question 60

Loops being slow

Answer 60

For loops are slow often because of a copy being made, e.g. subtracting median from every column of a df. But also vectorized functions are written in C code

Question 61

sample

Answer 61

sample(x, size, replace= FALSE)
rnorm(10)
sample.int(100,10, replace= FALSE)

Question 62

sort a dataframe by the order of the elements in B

Answer 62

x[order(x$B),]

Question 63

tilde and .x

Answer 63

We can either use

These two are the same:

starwars |> summarize(across(where(is.character), ~ length(unique(.x))))

starwars |> summarize(across(where(is.character), function(x) length(unique(x))))

Question 64

slice_max

Answer 64

Note that this is similar to using summarize to get max delay for each group, but this keeps the info for all columns

Also not it provides two or more if ties

flights |>
group_by(dest) |>
slice_max(arr_delay, n = 1) |>
relocate(dest)
#> # A tibble: 108 × 19
#> # Groups: dest [105]
#> dest year month day dep_time sched_dep_time dep_delay arr_time
#> <chr> <int> <int> <int> <int> <int> <dbl> <int>
#> 1 ABQ 2013 7 22 2145 2007 98 132
#> 2 ACK 2013 7 23 1139 800 219 1250
#> 3 ALB 2013 1 25 123 2000 323 229
#> 4 ANC 2013 8 17 1740 1625 75 2042
#> 5 ATL 2013 7 22 2257 759 898 121
#> 6 AUS 2013 7 10 2056 1505 351 2347
#> # ℹ 102 more rows
#> # ℹ 11 more variables: sched_arr_time <int>, arr_delay <dbl>, …</dbl></int></int></dbl></int></int></int></int></int></chr>

Question 65

Get first number from a string

Answer 65

parse_number() is a handy function that will extract the first number from a string, ignoring all other text.

Question 66

Example with connecting points in a group with geom,_line

Answer 66

billboard_longer |>
ggplot(aes(x = week, y = rank, group = track)) +
geom_line(alpha = 0.25) +
scale_y_reverse()

Question 67

Mapping within only one layer

Answer 67

ggplot(mpg, aes(x = displ, y = hwy)) +
geom_point(aes(color = class)) +
geom_smooth()

Question 68

Different data in one layer

Answer 68

ggplot(mpg, aes(x = displ, y = hwy)) +
geom_point() +
geom_point(
data = mpg |> filter(class == “2seater”),
color = “red”
) +
geom_point(
data = mpg |> filter(class == “2seater”),
shape = “circle open”, size = 3, color = “red”
)

Question 69

rearrange/reorder columnds

Answer 69

df %>% relocate(a, .after = c)
df %>% relocate(all_of(c(“n_group”,”gender”,”group”)), .before = sleepy)