Exam 2 review

Below are questions to help you study for Exam 2. These are some examples of the kinds of questions I might ask.

This is not a practice exam. There will be fewer questions on the actual exam.
The questions cover most, but not all, possible material for the exam.
The distribution of questions here is not necessarily reflective of the distribution of questions on the actual exam.

Strings and regular expressions

Example 1

Consider the following strings:

[1] "George Washington: February 22, 1732"
[2] "Thomas Jefferson: April 13, 1743"    
[3] "Abraham Lincoln: February 12, 1809"  
[4] "Theodore Roosevelt: October 27, 1858"

For each question below, fill in the R code to produce the desired output.

Solution:

str_extract(strings, ".+(?=:)")

[1] "George Washington"  "Thomas Jefferson"   "Abraham Lincoln"   
[4] "Theodore Roosevelt"

Solution:

str_extract(strings, "(?<=: ).+")

[1] "February 22, 1732" "April 13, 1743"    "February 12, 1809"
[4] "October 27, 1858"

Solution:

str_extract(strings, "(?<=\\s).+(?=:)")

[1] "Washington" "Jefferson"  "Lincoln"    "Roosevelt"

Solution:

str_extract(strings, "(?<=, ).+")

[1] "1732" "1743" "1809" "1858"

Example 2

Consider the following strings:

strings

 [1] "apple"           "banana"          "canteloupe"      "durian"         
 [5] "eggplant"        "french fries"    "goat cheese"     "pizza"          
 [9] "99 red balloons" "101 dalmatians"  "route 66"

For each question below, fill in the R code to produce the desired output.

Solution:

str_subset(strings, "\\d+")

[1] "99 red balloons" "101 dalmatians"  "route 66"

Solution:

str_subset(strings, "^\\d+")

[1] "99 red balloons" "101 dalmatians"

Solution:

str_subset(strings, "a")

[1] "apple"           "banana"          "canteloupe"      "durian"         
[5] "eggplant"        "goat cheese"     "pizza"           "99 red balloons"
[9] "101 dalmatians"

Solution:

str_subset(strings, "\\s")

[1] "french fries"    "goat cheese"     "99 red balloons" "101 dalmatians" 
[5] "route 66"

Solution:

str_subset(strings, "(.)\\1{1}")

[1] "apple"           "eggplant"        "goat cheese"     "pizza"          
[5] "99 red balloons" "route 66"

Example 3

Suppose you receive a vector containing the following email addresses:

emails <- c("evansc@wfu.edu",
            "dalzelnm@wfu.edu",
            "heplersa@wfu.edu")

Fill in the following code to produce the desired output:

Solution:

str_extract(emails, "^.+(?=@)")

[1] "evansc"   "dalzelnm" "heplersa"

Example 4

You have a folder with the following filenames:

filenames <- c(".bash_profile",
               "workspace.doc",
               "img0912.jpg",
               "updated_img0912.png",
               "documentation.html",
               "favicon.gif",
               "access.lock")

Fill in the following code to produce the desired output:

Solution:

str_subset(filenames, "(gif|png|jpg)")

[1] "img0912.jpg"         "updated_img0912.png" "favicon.gif"

Example 5

Consider the following string:

ex_str <- "The kurtosis is defined by the equation $\\mu_4/\\sigma^4$, 
where $\\mu_4$ is the fourth central moment 
and $\\sigma$ is the standard deviation."

Fill in the following code to produce the desired output:

Solution:

str_extract_all(ex_str, "\\$[^\\$]+\\$")

[[1]]
[1] "$\\mu_4/\\sigma^4$" "$\\mu_4$"           "$\\sigma$"

Example 6

Here is another small dataset, ex_df, which has a single column (faculty):

ex_df

                                                faculty
1                       Rob Erhardt, Professor: MAN 343
2          Daniel Beavers, Associate Professor: MAN 337
3                 Kenneth Berenhaut, Professor: MAN 379
4          Leonardo Cella, Assistant Professor: MAN 344
5 Lucy D'Agostino McGowan, Assistant Professor: MAN 342
6 Nicole Dalzell, Associate Teaching Professor: MAN 338

Write code to produce the following output using ex_df:

Solution:

Option 1:

ex_df |>
  mutate(name = str_extract(faculty, ".+(?=,)"),
         position = str_extract(faculty, "(?<=, ).+(?=:)"),
         office = str_extract(faculty, "(?<=: ).+")) |>
  select(-faculty)

                     name                     position  office
1             Rob Erhardt                    Professor MAN 343
2          Daniel Beavers          Associate Professor MAN 337
3       Kenneth Berenhaut                    Professor MAN 379
4          Leonardo Cella          Assistant Professor MAN 344
5 Lucy D'Agostino McGowan          Assistant Professor MAN 342
6          Nicole Dalzell Associate Teaching Professor MAN 338

Option 2:

ex_df |>
  separate_wider_regex(faculty,
                       patterns = c(
                         name = ".+",
                         ", ",
                         position = ".+",
                         ": ",
                         office = ".+"
                       ))

# A tibble: 6 × 3
  name                    position                     office 
  <chr>                   <chr>                        <chr>  
1 Rob Erhardt             Professor                    MAN 343
2 Daniel Beavers          Associate Professor          MAN 337
3 Kenneth Berenhaut       Professor                    MAN 379
4 Leonardo Cella          Assistant Professor          MAN 344
5 Lucy D'Agostino McGowan Assistant Professor          MAN 342
6 Nicole Dalzell          Associate Teaching Professor MAN 338

Web scraping

Example 1

Consider the following portion of a web page:

STA 279

Welcome to STA 279! This is a 16-week course offered at Wake Forest University, which meets 3 hours each week. The syllabus can be found here

Below are some useful dates for the semester:

Exam 1	March 1
Exam 2	April 12

Now here is the source HTML for that simple web page:

<h1> STA 279 </h1>

<p> 
  Welcome to STA 279! 
  This is a 16-week course offered at Wake Forest University, 
  which meets 3 hours each week. 
  The syllabus can be found 
  <a href='https://sta279-f25.github.io/about/'>here</a> 
</p>

<p>
  Below are some useful dates for the semester:
   <table>
  <tr>
    <td>Exam 1</td>
    <td>March 1</td>
  </tr>
  <tr>
    <td>Exam 2</td>
    <td>April 12</td>
  </tr>
</table> 
</p>

Suppose that the url for this page is “https://fake_web_page.html”. Write code that will import the table of exam dates into R, as a data frame.

Solution:

read_html("https://fake_web_page.html") |>
  html_element("table") |>
  html_table()

Suppose that the url for this page is “https://fake_web_page.html”. What will the following code return?

read_html("https://fake_web_page.html") |>
  html_element("h1") |>
  html_text2() |>
  str_extract("\\d+")

Solution: 279

Example 2

Consider the following short portion of a web page, with URL https://fake_page.html.

<div class="mesa">
  <span>
    <p> Hello! </p>
  </span>
  
  <p class="small"> Welcome to STA 279! </p>
  
  <span>
    <p class="small"> Room: Carswell 102 </p>
  </span>
  
  <span>
    <h3 class="small"> Grading policy </h3>
  </span>
  
  <span>
    <p class="small"> Exam dates </p>
  </span>
</div>

Write code in R that will extract the following string from this web page:

[1] "Carswell 102"

Solution:

read_html("https://fake_page.html") |>
  html_element("span > p.small") |>
  html_text2() |>
  str_extract("(?<=: ).+")

Using your code from the previous question, write a short function with the following specifications:
- Input: url the URL for a web page with the same format as above
- Output: the building and room number for the class, scraped from the web page

Solution:

get_room <- function(url){
  read_html(url) |>
  html_element("span > p.small") |>
  html_text2() |>
  str_extract("(?<=: ).+")
}

Suppose that we have the web pages for 20 different classes at Wake Forest, all of which have the same format as shown above, but which take places in different rooms. The URLS are stored in a vector:

class_urls

 [1] "https://fake_page_1.html"  "https://fake_page_2.html" 
 [3] "https://fake_page_3.html"  "https://fake_page_4.html" 
 [5] "https://fake_page_5.html"  "https://fake_page_6.html" 
 [7] "https://fake_page_7.html"  "https://fake_page_8.html" 
 [9] "https://fake_page_9.html"  "https://fake_page_10.html"
[11] "https://fake_page_11.html" "https://fake_page_12.html"
[13] "https://fake_page_13.html" "https://fake_page_14.html"
[15] "https://fake_page_15.html" "https://fake_page_16.html"
[17] "https://fake_page_17.html" "https://fake_page_18.html"
[19] "https://fake_page_19.html" "https://fake_page_20.html"

Using your answers to the previous questions, iterate over this vector of URLs to produce a vector containing the building and room number for each class.

Solution:

map_chr(class_urls, get_room)

I’m using map_chr here since I want to return a vector rather than a list. If I asked this question on the exam, I would also accept a solution with map in place of map_chr.

Polite web scraping

Why is it important to be polite when scraping from a website?
What are the features of polite web scraping that we discussed in class?

SQL

Here is a rough summary of what you should know from the SQL unit:

Pieces of a SQL query you should know:
- SELECT
- FROM
- JOIN (an inner join) and LEFT JOIN (a left join)
- WHERE
- GROUP BY
- HAVING
- ORDER BY
- LIMIT
You should also know the order in which queries must be written
Other functions to know for queries:
- DISTINCT
- Summary functions like SUM, AVG, MIN, MAX, COUNT
- LIKE, and the wildcards % and _
- CASE
The function of primary and foreign keys, and the constraints they place on tables
How to read and interpret an ERD
You should be able to read and interpet the results of DESCRIBE, ALTER TABLE, and INSERT INTO statements. You will not have to write these statements
Subqueries

Practice with SQL queries

Here is a small table in a SQL database. The name of the table is small_penguins.

     species island bill_length_mm bill_depth_mm
1     Gentoo Biscoe           43.3          14.0
2     Gentoo Biscoe           59.6          17.0
3     Adelie  Dream           39.7          17.9
4     Adelie  Dream           39.2          21.1
5  Chinstrap  Dream           50.8          19.0
6     Gentoo Biscoe           49.9          16.1
7  Chinstrap  Dream           50.7          19.7
8     Gentoo Biscoe           47.3          15.3
9     Gentoo Biscoe           49.3          15.7
10    Adelie  Dream           37.5          18.9

In each of the following examples, write a SQL query to produce the desired output from small_penguins.

Output:

    species island n
1    Adelie  Dream 3
2 Chinstrap  Dream 2
3    Gentoo Biscoe 5

SELECT species, island, COUNT(*) as n
FROM small_penguins
GROUP BY species, island;

3 records
species	island	n
Adelie	Dream	3
Chinstrap	Dream	2
Gentoo	Biscoe	5

Output:

  island   species mean_length
1 Biscoe    Gentoo       49.88
2  Dream    Adelie       38.80
3  Dream Chinstrap       50.75

SELECT island, species, AVG(bill_length_mm) AS mean_length
FROM small_penguins
GROUP BY island, species;

3 records
island	species	mean_length
Biscoe	Gentoo	49.88
Dream	Adelie	38.80
Dream	Chinstrap	50.75

Output:

     species island bill_length_mm bill_depth_mm bill_ratio
1     Gentoo Biscoe           43.3          14.0   3.092857
2     Gentoo Biscoe           59.6          17.0   3.505882
3     Adelie  Dream           39.7          17.9   2.217877
4     Adelie  Dream           39.2          21.1   1.857820
5  Chinstrap  Dream           50.8          19.0   2.673684
6     Gentoo Biscoe           49.9          16.1   3.099379
7  Chinstrap  Dream           50.7          19.7   2.573604
8     Gentoo Biscoe           47.3          15.3   3.091503
9     Gentoo Biscoe           49.3          15.7   3.140127
10    Adelie  Dream           37.5          18.9   1.984127

SELECT species, island, bill_length_mm, bill_depth_mm,
  bill_length_mm/bill_depth_mm AS bill_ratio
FROM small_penguins;

Displaying records 1 - 10
species	island	bill_length_mm	bill_depth_mm	bill_ratio
Gentoo	Biscoe	43.3	14.0	3.092857
Gentoo	Biscoe	59.6	17.0	3.505882
Adelie	Dream	39.7	17.9	2.217877
Adelie	Dream	39.2	21.1	1.857820
Chinstrap	Dream	50.8	19.0	2.673684
Gentoo	Biscoe	49.9	16.1	3.099379
Chinstrap	Dream	50.7	19.7	2.573604
Gentoo	Biscoe	47.3	15.3	3.091503
Gentoo	Biscoe	49.3	15.7	3.140127
Adelie	Dream	37.5	18.9	1.984127

Output:

  species island bill_length_mm bill_depth_mm
1  Adelie  Dream           39.7          17.9
2  Adelie  Dream           39.2          21.1
3  Adelie  Dream           37.5          18.9

SELECT species, island, bill_length_mm, bill_depth_mm
FROM small_penguins
WHERE species = 'Adelie' AND island = 'Dream';

3 records
species	island	bill_length_mm	bill_depth_mm
Adelie	Dream	39.7	17.9
Adelie	Dream	39.2	21.1
Adelie	Dream	37.5	18.9

Joins

In each of the following questions, write a SQL query to produce the desired output from the two input tables

df1

df2

Output:

SELECT df1.id, x, y
FROM df1 
LEFT JOIN df2 ON df1.id = df2.id;

3 records
id	x	y
1	7	10
2	9	12
3	13	NA

df1

df2

Output:

  id x  y
1  1 7 10
2  2 9 12

SELECT df1.id, x, y
FROM df1 
JOIN df2 ON df1.id = df2.id;

2 records
id	x	y
1	7	10
2	9	12

More practice with SQL queries

The nycflights13 package in R contains several datasets describing flights from NY airports in 2013, including the airports and weather tables.

The airports table has one row for each different airport in the database. There are a total of 1458 rows and 8 columns in the airports table; here are a few random rows from the table, showing the relevant columns for this question:

  faa                      name
1 AOS   Amook Bay Seaplane Base
2 2H0     Shelby County Airport
3 CEF Westover Arb Metropolitan
4 AVW           Marana Regional
5 DLF              Laughlin Afb

The weather table has one row for each date-time, for each of the three NY airports (EWR, JFK, LGA), and records a variety of weather variables at each time for each location. Here are a few random rows from the table, showing the relevant columns for this question:

  origin month  temp humid
1    EWR    12 24.98 62.55
2    LGA     2 33.98 88.42
3    LGA     5 57.92 71.90
4    EWR     6 64.04 93.28
5    LGA     4 42.98 88.85

The weather table uses a 3-character FAA code to identify each airport. It does not include the full name for the airports.

We would like to produce a table with the mean temperature, and humidity for each month, for each of the 3 NYC airports in the database. The output table should also include the full name of each airport. Here is what the output table looks like:

   origin                name month mean_temp mean_humid
1     EWR Newark Liberty Intl     1  35.56216   62.12451
2     EWR Newark Liberty Intl     2  34.26332   63.33558
3     EWR Newark Liberty Intl     3  40.11865   57.91382
4     EWR Newark Liberty Intl     4  52.97750   54.23513
5     EWR Newark Liberty Intl     5  63.32024   64.07371
6     EWR Newark Liberty Intl     6  73.26725   67.00229
7     EWR Newark Liberty Intl     7  80.70300   66.67641
8     EWR Newark Liberty Intl     8  74.53748   65.75175
9     EWR Newark Liberty Intl     9  67.30478   64.03556
10    EWR Newark Liberty Intl    10  59.77821   65.31928
11    EWR Newark Liberty Intl    11  44.57734   56.52011
12    EWR Newark Liberty Intl    12  37.95008   69.72141
13    JFK John F Kennedy Intl     1  35.38555   61.66162
14    JFK John F Kennedy Intl     2  34.19246   62.55848
15    JFK John F Kennedy Intl     3  39.54472   58.02222
16    JFK John F Kennedy Intl     4  50.14270   60.05764
17    JFK John F Kennedy Intl     5  59.31476   70.98926
18    JFK John F Kennedy Intl     6  69.95825   73.76147
19    JFK John F Kennedy Intl     7  78.73492   71.74469
20    JFK John F Kennedy Intl     8  73.81878   67.84420
21    JFK John F Kennedy Intl     9  66.89775   65.25182
22    JFK John F Kennedy Intl    10  59.80195   65.06362
23    JFK John F Kennedy Intl    11  45.13419   56.79961
24    JFK John F Kennedy Intl    12  38.60487   68.31341
25    LGA          La Guardia     1  35.95927   59.15609
26    LGA          La Guardia     2  34.35612   60.68603
27    LGA          La Guardia     3  39.97652   56.55947
28    LGA          La Guardia     4  52.11450   53.15060
29    LGA          La Guardia     5  62.75000   61.49618
30    LGA          La Guardia     6  73.32650   61.22686
31    LGA          La Guardia     7  80.76425   62.26404
32    LGA          La Guardia     8  75.04825   60.34574
33    LGA          La Guardia     9  67.91125   59.67743
34    LGA          La Guardia    10  60.63244   60.13875
35    LGA          La Guardia    11  45.26093   53.49296
36    LGA          La Guardia    12  38.76976   63.62769

Write a SQL query to produce the desired results.

Solution:

SELECT weather.origin, name, month, 
  AVG(temp) AS mean_temp, AVG(humid) AS mean_humid
FROM weather
JOIN airports ON weather.origin = airports.faa
GROUP BY origin, name, month
ORDER BY origin, month;

Displaying records 1 - 10
origin	name	month	mean_temp	mean_humid
EWR	Newark Liberty Intl	1	35.56216	62.12451
EWR	Newark Liberty Intl	2	34.26332	63.33558
EWR	Newark Liberty Intl	3	40.11865	57.91382
EWR	Newark Liberty Intl	4	52.97750	54.23512
EWR	Newark Liberty Intl	5	63.32024	64.07371
EWR	Newark Liberty Intl	6	73.26725	67.00229
EWR	Newark Liberty Intl	7	80.70300	66.67641
EWR	Newark Liberty Intl	8	74.53748	65.75175
EWR	Newark Liberty Intl	9	67.30478	64.03556
EWR	Newark Liberty Intl	10	59.77821	65.31928

How would you modify your above query to keep only the rows in the results for which the mean temperature is at least 60?

Solution:

SELECT weather.origin, name, month, 
  AVG(temp) AS mean_temp, AVG(humid) AS mean_humid
FROM weather
JOIN airports ON weather.origin = airports.faa
GROUP BY origin, name, month
HAVING mean_temp >= 60
ORDER BY origin, month;

Displaying records 1 - 10
origin	name	month	mean_temp	mean_humid
EWR	Newark Liberty Intl	5	63.32024	64.07371
EWR	Newark Liberty Intl	6	73.26725	67.00229
EWR	Newark Liberty Intl	7	80.70300	66.67641
EWR	Newark Liberty Intl	8	74.53748	65.75175
EWR	Newark Liberty Intl	9	67.30478	64.03556
JFK	John F Kennedy Intl	6	69.95825	73.76147
JFK	John F Kennedy Intl	7	78.73492	71.74469
JFK	John F Kennedy Intl	8	73.81878	67.84420
JFK	John F Kennedy Intl	9	66.89775	65.25182
LGA	La Guardia	5	62.75000	61.49618

How would you modify your above query to keep only the rows for which the mean temperature is the highest average temperature?

Solution: This is quite tricky, and requires us to combine some queries.

Assuming that we want to get the highest average monthly temperature for each airport, a partial answer is the following, though note that this only gives us the origin and max mean temperature columns:

SELECT origin, MAX(mean_temp)
FROM 
  (
  SELECT weather.origin, name, month, 
    AVG(temp) AS mean_temp, AVG(humid) AS mean_humid
  FROM weather
  JOIN airports ON weather.origin = airports.faa
  GROUP BY origin, name, month
  ORDER BY origin, month
  ) 
GROUP BY origin;

3 records
origin	max(mean_temp)
EWR	80.70300
LGA	80.76425
JFK	78.73492

This partial solution uses a subquery, and this is about the level of difficulty I would put on the exam.

A full solution is beyond the scope of the exam, and I think requires us to use either multiple subqueries, or to create an intermediate view/table. Here is one solution which uses two subqueries:

SELECT weather.origin, name, month, 
    AVG(temp) AS mean_temp, AVG(humid) AS mean_humid
FROM weather
JOIN airports ON weather.origin = airports.faa
GROUP BY origin, name, month
HAVING (origin, mean_temp) IN
  (SELECT origin, MAX(mean_temp)
    FROM 
    (
      SELECT weather.origin, name, month, 
        AVG(temp) AS mean_temp, AVG(humid) AS mean_humid
      FROM weather
      JOIN airports ON weather.origin = airports.faa
      GROUP BY origin, name, month
      ORDER BY origin, month
  ) 
  GROUP BY origin
 );

3 records
origin	name	month	mean_temp	mean_humid
EWR	Newark Liberty Intl	7	80.70300	66.67641
JFK	John F Kennedy Intl	7	78.73492	71.74469
LGA	La Guardia	7	80.76425	62.26404

However, the problem with this solution is that it requires us to create the results from question 27 twice, which is quite inefficient. Instead, we could store the results from question 27, then use these results. Here is what those queries might look like (again, I don’t expect you to be able to do this part for the exam):

CREATE VIEW monthly_weather AS
  (SELECT weather.origin, name, month, 
    AVG(temp) AS mean_temp, AVG(humid) AS mean_humid
  FROM weather
  JOIN airports ON weather.origin = airports.faa
  GROUP BY origin, name, month);

SELECT * 
FROM monthly_weather
WHERE (origin, mean_temp) IN
  (SELECT origin, MAX(mean_temp)
   FROM monthly_weather
   GROUP BY origin);

3 records
origin	name	month	mean_temp	mean_humid
JFK	John F Kennedy Intl	7	78.73492	71.74469
LGA	La Guardia	7	80.76425	62.26404
EWR	Newark Liberty Intl	7	80.70300	66.67641

Conceptual questions

What are the differences between SELECT in a SQL query, and the select function from dplyr in R?

Solution: The select function from the dplyr package has one function: selecting columns from the data frame. We don’t use the select function to create new columns or summary statistics – those roles are performed by, respectively, the mutate and summarize functions. However, in SQL SELECT does all of this. Choosing existing columns, creating a new column, or creating aggregates of existing columns all happen with SELECT.

Suppose you are managing a database for a hospital. List four different tables which might be included in the hospital database. For each table:

Describe what information would be included (what would a row represent? what would be some of the columns?)
Is there a natural primary key for the table? What column or set of columns would form that primary key?
Are there any columns in the table that would be foreign keys for other tables in the database?

Solution: There is not just one right answer here. Here is one possible answer.

Tables:

Patients: information on each patient that has been treated at the hospital
- Primary key:
  - MRN (medical record number), a unique identifier assigned to each patient in the healthcare organization
- Other columns
  - Date of birth
  - Sex
  - Insurance
  - Address
  - Phone number
Staff: information on doctors, nurses, etc. who have worked at the hospital
- Primary key:
  - ID employee ID
- Other columns
  - Role
  - Employment status
  - Age
  - Area(s) of specialty
  - Department
Pharmacy: information on drugs prescribed to patients in the hospital
- Foreign keys:
  - PatientID refers to MRN in the Patients table. Patient to whom drug was prescribed
  - StaffID refers to ID in the Staff table. Staff member who prescribed the drug
- Other columns
  - Date prescribed
  - Drug
  - Dosage
Visits: information on each visit
- Foreign keys:
  - PatientID refers to MRN in the Patients table
- Other columns
  - Check-in time
  - Check-out time
  - Reason for visit

Practice with function scoping

What will be the output of the following code?

x <- 10
test_fun <- function(x = 11){
  return(x)
}
test_fun()

[1] 11

[1] 10

What will be the output of the following code?

x <- 10
test_fun <- function(y = 11){
  return(x + 1)
}
test_fun()

[1] 11

[1] 10

What will be the output of the following code?

x <- 10
test_fun <- function(y = 11){
  x <- x + 1
  return(x + 1)
}
test_fun()

[1] 12

[1] 10

What will be the output of the following code?

x <- 10
test_fun <- function(x = 11){
  x <- x + 1
  return(x + 1)
}
test_fun()

[1] 13

[1] 10

What will be the output of the following code?

x <- 10
test_fun <- function(x = 11){
  x <- x + 1
  return(x + 1)
}
x <- test_fun(x)
x

[1] 12

Practice with lists

Create a list x in R such that:

x[[1]] returns the function mean
x[[2]] returns the function sd
x[[3]][[1]] returns the vector c(0, 1, 2)
x[[3]][[2]] returns a function which calculates the cube root of a vector

Solution:

x <- list(mean, sd, list(c(0, 1, 2), 
                         function(y) y^(1/3)))

x[[1]](1:10)

[1] 5.5

x[[2]](1:10)

[1] 3.02765

x[[3]][[1]]

[1] 0 1 2

x[[3]][[2]](c(8, 27, 64))

[1] 2 3 4