Last week you were asked:
How much money have tuna purse seiners made since 2000 when fishing for bigeye tuna (Thunnus obesus) in the Eastern Pacific Ocean?
We made some simplifying assumptions and got some values (a total of 3,070 M USD since 2000, or about 127 M USD per year). You are now tasked with coming up with more refined estimates. For example, we will account for the fact that the price of fish varies every year.
How we will approach this:
This will require three pipelines:
Pipelines 1 and 3 contain tools covered this week. You should already be familiar with pipeline 2.
Post-it up
Economic and Development Indicators and Statistics: Tuna Fishery of the Western and Central Pacific Ocean 2024EVR628/data/raw folder and proceed to extract itCompendium of Economic and Development Statistics 2024 and study the Contents tabPost-it down
Post-it up
readxl package: install.packages("readxl")tuna_analysis_prices.R1readxl, janitor, and tidyverse, load them at the top of your script using library()?read_excel() to look at the documentation for the functionread_excel() to create a new object called tuna_prices and read the price data we need2. Immediately pipe it into clean_names.Post-it down
library(readxl)
library(janitor)
library(tidyverse)
tuna_prices <- read_excel(path = "data/raw/Economic-and-Development-Indicators-and-Statistics-Tuna-Fishery-of-the-Western-and-Central-Pacific-Ocean-2024-32765/Compendium of Economic and Development Statistics 2024.xlsx",
sheet = "B. Prices",
range = "A35:E63",
na = "na") |>
clean_names()Be prepared to discuss the following points:
Post-it up
tuna_prices using colnames() in your console.Post-it down
See an example of my description below
colnames(tuna_prices)[1] "year" "japan_fresha" "japan_frozenb" "us_freshc"
[5] "us_frozend" dim(tuna_prices)[1] 28 5tuna_prices |>
filter_all(any_vars(is.na(.)))# A tibble: 4 × 5
year japan_fresha japan_frozenb us_freshc us_frozend
<dbl> <dbl> <dbl> <dbl> <dbl>
1 1997 8204. 8169. NA NA
2 1998 7703. 6320. NA NA
3 1999 8809. 9093. NA NA
4 2000 9198. 8557. NA NA# The final data set should have two columns: year and price. Since we have four
# prices (two markets, two presentations), I will use the average price per year.
# The tidy data set should therefore have four columns: year, market,
# presentation, and price.Post-it up
pivot_* function. What does it say about cases where names_to is of length > 1?names_sep argument?pivot_* function to reshape your data and save them to a new object called tidy_tuna_prices3tidy_tuna_prices <- tuna_prices |>
pivot_longer(cols = 2:5,
names_to = c("market", "presentation"),
names_sep = "_",
values_to = "price",
values_drop_na = T)
tidy_tuna_prices# A tibble: 104 × 4
year market presentation price
<dbl> <chr> <chr> <dbl>
1 1997 japan fresha 8204.
2 1997 japan frozenb 8169.
3 1998 japan fresha 7703.
4 1998 japan frozenb 6320.
5 1999 japan fresha 8809.
6 1999 japan frozenb 9093.
7 2000 japan fresha 9198.
8 2000 japan frozenb 8557.
9 2001 japan fresha 8260.
10 2001 japan frozenb 5983.
# ℹ 94 more rowsPost-it down
presentation
Note that the values in the presentation column are not ideal. They end in a, b, c, and d due to footnotes included in Excel. For now this doesn’t matter because we will quickly remove them. We’ll cover some text wrangling in Week 9.
Post-it up
tidy_tuna_prices to get the mean price per year5tidy_tuna_prices <- tuna_prices |>
pivot_longer(cols = 2:5,
names_to = c("market", "presentation"),
names_sep = "_",
values_to = "price",
values_drop_na = T) |>
group_by(year) |>
summarize(price = mean(price))
tidy_tuna_prices# A tibble: 28 × 2
year price
<dbl> <dbl>
1 1997 8186.
2 1998 7011.
3 1999 8951.
4 2000 8877.
5 2001 5633.
6 2002 5342.
7 2003 5285.
8 2004 5739.
9 2005 5554.
10 2006 5177.
# ℹ 18 more rowsPost-it down
Note: You can copy-paste and modify your code from last week, but make sure your code is organized.
Post-it up
BET) caught by the purse seine fleet (PS) since 2000Post-it down
# Load the data
tuna_data <- read_csv("data/raw/CatchByFlagGear/CatchByFlagGear1918-2023.csv") |>
# Clean column names
clean_names() |>
# Rename some columns
rename(year = ano_year,
flag = bandera_flag,
gear = arte_gear,
species = especies_species,
catch = t)
ps_tuna_data <- tuna_data |>
filter(species == "BET", # Retain BET values only
gear == "PS", # Retain PS values only
year >= 2000) |> # Retain data from 2000
group_by(year) |> # Specify that I am grouping by year
# Tell summarize that I want to collapse the catch column by summing all its values
summarize(catch = sum(catch))
ps_tuna_data# A tibble: 24 × 2
year catch
<dbl> <dbl>
1 2000 95283
2 2001 60518
3 2002 57422
4 2003 53051
5 2004 65471
6 2005 67895
7 2006 83837
8 2007 63451
9 2008 75028
10 2009 76800
# ℹ 14 more rowsPost-it up
tuna_revenuesPost-it down
tuna_revenues <- ps_tuna_data |>
left_join(tidy_tuna_prices, by = join_by(year)) |>
mutate(revenue = price * catch / 1e6)
tuna_revenues# A tibble: 24 × 4
year catch price revenue
<dbl> <dbl> <dbl> <dbl>
1 2000 95283 8877. 846.
2 2001 60518 5633. 341.
3 2002 57422 5342. 307.
4 2003 53051 5285. 280.
5 2004 65471 5739. 376.
6 2005 67895 5554. 377.
7 2006 83837 5177. 434.
8 2007 63451 5054. 321.
9 2008 75028 5636. 423.
10 2009 76800 6175. 474.
# ℹ 14 more rowssum(tuna_revenues$revenue)[1] 11752.32mean(tuna_revenues$revenue)[1] 489.68# Build plot
ggplot(data = tuna_revenues, # Specify my data
mapping = aes(x = year, y = revenue)) + # And my aesthetics
geom_line(linetype = "dashed") + # Add a dashed line
geom_point() + # With points on top
labs(x = "Year", # Add some labels
y = "Revenue (M USD)",
title = "Annual revenue from fishing bigeye tuna by purse seine vessels",
caption = "Data come from the IATTC") +
# Modify the theme
theme_minimal(base_size = 14, # Font size 14
base_family = "Times") # Font family Times
The following four exercises use data that are built directly in R. You will need to copy and paste the provided code in your console (or R script) to make sure the objects appear in your environment.
Scenario: You are a TA and have been given grade data where each row represents a student and columns represent their scores on assignments 1-4. You need to calculate the mean grade for each student.6
# Create the dataset
student_scores <- tribble(
~student_id, ~assignment_1, ~assignment_2, ~assignment_3, ~assignment_4,
"S001", 85, 92, 78, 88,
"S002", 91, 89, 95, 82,
"S003", 76, 84, 91, 79,
"S004", 88, 93, 87, 94
)
student_scores# A tibble: 4 × 5
student_id assignment_1 assignment_2 assignment_3 assignment_4
<chr> <dbl> <dbl> <dbl> <dbl>
1 S001 85 92 78 88
2 S002 91 89 95 82
3 S003 76 84 91 79
4 S004 88 93 87 94How I did it:
student_scores_long <- student_scores |>
pivot_longer(cols = starts_with("assignment"), # All columns from assignment_1 to assignment_4
names_to = "assignment", # Create new column called "assignment" with column names
values_to = "score") # Create new column called "score" with the values
student_means <- student_scores_long |>
group_by(student_id) |> # Group rows by student_id
summarize(mean_grade = mean(score)) # Calculate mean score for each student
student_means # View the result# A tibble: 4 × 2
student_id mean_grade
<chr> <dbl>
1 S001 85.8
2 S002 89.2
3 S003 82.5
4 S004 90.5Scenario: You have hurricane exposure data from different Florida counties. You are asked to build a figure showing the relationship between pressure and wind speed. Modify the data as needed and build a figure.7
# Create the dataset
hurricane_data <- tribble(
~county, ~metric, ~measurement,
"Miami-Dade", "pressure", 950,
"Miami-Dade", "precipitation", 12.5,
"Miami-Dade", "wind_speed", 85,
"Broward", "pressure", 955,
"Broward", "precipitation", 8.3,
"Broward", "wind_speed", 72,
"Palm Beach", "pressure", 960,
"Palm Beach", "precipitation", 6.1,
"Palm Beach", "wind_speed", 68,
"Monroe", "pressure", 945,
"Monroe", "precipitation", 15.2,
"Monroe", "wind_speed", 95
)
hurricane_data# A tibble: 12 × 3
county metric measurement
<chr> <chr> <dbl>
1 Miami-Dade pressure 950
2 Miami-Dade precipitation 12.5
3 Miami-Dade wind_speed 85
4 Broward pressure 955
5 Broward precipitation 8.3
6 Broward wind_speed 72
7 Palm Beach pressure 960
8 Palm Beach precipitation 6.1
9 Palm Beach wind_speed 68
10 Monroe pressure 945
11 Monroe precipitation 15.2
12 Monroe wind_speed 95 How I did it:
hurricane_wide <- hurricane_data |>
pivot_wider(names_from = metric, # Use values in "metric" column as new column names
values_from = measurement) # Use values in "measurement" column to fill new columns
hurricane_wide # Now each county is a row with separate columns for pressure, precipitation, wind_speed# A tibble: 4 × 4
county pressure precipitation wind_speed
<chr> <dbl> <dbl> <dbl>
1 Miami-Dade 950 12.5 85
2 Broward 955 8.3 72
3 Palm Beach 960 6.1 68
4 Monroe 945 15.2 95ggplot(hurricane_wide,
aes(x = pressure, y = wind_speed)) +
geom_point() +
labs(x = "Pressure",
y = "Wind speed")
Scenario: You have two datasets - one with student information and another with their enrollment details. You need to be able to identify the names of students taking stats courses.
# Create the datasets
students <- tribble(
~student_id, ~name, ~age,
"S001", "Alice Johnson", 20,
"S002", "Bob Smith", 22,
"S003", "Carol Davis", 19,
"S004", "David Wilson", 21,
"S005", "Eva Brown", 23
)
enrollments <- tribble(
~stdt_identifier, ~course, ~credits,
"S001", "Statistics", 3,
"S001", "Biology", 4,
"S002", "Statistics", 3,
"S003", "Chemistry", 4,
"S004", "Statistics", 3,
"S004", "Physics", 4,
"S006", "Math", 3
)
students# A tibble: 5 × 3
student_id name age
<chr> <chr> <dbl>
1 S001 Alice Johnson 20
2 S002 Bob Smith 22
3 S003 Carol Davis 19
4 S004 David Wilson 21
5 S005 Eva Brown 23enrollments# A tibble: 7 × 3
stdt_identifier course credits
<chr> <chr> <dbl>
1 S001 Statistics 3
2 S001 Biology 4
3 S002 Statistics 3
4 S003 Chemistry 4
5 S004 Statistics 3
6 S004 Physics 4
7 S006 Math 3How I did it:
combined_data <- students |>
left_join(enrollments, by = join_by(student_id == stdt_identifier)) # Keep all students, match different column names
combined_data # View the combined data# A tibble: 7 × 5
student_id name age course credits
<chr> <chr> <dbl> <chr> <dbl>
1 S001 Alice Johnson 20 Statistics 3
2 S001 Alice Johnson 20 Biology 4
3 S002 Bob Smith 22 Statistics 3
4 S003 Carol Davis 19 Chemistry 4
5 S004 David Wilson 21 Statistics 3
6 S004 David Wilson 21 Physics 4
7 S005 Eva Brown 23 <NA> NAstatistics_students <- combined_data |>
filter(course == "Statistics") |> # Keep only rows where course equals "Statistics"
select(name, course) # Keep only the name and course columns
statistics_students # View the result# A tibble: 3 × 2
name course
<chr> <chr>
1 Alice Johnson Statistics
2 Bob Smith Statistics
3 David Wilson StatisticsScenario: You have swimming data from beachgoers and bull shark detection data from acoustic telemetry during fourth of July. The swimming data tell you when someone entered and left the water. The shark detection data tells you which sharks were detected within the acoustic array in front of the beach, and the time of detection. Who was in the water while a shark was nearby?8
# Create the datasets
swimming_data <- tribble(
~name, ~swim_start, ~swim_end,
"Alice", "2024-07-04 10:30:00", "2024-07-04 11:15:00",
"Bob", "2024-07-04 10:45:00", "2024-07-04 11:30:00",
"Carol", "2024-07-04 11:00:00", "2024-07-04 11:45:00",
"David", "2024-07-04 11:20:00", "2024-07-04 12:00:00",
"Eva", "2024-07-04 12:10:00", "2024-07-04 12:45:00"
) |>
mutate(swim_start = as_datetime(swim_start),
swim_end = as_datetime(swim_end))
shark_detections <- tribble(
~shark_id, ~detection_time,
"SH001", "2024-07-04 09:40:00",
"SH002", "2024-07-04 11:25:00",
"SH003", "2024-07-04 11:35:00"
) |>
mutate(detection_time = as_datetime(detection_time))
swimming_data# A tibble: 5 × 3
name swim_start swim_end
<chr> <dttm> <dttm>
1 Alice 2024-07-04 10:30:00 2024-07-04 11:15:00
2 Bob 2024-07-04 10:45:00 2024-07-04 11:30:00
3 Carol 2024-07-04 11:00:00 2024-07-04 11:45:00
4 David 2024-07-04 11:20:00 2024-07-04 12:00:00
5 Eva 2024-07-04 12:10:00 2024-07-04 12:45:00shark_detections# A tibble: 3 × 2
shark_id detection_time
<chr> <dttm>
1 SH001 2024-07-04 09:40:00
2 SH002 2024-07-04 11:25:00
3 SH003 2024-07-04 11:35:00How I did it:
swimmer_shark_overlap <- shark_detections |>
inner_join(swimming_data, # Note that I am using an inner join. Play with inner, left, and right to see what happens
by = join_by(between(detection_time, swim_start, swim_end))) # Find swimmers in water during shark detections
swimmer_shark_overlap # View all the overlap data# A tibble: 5 × 5
shark_id detection_time name swim_start swim_end
<chr> <dttm> <chr> <dttm> <dttm>
1 SH002 2024-07-04 11:25:00 Bob 2024-07-04 10:45:00 2024-07-04 11:30:00
2 SH002 2024-07-04 11:25:00 Carol 2024-07-04 11:00:00 2024-07-04 11:45:00
3 SH002 2024-07-04 11:25:00 David 2024-07-04 11:20:00 2024-07-04 12:00:00
4 SH003 2024-07-04 11:35:00 Carol 2024-07-04 11:00:00 2024-07-04 11:45:00
5 SH003 2024-07-04 11:35:00 David 2024-07-04 11:20:00 2024-07-04 12:00:00at_risk_swimmers <- swimmer_shark_overlap |>
group_by(name) |> # Keep only the columns we want to see
summarize(n_sharks_near = n_distinct(shark_id))
at_risk_swimmers # These swimmers were in the water when sharks were detected# A tibble: 3 × 2
name n_sharks_near
<chr> <int>
1 Bob 1
2 Carol 2
3 David 2I would typically suggest to overwrite whatever we had last week in tuna_analysis.R because GitHub would keep a version, but I understand you might want to keep the script as is↩︎
Hint: You will need to specify a file path, a sheet, and a range of cells.↩︎
Hint: Your names_to argument should be a character vector of with two items. names_ _sep should be inspired by our clever use of snake_case.↩︎
Hint: If you have 112 rows, remember you can use values_drop_na = T↩︎
Hint: You will use group_by() and summarize(), as well as |>↩︎
Hint: Use pivot_longer() to transform this data so that each row represents one student-assignment-score combination, then use group_by() and summarize() to calculate the mean grade for each student.↩︎
Hint: Use pivot_wider() to transform this data so that each row represents a county and each metric becomes its own column.↩︎
Hint: Look at the documentation for join_by(). What does it say about “Overlap helpers”? You’ll want to use the between() function.↩︎