### Creating objects in R ### Challenge ## ## What are the values after each statement in the following? ## ## mass <- 47.5 # mass? ## age <- 122 # age? ## mass <- mass * 2.0 # mass? ## age <- age - 20 # age? ## mass_index <- mass/age # mass_index? ### Vectors and data types ## ## We’ve seen that atomic vectors can be of type character, numeric, integer, and ## ## logical. But what happens if we try to mix these types in a single ## ## vector? ## ## ## What will happen in each of these examples? (hint: use `class()` to ## ## check the data type of your object) ## num_char <- c(1, 2, 3, "a") ## ## num_logical <- c(1, 2, 3, TRUE) ## ## char_logical <- c("a", "b", "c", TRUE) ## ## tricky <- c(1, 2, 3, "4") ## ## ## Why do you think it happens? ## ## ## You've probably noticed that objects of different types get ## ## converted into a single, shared type within a vector. In R, we call ## ## converting objects from one class into another class ## ## _coercion_. These conversions happen according to a hierarchy, ## ## whereby some types get preferentially coerced into other types. Can ## ## you draw a diagram that represents the hierarchy of how these data ## ## types are coerced? ### Challenge (optional) ## ## * Can you figure out why `"four" > "five"` returns `TRUE`? ## ### Challenge ## 1. Using this vector of length measurements, create a new vector with the NAs ## removed. ## ## lengths <- c(10,24,NA,18,NA,20) ## ## 2. Use the function `median()` to calculate the median of the `lengths` vector. ### Presentation of the survey data ## download.file("https://ndownloader.figshare.com/files/2292169", ## "data/portal_data_joined.csv") ## Challenge ## Based on the output of `str(surveys)`, can you answer the following questions? ## * What is the class of the object `surveys`? ## * How many rows and how many columns are in this object? ## * How many species have been recorded during these surveys? ## Indexing and subsetting data frames ### Challenges: ### ### 1. Create a `data.frame` (`surveys_200`) containing only the ### observations from row 200 of the `surveys` dataset. ### ### 2. Notice how `nrow()` gave you the number of rows in a `data.frame`? ### ### * Use that number to pull out just that last row in the data frame ### * Compare that with what you see as the last row using `tail()` to make ### sure it's meeting expectations. ### * Pull out that last row using `nrow()` instead of the row number ### * Create a new data frame object (`surveys_last`) from that last row ### ### 3. Use `nrow()` to extract the row that is in the middle of the ### data frame. Store the content of this row in an object named ### `surveys_middle`. ### ### 4. Combine `nrow()` with the `-` notation above to reproduce the behavior of ### `head(surveys)` keeping just the first through 6th rows of the surveys ### dataset. ### Factors sex <- factor(c("male", "female", "female", "male")) f <- factor(c(1990, 1983, 1977, 1998, 1990)) as.numeric(f) # wrong! and there is no warning... as.numeric(as.character(f)) # works... as.numeric(levels(f))[f] # The recommended way. ## bar plot of the number of females and males captured during the experiment: plot(surveys$sex) ## Challenges ## ## * Rename "F" and "M" to "female" and "male" respectively. ## * Now that we have renamed the factor level to "missing", can you recreate the ## barplot such that "missing" is last (after "male") ## ## Challenge: ## ## There are a few mistakes in this hand-crafted `data.frame`, ## ## can you spot and fix them? Don't hesitate to experiment! ## animal_data <- data.frame(animal=c(dog, cat, sea cucumber, sea urchin), ## feel=c("furry", "squishy", "spiny"), ## weight=c(45, 8 1.1, 0.8)) ## ## Challenge: ## ## Can you predict the class for each of the columns in the following ## ## example? ## ## Check your guesses using `str(country_climate)`: ## ## * Are they what you expected? Why? why not? ## ## * What would have been different if we had added `stringsAsFactors = FALSE` ## ## to this call? ## ## * What would you need to change to ensure that each column had the ## ## accurate data type? ## country_climate <- data.frame(country=c("Canada", "Panama", "South Africa", "Australia"), ## climate=c("cold", "hot", "temperate", "hot/temperate"), ## temperature=c(10, 30, 18, "15"), ## northern_hemisphere=c(TRUE, TRUE, FALSE, "FALSE"), ## has_kangaroo=c(FALSE, FALSE, FALSE, 1)) ### Manipulating and analyzing data with dplyr ## ## Pipes Challenge: ## ## Using pipes, subset the data to include individuals collected ## ## before 1995, and retain the columns `year`, `sex`, and `weight.` ## ## ## Mutate Challenge: ## ## Create a new data frame from the survey data that meets the following ## ## criteria: contains only the `species_id` column and a column that ## ## contains values that are half the `hindfoot_length` values (e.g. a ## ## new column `hindfoot_half`). In this `hindfoot_half` column, there are ## ## no NA values and all values are < 30. ## ## ## Hint: think about how the commands should be ordered to produce this data frame! ## ## ## Tally Challenges: ## ## 1. How many individuals were caught in each `plot_type` surveyed? ## ## ## 2. Use `group_by()` and `summarize()` to find the mean, min, and ## ## max hindfoot length for each species (using `species_id`). ## ## ## 3. What was the heaviest animal measured in each year? Return the ## ## columns `year`, `genus`, `species_id`, and `weight`. ## ## ## 4. You saw above how to count the number of individuals of each `sex` using a ## ## combination of `group_by()` and `tally()`. How could you get the same result ## ## using `group_by()` and `summarize()`? Hint: see `?n`. ## ## ### Create the dataset for exporting: ## ## Start by removing observations for which the `species_id`, `weight`, ## ## `hindfoot_length`, or `sex` data are missing: ## surveys_complete <- surveys %>% ## filter(species_id != "", # remove missing species_id ## !is.na(weight), # remove missing weight ## !is.na(hindfoot_length), # remove missing hindfoot_length ## sex != "") # remove missing sex ## ## ## Now remove rare species in two steps. First, make a list of species which ## ## appear at least 50 times in our dataset: ## species_counts <- surveys_complete %>% ## group_by(species_id) %>% ## tally %>% ## filter(n >= 50) %>% ## select(species_id) ## ## ## Second, keep only those species: ## surveys_complete <- surveys_complete %>% ## filter(species_id %in% species_counts$species_id) ## ### Data Visualization with ggplot2 ## install.packages("hexbin") ## surveys_plot + ## geom_hex() ## ## Challenges: ## ## Start with the boxplot we created: ## ggplot(data = surveys_complete, aes(x = species_id, y = hindfoot_length)) + ## geom_boxplot(alpha = 0) + ## geom_jitter(alpha = 0.3, color = "tomato") ## ## ## 1. Replace the box plot with a violin plot; see `geom_violin()`. ## ## ## 2. Represent weight on the log10 scale; see `scale_y_log10()`. ## ## ## 3. Create boxplot for `hindfoot_length`. ## ## ## 4. Add color to the datapoints on your boxplot according to the ## ## plot from which the sample was taken (`plot_id`). ## ## Hint: Check the class for `plot_id`. Consider changing the class ## ## of `plot_id` from integer to factor. Why does this change how R ## ## makes the graph? ## ## ## Plotting time series challenge: ## ## Use what you just learned to create a plot that depicts how the ## ## average weight of each species changes through the years. ## ## ## Final plotting challenge: ## ## With all of this information in hand, please take another five ## ## minutes to either improve one of the plots generated in this ## ## exercise or create a beautiful graph of your own. Use the RStudio ## ## ggplot2 cheat sheet for inspiration: ## ## https://www.rstudio.com/wp-content/uploads/2015/08/ggplot2-cheatsheet.pdf ## ## SQL databases and R library(dplyr) mammals <- src_sqlite("data/portal_mammals.sqlite") mammals tbl(mammals, sql("SELECT year, species_id, plot_id FROM surveys")) surveys <- tbl(mammals, "surveys") surveys %>% select(year, species_id, plot_id) ### Challenge ## Write a query that returns the number of rodents observed in each ## plot in each year. ## Hint: Connect to the species table and write a query that joins ## the species and survey tables together to exclude all ## non-rodents. The query should return counts of rodents by year. ## Optional: Write a query in SQL that will produce the same ## result. You can join multiple tables together using the following ## syntax where foreign key refers to your unique id (e.g., ## `species_id`): ## SELECT table.col, table.col ## FROM table1 JOIN table2 ## ON table1.key = table2.key ## JOIN table3 ON table2.key = table3.key ## with dplyr syntax species <- tbl(mammals, "species") left_join(surveys, species) %>% filter(taxa == "Rodent") %>% group_by(taxa, year) %>% tally %>% collect() ## with SQL syntax query <- paste(" SELECT a.year, b.taxa,count(*) as count FROM surveys a JOIN species b ON a.species_id = b.species_id AND b.taxa = 'Rodent' GROUP BY a.year, b.taxa", sep = "" ) tbl(mammals, sql(query)) ### Challenge ## Write a query that returns the total number of rodents in each ## genus caught in the different plot types. ## Hint: Write a query that joins the species, plot, and survey ## tables together. The query should return counts of genus by plot ## type. genus_counts <- left_join(surveys, plots) %>% left_join(species) %>% group_by(plot_type, genus) %>% tally %>% collect() ## if you haven't downloaded the csv files yet, you can do: download.file("https://ndownloader.figshare.com/files/3299474", "data/plots.csv") download.file("https://ndownloader.figshare.com/files/3299483", "data/species.csv") download.file("https://ndownloader.figshare.com/files/2292172", "data/surveys.csv") species <- read.csv("data/species.csv") surveys <- read.csv("data/surveys.csv") plots <- read.csv("data/plots.csv") my_db_file <- "portal-database.sqlite" my_db <- src_sqlite(my_db_file, create = TRUE) my_db ### Challenge ## Add the remaining species table to the my_db database and run some ## of your queries from earlier in the lesson to verify that you ## have faithfully recreated the mammals database.