Joins

Overview

Teaching: 15 min
Exercises: 10 min

Questions

• How do I bring data together from separate tables?

Objectives

• Employ joins to combine data from two tables.

• Apply functions to manipulate individual values.

• Employ aliases to assign new names to tables and columns in a query.

Joins

To combine data from two tables we use the SQL JOIN command, which comes after the FROM command.

The JOIN command on its own will result in a cross product, where each row in the first table is paired with each row in the second table. Usually this is not what is desired when combining two tables with data that is related in some way.

For that, we need to tell the computer which columns provide the link between the two tables using the word ON. What we want is to join the data with the same species id.

SELECT *
FROM surveys
JOIN species
ON surveys.species_id = species.species_id;

ON is like WHERE, it filters things out according to a test condition. We use the table.colname format to tell the manager what column in which table we are referring to.

The output of the JOIN command will have columns from the first table plus the columns from the second table. For the above command, the output will be a table that has the following column names:

Alternatively, we can use the word USING, as a short-hand. USING only works on columns which share the same name. In this case we are telling the manager that we want to combine surveys with species and that the common column is species_id.

SELECT *
FROM surveys
JOIN species
USING (species_id);

The output will only have one species_id column

We often won’t want all of the fields from both tables, so anywhere we would have used a field name in a non-join query, we can use table.colname.

For example, what if we wanted information on when individuals of each species were captured, but instead of their species ID we wanted their actual species names.

SELECT surveys.year, surveys.month, surveys.day, species.genus, species.species
FROM surveys
JOIN species
ON surveys.species_id = species.species_id;

Many databases, including SQLite, also support a join through the WHERE clause of a query.
For example, you may see the query above written without an explicit JOIN.

SELECT surveys.year, surveys.month, surveys.day, species.genus, species.species
FROM surveys, species
WHERE surveys.species_id = species.species_id;

For the remainder of this lesson, we’ll stick with the explicit use of the JOIN keyword for joining tables in SQL.

Challenge:

• Write a query that returns the genus, the species name, and the weight of every individual captured at the site

Solution

SELECT species.genus, species.species, surveys.weight
FROM surveys
JOIN species
ON surveys.species_id = species.species_id;

Different join types

We can count the number of records returned by our original join query.

SELECT COUNT(*)
FROM surveys
JOIN species
USING (species_id);

Notice that this number is smaller than the number of records present in the survey data.

SELECT COUNT(*) FROM surveys;

This is because, by default, SQL only returns records where the joining value is present in the joined columns of both tables (i.e. it takes the intersection of the two join columns). This joining behaviour is known as an INNER JOIN. In fact the JOIN command is simply shorthand for INNER JOIN and the two terms can be used interchangably as they will produce the same result.

We can also tell the computer that we wish to keep all the records in the first table by using the command LEFT OUTER JOIN, or LEFT JOIN for short.

Challenge:

• Re-write the original query to keep all the entries present in the surveys table. How many records are returned by this query?

Solution

SELECT * FROM surveys
LEFT JOIN species
USING (species_id);

Challenge:

• Count the number of records in the surveys table that have a NULL value in the species_id column.

Solution

SELECT COUNT(*)
FROM surveys
WHERE species_id IS NULL;

Remember: In SQL a NULL value in one table can never be joined to a NULL value in a second table because NULL is not equal to anything, not even itself.

Combining joins with sorting and aggregation

Joins can be combined with sorting, filtering, and aggregation. So, if we wanted average mass of the individuals on each different type of treatment, we could do something like

SELECT plots.plot_type, AVG(surveys.weight)
FROM surveys
JOIN plots
ON surveys.plot_id = plots.plot_id
GROUP BY plots.plot_type;

Challenge:

• Write a query that returns the number of animals caught of each genus in each plot. Order the results by plot number (ascending) and by descending number of individuals in each plot.

Solution

SELECT surveys.plot_id, species.genus, COUNT(*) AS number_indiv
FROM surveys
JOIN species
ON surveys.species_id = species.species_id
GROUP BY species.genus, surveys.plot_id
ORDER BY surveys.plot_id ASC, number_indiv DESC;

Challenge:

• Write a query that finds the average weight of each rodent species (i.e., only include species with Rodent in the taxa field).

Solution

SELECT surveys.species_id, AVG(surveys.weight)
FROM surveys
JOIN species
ON surveys.species_id = species.species_id
WHERE species.taxa = 'Rodent'
GROUP BY surveys.species_id;

Functions IFNULL and NULLIF and more

SQL includes numerous functions for manipulating data. You’ve already seen some of these being used for aggregation (SUM and COUNT) but there are functions that operate on individual values as well. Probably the most important of these are IFNULL and NULLIF. IFNULL allows us to specify a value to use in place of NULL.

We can represent unknown sexes with 'U' instead of NULL:

SELECT species_id, sex, IFNULL(sex, 'U')
FROM surveys;

The lone “sex” column is only included in the query above to illustrate where IFNULL has changed values; this isn’t a usage requirement.

Challenge:

• Write a query that returns 30 instead of NULL for values in the hindfoot_length column.

Solution

SELECT hindfoot_length, IFNULL(hindfoot_length, 30)
FROM surveys;

Challenge:

• Write a query that calculates the average hind-foot length of each species, assuming that unknown lengths are 30 (as above).

Solution

SELECT species_id, AVG(IFNULL(hindfoot_length,30))
FROM surveys
GROUP BY species_id;

IFNULL can be particularly useful in JOIN. When joining the species and surveys tables earlier, some results were excluded because the species_id was NULL in the surveys table. We can use IFNULL to include them again, re-writing the NULL to a valid joining value:

SELECT surveys.year, surveys.month, surveys.day, species.genus, species.species
FROM surveys
JOIN species
ON IFNULL(surveys.species_id,'AB') = species.species_id;

Challenge:

• Write a query that returns the number of animals caught of each genus in each plot, using IFNULL to assume that unknown species are all of the genus “Rodent”.

Solution

SELECT plot_id, IFNULL(genus, 'Rodent') AS genus2, COUNT(*)
FROM surveys 
LEFT JOIN species
ON surveys.species_id=species.species_id
GROUP BY plot_id, genus2;

The inverse of IFNULL is NULLIF. This returns NULL if the first argument is equal to the second argument. If the two are not equal, the first argument is returned. This is useful for “nulling out” specific values.

We can “null out” plot 7:

SELECT species_id, plot_id, NULLIF(plot_id, 7)
FROM surveys;

Some more functions which are common to SQL databases are listed in the table below:

Finally, some useful functions which are particular to SQLite are listed in the table below:

Challenge:

Write a query that returns genus names (no repeats), sorted from longest genus name down to shortest.

Solution

SELECT DISTINCT genus
FROM species
ORDER BY LENGTH(genus) DESC;

As we saw before, aliases make things clearer, and are especially useful when joining tables.

SELECT surv.year AS yr, surv.month AS mo, surv.day AS day, sp.genus AS gen, sp.species AS sp
FROM surveys AS surv
JOIN species AS sp
ON surv.species_id = sp.species_id;

To practice we have some optional challenges for you.

Challenge (optional):

SQL queries help us ask specific questions which we want to answer about our data. The real skill with SQL is to know how to translate our scientific questions into a sensible SQL query (and subsequently visualize and interpret our results).

Have a look at the following questions; these questions are written in plain English. Can you translate them to SQL queries and give a suitable answer?

1. How many plots from each type are there?

2. How many specimens are of each sex are there for each year?

3. How many specimens of each species were captured in each type of plot?

4. What is the average weight of each taxa?

5. What are the minimum, maximum and average weight for each species of Rodent?

6. What is the average hindfoot length for male and female rodent of each species? Is there a Male / Female difference?

7. What is the average weight of each rodent species over the course of the years? Is there any noticeable trend for any of the species?

Proposed solutions:

1. Solution:

   SELECT plot_type, COUNT(*) AS num_plots
   FROM plots
   GROUP BY plot_type;
   

1. Solution:

   SELECT year, sex, COUNT(*) AS num_animal
   FROM surveys
   WHERE sex IS NOT NULL
   GROUP BY sex, year;
   

1. Solution:

   SELECT species_id, plot_type, COUNT(*) 
   FROM surveys 
   JOIN plots USING(plot_id) 
   WHERE species_id IS NOT NULL 
   GROUP BY species_id, plot_type;
   

1. Solution:

   SELECT taxa, AVG(weight) 
   FROM surveys 
   JOIN species ON species.species_id = surveys.species_id
   GROUP BY taxa;
   

1. Solution:

   SELECT surveys.species_id, MIN(weight), MAX(weight), AVG(weight) FROM surveys 
   JOIN species ON surveys.species_id = species.species_id 
   WHERE taxa = 'Rodent' 
   GROUP BY surveys.species_id;
   

1. Solution:

   SELECT surveys.species_id, sex, AVG(hindfoot_length)
   FROM surveys JOIN species ON surveys.species_id = species.species_id 
   WHERE (taxa = 'Rodent') AND (sex IS NOT NULL) 
   GROUP BY surveys.species_id, sex;
   

1. Solution:

   SELECT surveys.species_id, year, AVG(weight) as mean_weight
   FROM surveys 
   JOIN species ON surveys.species_id = species.species_id 
   WHERE taxa = 'Rodent' GROUP BY surveys.species_id, year;
   

Key Points

• Use the JOIN command to combine data from two tables—the ON or USING keywords specify which columns link the tables.

• Regular JOIN returns only matching rows. Other join commands provide different behavior, e.g., LEFT JOIN retains all rows of the table on the left side of the command.

• IFNULL allows you to specify a value to use in place of NULL, which can help in joins

• NULLIF can be used to replace certain values with NULL in results

• Many other functions like IFNULL and NULLIF can operate on individual values.