03-sql-joins-aliases.md

layout	root	title	minutes
lesson	.	Joins and aliases	30

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 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 codes.

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 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:

record_id	month	day	year	plot_id	species_id	sex	hindfoot_length	weight	species_id	genus	species	taxa
...
96	8	20	1997	12	DM	M	36	41	DM	Dipodomys	merriami	Rodent
...

Alternatively, we can use the word USING, as a short-hand. 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

record_id	month	day	year	plot_id	species_id	sex	hindfoot_length	weight	genus	species	taxa
...
96	8	20	1997	12	DM	M	36	41	Dipodomys	merriami	Rodent
...

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;

year	month	day	genus	species
...
1977	7	16	Neotoma	albigula
1977	7	16	Dipodomys	merriami
...

Challenge:

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

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 genus of the animals caught in each plot in descending order.

Challenge:

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

Functions

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') AS non_null_sex
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.

Challenge:

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

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. 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 surveys.species_id = IFNULL(species.species_id, 'AB');

Challenge:

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

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) AS partial_plot_id
FROM surveys;

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

Function	Description
ABS(n)	Returns the absolute (positive) value of the numeric expression n
LENGTH(s)	Returns the length of the string expression s
LOWER(s)	Returns the string expression s converted to lowercase
NULLIF(x, y)	Returns NULL if x is equal to y, otherwise returns x
ROUND(n) or ROUND(n, x)	Returns the numeric expression n rounded to x digits after the decimal point (0 by default)
TRIM(s)	Returns the string expression s without leading and trailing whitespace characters
UPPER(s)	Returns the string expression s converted to uppercase

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

Function	Description
IFNULL(x, y)	Returns x if it is non-NULL, otherwise returns y
RANDOM()	Returns a random integer between -9223372036854775808 and +9223372036854775807.
REPLACE(s, f, r)	Returns the string expression s in which every occurrence of f has been replaced with r
SUBSTR(s, x, y) or SUBSTR(s, x)	Returns the portion of the string expression s starting at the character position x (leftmost position is 1), y characters long (or to the end of s if y is omitted)

Challenge:

Write a query that returns genus names, sorted from longest genus name down to shortest.

Aliases

As queries get more complex names can get long and unwieldy. To help make things clearer we can use aliases to assign new names to things in the query.

We can alias both table names:

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

And column names:

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;

The AS isn't technically required, so you could do

SELECT surv.year yr
FROM surveys surv;

but using AS is much clearer so it is good style to include it.

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 is the percentage of each species in each taxa?

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

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

8. 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 ORDER BY num_plots DESC

2. Solution: SELECT year, sex, count(*) AS num_animal FROM surveys WHERE sex IS NOT null GROUP BY sex, year

3. Solution: SELECT species_id, plot_type, count(*) FROM surveys JOIN plots ON surveys.plot_id=plots.plot_id WHERE species_id IS NOT null GROUP BY species_id, plot_type

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

5. Solution: SELECT taxa, 100.0*count(*)/(SELECT count(*) FROM surveys) FROM surveys JOIN species ON surveys.species_id=species.species_id GROUP BY taxa

6. Solution: SELECT surveys.species_id, MIN(weight) as min_weight, MAX(weight) as max_weight, AVG(weight) as mean_weight FROM surveys JOIN species ON surveys.species_id=species.species_id WHERE taxa = 'Rodent' GROUP BY surveys.species_id

7. Solution: SELECT surveys.species_id, sex, AVG(hindfoot_length) as mean_foot_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

8. 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

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