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The AVG Function
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To find averages, we use the AVG function. The parameter for the function that is, what goes in the parentheses (. . .) is the expression you want to average. The expression must have a numeric value. The expression could be just the name of one of the numeric-valued columns or some function of a value, such as the length of a piece of text or the number of days between two dates. For example, we can find the average handicap for members of our club by averaging the values in the Handicap column, as in Listing 8-9.
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Listing 8-9. Return the Average Handicap SELECT AVG(Handicap) FROM Member
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As with the COUNT function, the AVG function includes only non-Null values for the handicap. We have 20 members in total, and 17 members with handicaps. If we sum all the handicaps, we get 287. The AVG function will take the total of the handicaps (287) and divide by the number of rows that have a value in the Handicap column (17). This is what we want. If we included the members without handicaps (by dividing by the total number of rows, 20), we would essentially be saying that these members have a handicap of 0 by default, which is not at all what we want in this case. It is not always so obvious whether you want the Null values considered. For example, say we have another database with a table called Student and a column called TestScore. If we enter test scores for students, and some of the students do not take the test, then we will have a Null in the TestScore column for that student. What do we mean by the average score We could take the average over all the students (divide the total score by the count of all students), which means the students who missed the test are effectively being counted as
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C HA PTE R 8 AGGREGA TE O PE RAT IO NS
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having scored 0. On the other hand, we might take the average of just those who participated in the test (divide by the number who took the test). People (especially academics!) will argue about such things. AVG(TestScore) will always give us just the average for those who took the test (which is what I personally think we want). If we want the average over all the students, including those with a Null mark (counted as 0), we can calculate the average by hand totaling the marks (using the SUM function) and dividing by the full count. This computed average is shown in Listing 8-10.
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Listing 8-10. Calculate an Average Where Null Values Are Counted As Zero SELECT SUM(TestMark)/COUNT(*) FROM Student
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The query in Listing 8-10 is preferable to entering a mark of 0 for students who missed the test. If we do that, then we can no longer distinguish students who took the test and got 0 from students who missed the test (and even academics will agree this distinction is useful). As with the COUNT function, the AVG function can also incorporate the keywords ALL and DISTINCT. Just be aware that ALL means all the non-Null values, as opposed to distinct non-Null values. It doesn t mean take an average over all the rows (including those with Nulls), as in our discussion about test scores. I find it quite hard to come up with examples of when you would want to just average over distinct values certainly none that apply to our club database. How do the different types of the fields used as a parameter to the AVG function affect the result The AVG function will accept only numeric types. We can t attempt to average FirstName or JoinDate (although we could use functions to average the length of members first names or the number of days since their join date). What result do we expect to get when we average the handicaps of our members The total of the handicaps is 287, and the number of people with handicaps is 17. If you divide these two numbers with a calculator, you get something like 16.88235. What will SQL give us That depends. When I try this in Access 2007, I get 16.88235. In SQL Server 2005, I get 16. In SQL Server (and some other implementations of SQL), the average function returns the same type as the numbers being averaged. In this case, the Handicap column is an INT type, and so AVG(Handicap) in SQL Server returns an integer. It also does an integer division (which means the result is truncated to 16 rather than rounded to 17). We do have control over how the result is calculated. If we want to get a noninteger result for our average, we can convert the Handicap value to a floating-point number before we do the average. To do this we can use the CONVERT function, described in 7.1 Another way to do this is just to multiply the handicap by 1.0: AVG(Handicap * 1.0). The SQL Server syntax using the CONVERT function is shown in Listing 8-11.
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1. Different versions of SQL will have different functions to do this. In Oracle, you might consider using the CAST function.
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