查找sql​​中两个组合列的平均值

2021-12-30 00:00:00 average sql count mysql

我想找到两列总数的平均值.我想计算 col1 的总数和 col2 的总数，然后找到平均值(它们在多少不同的行中).

I want to find the avg of the total of two columns. I want to count the total of col1 and the total of col2 then find the average(how many different rows they are in).

我设法在这个 sqlfiddle (另见下文)这是最好的方法吗?我最初认为我需要使用 avg 函数，但无法使用它来解决.

I have managed to come up with a solution in the this sqlfiddle (also see below) is this the best way? I initially thought I would need to use the avg function but couldn't work it out using this.

    CREATE TABLE test (
        id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
        uid INT,
        col1 INT,
        col2 INT
    ) DEFAULT CHARACTER SET utf8 ENGINE=InnoDB;

    INSERT INTO test (id, uid, col1, col2) VALUES
    (1,5,8,12),
    (2,1,2,3),
    (3,1,2,33),
    (4,5,25,50),
    (5,5,22,3);

    (
    SELECT ((sum(col1) + sum(col2))/count(*))
    FROM test
      WHERE uid=5
    )

推荐答案

根据定义，AVG(col1) = SUM(col1)/COUNT(*) 和 AVG(col2) =SUM(col2)/COUNT(*)，因此 (SUM(col1)+SUM(col2))/COUNT(*) = AVG(col1) + AVG(col2)).

By definition, AVG(col1) = SUM(col1)/COUNT(*) and AVG(col2) = SUM(col2)/COUNT(*), therefore (SUM(col1)+SUM(col2))/COUNT(*) = AVG(col1) + AVG(col2).

此外，加法的可交换性给了我们 (SUM(col1)+SUM(col2))/COUNT(*) = SUM(col1+col2)/COUNT(*) 因此 AVG(col1+col2).

Also, the commutativity of addition gives us (SUM(col1)+SUM(col2))/COUNT(*) = SUM(col1+col2)/COUNT(*) and hence AVG(col1+col2).

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