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How Can I Simulate a Lag Function in MySQL to Calculate Differences Between Successive Values?

Linda Hamilton
Release: 2025-01-18 00:56:08
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How Can I Simulate a Lag Function in MySQL to Calculate Differences Between Successive Values?

Simulating lag function in MySQL

Calculating the difference between consecutive values ​​in a time series (called a lag) can be implemented in MySQL using a clever workaround. To understand this technique, let us consider the table provided:

<code>| time                | company | quote |
+---------------------+---------+-------+
| 0000-00-00 00:00:00 | GOOGLE  |    40 |
| 2012-07-02 21:28:05 | GOOGLE  |    60 |
| 2012-07-02 21:28:51 | SAP     |    60 |
| 2012-07-02 21:29:05 | SAP     |    20 |</code>
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To simulate the lag function and calculate the difference in quotes, we adopt the following strategy:

  1. Initializes two variables, @quot has an initial value of -1, and @curr_quote is used to track the current quote value.
  2. Execute the following query to iterate through the table, updating the variables with each row:
<code class="language-sql">SET @quot=-1;
SELECT time, company, @quot AS lag_quote, @quot:=quote AS curr_quote
FROM stocks
ORDER BY company, time;</code>
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    The
  1. lag_quote column now holds the previous row's bid value, and curr_quote holds the current row's bid value.

To obtain the desired output format (company and quote differences):

<code>GOOGLE | 20
SAP    | 40</code>
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We use nested queries:

<code class="language-sql">SET @quot=0,@latest=0,company='';
SELECT B.*
FROM (
    SELECT A.time, A.change, IF(@comp<>A.company,1,0) AS LATEST, @comp:=A.company AS company
    FROM (
        SELECT time, company, quote-@quot AS change, @quot:=quote AS curr_quote
        FROM stocks
        ORDER BY company, time
    ) A
    ORDER BY company, time DESC
) B
WHERE B.LATEST=1;</code>
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This method effectively simulates a lag function by maintaining a session variable that keeps track of the previous row's value. While it looks computationally intensive, the nested queries are not related to each other, so it's still relatively efficient.

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