optimization for the calculation

Hello

Is there a faster way to calculate this rejection of common-mode...? I have een EEG signal (the first 64 channels) and they must be added and divide by the number of available channels, then the signal subtracte. See photo:

Best regards

Thijs Boeree

Hey Taylor,

Use a loop:

On my PC this routine takes ~ 50µs...

Tags: NI Software

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    (     ord_no          NUMBER(5),
         plan_start_dt     DATE,
         CONSTRAINT orders_pk PRIMARY KEY (ord_no)
    );
    
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    VALUES (1,To_Date('01/08/1980','mm/dd/yyyy'));
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    VALUES (3,To_Date('01/10/1980','mm/dd/yyyy'));
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    ,     orders.plan_start_dt
    ,     add_mdays(orders.plan_start_dt, -3) AS prep_date
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    Published by: user11033437 on February 2, 2010 08:55
    Fixed some typos in the last insert statements

    Published by: user11033437 on February 2, 2010 09:31 (fixed some syntax in the function)

    Hello

    Ah, referring to Oracle 8 and is not not able to test your own code makes me nostalgic for the good old days, when you have entered your cards and led to a window to the computer center and waited an hour for the work to be performed and then seen printing to find that you had made a typo.

    If you write functions, you should really test yourself. Like all codes, functions forge be written small not: write a line or two (or sometimes just a part of what would later become a single line), test, make sure it is running properly and repeat.
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    You need a function to get the desired results:

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    ,       o.plan_start_dt
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    FROM       orders     o
    ,       caln          l
    ,       caln          e
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    AND       e.shop_days     = l.shop_days - 3
    GROUP BY  o.ord_no
    ,            o.plan_start_dt
    ;
    

    It would be more effective (and somewhat simpler) If you've added a column (let's call it work_day) identified whether each line represents a work_day or not.
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    SELECT       o.ord_no
    ,       o.plan_start_dt
    ,       e.clndr_dt          AS prep_date
    FROM       orders     o
    ,       caln          l
    ,       caln          e
    WHERE       l.clndr_dt     = o.plan_start_dt
    AND       e.shop_days     = l.shop_days - 3
    AND       e.work_day     = 1
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    You can use the analytic LAG function to populate the work_day column.

    A function would certainly be useful, although perhaps slower.

    The function you have posted has some errors:
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    (b) when you SELECT in PL/SQL, as you do, you must SELECT a variable to store the results.
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    Given your current caln table, here's how I would write the function:

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         ( start_date     IN           DATE          DEFAULT     SYSDATE,
           mdays          IN           NUMBER          DEFAULT     1
         )
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    IS
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         --     will be before start_date).
         --     Work days do not include Saturdays, Sundays or holidays
         --     as indicated in the caln table.
    
         new_date     DATE;          -- to be returned
    BEGIN
    
         SELECT     MIN (t.clndr_dt)
         INTO     new_date
         FROM     caln     f     -- f stands for "from"
         ,     caln     t     -- t stands for "to"
         WHERE     f.clndr_dt     = TRUNC (start_date)
         AND     t.shop_days     = f.shop_days + TRUNC (mdays)
         ;
    
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    END     add_mdays;
    /
    SHOW ERRORS
    

    Production code forge be robust (which includes "fool-proofing").
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    I wish I could score questions such as 'Correct' or 'useful '; you get 10 points for sure.
    You posted CREATE TABLE and INSERT statements (without even be begged).
    You gave a clear description of the problem, including the expected results.
    The code is well formatted and easy to read.
    All around, one of the more thoughtful and well written questions I've seen.
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    Published by: Frank Kulash, February 2, 2010 13:10
    Added to my own version of the function.

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