best linear regression r2

Similar Questions

• non-linear regression

  Hello

  I need to make a non-linear regression in LabWindows but I don't not eat in order to find a function for that...

  I know that there s for linear regression (LinFit) function, but not could find one for non-linear... Is there such function CVI?

  Or maybe someone has an algorithm to do this...

  Any help is very appreciated!

  Thank you

  ZZippo

  Hi ZZippo,

  as long as you have the AdvancedAnalysisLibrary there are three related functions for nonlinear curve fit:

  NonLinearFit (-);

  NonLinearFitWithMaxIters (, 100);

  NonLinearFitWithWeight (, 100);

  the last two offer more options than the first.

  If you do not have this library (i.e. the full version of the CVI instead of the basic version), you can use the following implementation of Levenberg-Marquardt:http://www.ics.forth.gr/~lourakis/levmar/

• Line of best fit / linear regression

  I am trying ot get a line to better adapt to a set of data in oracle
  
  Let's say I have the following dataset.
  Since there are a lot of 0 in the terms I want to ignore these terms
  Use only terms that have values for a line of regression.
  
  Any help will be appreciated.
  Thank you.
  

         TERMS            Count
  TERM_0801     78
  TERM_0802     58
  TERM_0803     0
  TERM_0804     82
  TERM_0805     0
  TERM_0806     0
  TERM_0807     32
  TERM_0808     0
  TERM_0901     92
  TERM_0902     0
  TERM_0903     0
  TERM_0904     56
  TERM_0905     0
  TERM_0906     0
  TERM_0907     0
  TERM_0908     0
  TERM_1001     85
  TERM_1002     0
  TERM_1003     0
  TERM_1004     67

  Oracle Database 10g Enterprise Edition Release 10.2.0.5.0 - 64bi
  
  Published by: Chloe_19 on July 24, 2012 23:14

  Maybe:

  with
  the_data(terms,count) as
  (select 'TERM_0801',78 from dual union all
   select 'TERM_0802',58 from dual union all
   select 'TERM_0803',0 from dual union all
   select 'TERM_0804',82 from dual union all
   select 'TERM_0805',0 from dual union all
   select 'TERM_0806',0 from dual union all
   select 'TERM_0807',32 from dual union all
   select 'TERM_0808',0 from dual union all
   select 'TERM_0901',92 from dual union all
   select 'TERM_0902',0 from dual union all
   select 'TERM_0903',0 from dual union all
   select 'TERM_0904',56 from dual union all
   select 'TERM_0905',0 from dual union all
   select 'TERM_0906',0 from dual union all
   select 'TERM_0907',0 from dual union all
   select 'TERM_0908',0 from dual union all
   select 'TERM_1001',85 from dual union all
   select 'TERM_1002',0 from dual union all
   select 'TERM_1003',0 from dual union all
   select 'TERM_1004',67 from dual
  )
  select regr_slope(cnt,rn) r_slope,
         regr_intercept(cnt,rn) r_intercept,
         regr_count(cnt,rn) r_count,
         regr_r2(cnt,rn) r_r2,
         regr_avgx(cnt,rn) r_avgx,
         regr_avgy(cnt,rn) r_avgy,
         regr_sxx(cnt,rn) r_sxx,
         regr_syy(cnt,rn) r_syy,
         regr_sxy(cnt,rn) r_sxy
    from (select count cnt,row_number() over (order by terms) rn
            from the_data
         )
   where cnt != 0
  
   R_SLOPE          | R_INTERCEPT      | R_COUNT | R_R2                | R_AVGX | R_AVGY | R_SXX | R_SYY  | R_SXY
  0,241071428571429 | 66,5803571428571 |       8 | 0,00723884549185754 |      9 |  68,75 |   336 | 2697,5 |    81
  

  Concerning

  Etbin

• PLSQL - prediction of linear regression.

  Hey, Chaps.

  I'm puzzled!  I hope you can help...

  SQL

  WITH tblData AS
  (
      SELECT  1 AS known_ys, TO_DATE('01/01/2015', 'DD/MM/YYYY') AS known_xs FROM DUAL UNION ALL
      SELECT  2, TO_DATE('02/01/2015', 'DD/MM/YYYY') FROM DUAL UNION ALL
      SELECT  3, TO_DATE('03/01/2015', 'DD/MM/YYYY') FROM DUAL UNION ALL
      SELECT  4, TO_DATE('04/01/2015', 'DD/MM/YYYY') FROM DUAL
  )
      SELECT  *
        FROM  tblData
  ;
  
  

  What I'm trying to do, is provide by using my known_ys what happens on 01/05/2015 (known_xs).  The answer is of course 5

  In Excel, I'd FORECAST ("01/05/2015", known_ys, knownxs) and it spits out 5.  Unfortunately my data are so large in Excel (and other stats packages) I need to push the calculation in the Oracle server.

  I spent 30 minutes to read the Sub and get more confused -anyone know (and ideally give an example) how to proceed?

  Created this just now that works but it is a little "garbage."

  WITH tblData AS
  (
      SELECT  1 AS known_ys, TO_DATE('01/01/2015', 'DD/MM/YYYY') AS known_xs FROM DUAL UNION ALL
      SELECT  2, TO_DATE('02/01/2015', 'DD/MM/YYYY') FROM DUAL UNION ALL
      SELECT  4, TO_DATE('03/01/2015', 'DD/MM/YYYY') FROM DUAL UNION ALL
      SELECT  8, TO_DATE('04/01/2015', 'DD/MM/YYYY') FROM DUAL
  )
      SELECT  last_value,
              REGR_SLOPE(known_ys, (known_xs - SYSDATE)) AS slope,
              last_value + REGR_SLOPE(known_ys, (known_xs - SYSDATE)) AS nxt_value
        FROM  tblData, 
              (SELECT FIRST_VALUE(known_ys) OVER (PARTITION BY 1 ORDER BY known_xs DESC) AS last_value FROM tblData)
          
    GROUP BY  last_value
  ;
  

  Something like this (no PL/SQL required):

  WITH tblData AS

  (

  SELECT 1 AS known_ys, TO_DATE('01/01/2015', 'DD/MM/YYYY') AS known_xs DE DOUBLE UNION ALL

  SELECT 2, TO_DATE (FEBRUARY 1, 2015 ',' DD/MM/YYYY "") OF THE WHOLE UNION DOUBLE

  CHOOSE 3, TO_DATE (MARCH 1, 2015 ',' DD/MM/YYYY ') FROM DUAL UNION ALL

  SELECT 4, TO_DATE (APRIL 1, 2015 ',' DD/MM/YYYY ') FROM DUAL

  )

  SELECT REGR_INTERCEPT (known_ys, known_xs - TO_DATE (May 1, 2015 ',' DD/MM/YYYY ')) x

  OF tblData

  ;

  SQL > /.

  X

  ----------

  5

  http://docs.Oracle.com/database/121/SQLRF/functions165.htm#SQLRF00696

  REGR_INTERCEPT predicted where the graph will cross the axis, so I subtracted TO_DATE(5/01/2015,'DD/MM/YYYY') get a data type number, 0 being equivalent to the date that I was interested.

  REGR_ features * supports window clauses, you can perform calculations on multiple series of data in the same table with a single pass in the data.

• How to choose between a linear and logistic model in the regression of the glm?

  Hello

  documentation, I learned that the algorithm of glm for the linear regression model supports linear and logistic regression. However, I can't find any switch or parameter that selects one or the other. How can I tell ODM what kind of regression I want?

  Thanks in advance!

  Hi Bernd,

  Here are some additional tips.

  Hope this is what you are looking for.

  THX, mark

  If you specify a classification, it is logistics. If you specify the regression, linear regression. Logistic regression only we offer is the variance and binary logistic regression canonical link. If you have a target with values in [0,..., 100] and you want to use logistic regression of glm, and then create a view that selects a share point, for example, 50. There is no parameter variation algorithm.

  Create view bin_v as

  to select which target< 50="" then="" 0="" else="" 1="" end="">

  var1, var2,... varn my_data.

  Otherwise if you want linear regression, specify regression and let the data it is.

• How to use BIEE to analysis of the data. as the regression

  I want to know information about the feature of oracle BIEE. and I need to regression analysis or other logistical methods in my data.how to solve it thanks!

  Hi user,

  Take a look: http://kpipartners.blogspot.com/2009/04/linear-regression.html

  Thank you
  Saichand.v

• HP 10BII + statistics PN questions?

  After getting weird with the HP 10BII PN + statistics (unexpected) results, I dug out some old books with sample data and checked I get different results on the 10BII + on other machines.

  Examples of data of HP - 34 c Manual (p. 38):

  x / y: 696/1945, 994/1950, 1330/1955, 1512, 1960, 1965-1750, 2162/1970, 2243/1971, 2382/1972, 2482/1973.

  I enter all 9 data and even on the 10BII +, I check the linear regression and then use:

  RS-5, RS - K (Swap) to get the slope, m - HP 10BII + returns 0.01612 [should be 61.1612]

  RS-6, RS - K (Swap) to get the y originally b - HP 10BII + return 1934.1695 [should be-118, 290.6295]

  Maybe I'm missing something basic, but the 10BII + gives the answers that appear all just bad. I checked the "should be" translated on a HP - 34 c, WP 34s and HP - 15 c and they agree.

  I read (and reread) the 10BII User Guide + to ensure the correct keystrokes, and also made C.ALL to ensure no conflict of memory (e.g. too many CFlows, etc.). Curiously, linear estimates appear to be OK when I interpolate w/samples.

  What happens to Tim?

  but here's an interesting result...

  If the order of the columns is changed (x and y values are reversed)...

  X Y
  [[1945 696.]
  [1950 994.]
  [1955 1330.]
  [1960 1512.]
  [1750 1965.]
  [2162 1970.]
  [1971 2243.]
  [1972 2382.]
  [1973 2482.]]

  Now run a linear regression model... and the result is...

  '(-118242.173643) +(61.1364341085*X)'

  the issue seems to be the way in which data is entered into the calculator...

• PREMIUM HP: HP PREMIUM: improve 2 VAR APP with interception of axis Y

  Here the desired improvement:

  With MS Excel, in linear regression is chosen, it is possible to define the intercept with the y-axis.

  Its is interesting when you know that the correct answer is y = M * X (for example: d = V * t or A = k * C (beer-Lambert),...)

  So, I suggest to add this new setting with linear, polynomial, and exponential regression.

  Thank you.

  Simpler: F1 (X) = Statistics_2VAR Sigma (XY) / Statistics_2VAR. Sigma(x2)

  Then 'User Defined' and fit: F1 (X)

• NoValue handling

  I'm relatively new to VBS and difficulties to come with a solution. I have a sine wave which I need to do some calculations on. I need to cut a window each cycle to +-1000N. To calculate the linear regression on each of these sections. Than the average slopes that positive/negative. I'm looking to produce 2 plots, one to the positive terminal and the other for negative in the end. So far, I was able to cut the data and insert the NoValues. I'm looking for assistance under each of the linear regression between the NoValues data sections.

  Dim oChn, K, currentValue lastValue, I have

  Calculate call ("Ch(""[1]/Total)" "") = Ch(""[1]/Filtered_Total_TR_Load"") + CTNV (Ch(""[1]/Filtered_Total_TR_Load"") > 1000) ", NULL, NULL," ")"
  Calculate call ("Ch(""[1]/Total)" "") = Ch(""[1]/Total"") + CTNV (Ch(""[1]/Filtered_Total_TR_Load"")<>

  lastValue = 0

  Set oChn = data. GetChannel ("Total")

  For K = 1 To oChn.Size
  currentValue = oChn (K)

  If (IsNull (currentValue)) then
  oChn (K) = lastValue

  On the other
  lastValue = currentValue

  End If

  Next

  Hi mindmaster,.

  What version of DIAdem do you use?  If you have DIAdem 2015, then you can use the new ChnEvent... functions () to find the beaches of the line of all your tracks positive and negative.  From there you can loop over each of them, extraction ranging from the line to a new channel and the execution of the function of linear regression from the ANALYSIS Panel on this subset extracted to calculate the slope of this cycle.  You can then save the values of slope in the new 'Cycle number' and 'Positive slope' channels 'negative slope", which you can then average on all cycles.

  Even if you have DIAdem 2014, I can show you the pre-release version hidden in the ChnEvent... functions () which will allow the first step of the preceding analysis.  If you have an even older version of DIAdem, you must use the ChnFind() function in a loop to determine the row of slopes of positive and negative of each cycle ranges.

  Brad Turpin

  Tiara Product Support Engineer

  National Instruments

• Function MATLAB Detrend in DIAdem

  Hello

  I'm trying to pre process certain data for the treatment of the fft in Matlab, I would use the Detrend function and I was wondering if it's something similar in tiara.

  I use DIAdem 2012

  Thank you

  Essentially in Matlab Detrend on a simple matrix subtracts the linear regression of each data value line in a table.

  For example, if you have a table a [2,3,5,7,5] and you had to he relaxes in MatLab, it would equal detA [-04 - 0.4, 0.6, 1.6, -1,4].

  If you subtract A detA you will get A-detA [2.4, 3.4, 4.4, 5.4, 6.4] which is exaclty same find the linera for optimal adjustment to A line.

• noise of biaxial precision accelerometer

  Hello

  I do calibration on accelerometer analog biaxial acceleration providing the angle of the sensor measurement.

  Based on spec sheet of the analog, the sensor should be in equation linear V = Ka + B

  V = output voltage

  K = linear sensitivity

  a = acceleration of axis

  B = polarization voltage.

  However, I turn the sensor with a known angle, I get the tension. Draw the graph of the voltage compared to acceleration, I'm not able to obtain a linear curve V = Ka + B, but not linear graph.

  How can I determine the non-linear error and the resolution on the graph newly traced in labview? I collected data. and I want to do a regression analysis. How can I do it in labview?

  After getting the non-linear regression, how will I know my new model (new equation) improved in error and resolution?

  Can anyone help ple

  your kindness is much appreciated

  Concerning

  OK, this description makes sense.  Probably nothing wrong with your Setup.  If you want to create an equation using data points you acquired during the tilt in the X direction and Y?  The function Arctan not serve the purpose?  Divide all values of all values of Ay Ax and feed the ATAN function table.  I used three-point line in the following example, and the output is linear.

  

• Levenberg-Marquardt optimization question

  Hello

  I am trying to use the Levenberg-Marquardt algorithm, using a technique of optimization error percentage instead of the technique of the absolute error. AFAIK the Levenberg Marquardt in Labview algorithm uses the technique of the absolute error (i.e. Y - f(x)) when calculating on the errors. How can I change this option to use the technique of percentage error, when the error = [(Y - f(x)) / y] * 100? I changed the calculation of the error in the algorithm to calculate the error in this way, but the results were not satisfactory. I wonder if I forget something or there at - it much more to what meets the eye? I know that in other programs like Mathematica you can choose how you want the error calculated, but I do not see this option in the Labview vi. Any help appreciated.

  Thank you

  LMS

  Additional information:

  I'm using Labview version 8.2

  I changed the Subvi ABX to calculate the error differently. The ABX vi contains a Subvi called gradient and function of LM. The output of this Subvi calculates better suited like Y - the error. I changed this calculation to be [((ajustement Y-meilleur) / y) * 100] to obtain the percentage of error instead. This affects alpha, beta and chisqr. I think I may need to make other changes to the algorithm to implement the technique of error in percentage but I don't know what changes to make. The results that I have received the above changes were worse than the results I got from the method of the absolute error (error = Y – the better suited). This should not be the case. I implement the VI of the Levenberg-Marquardt algorithm format and not the string of the formula. Any help appreciated.

  Thank you

  LMS

  Hi LMS.

  I don't know if I understand.

  However, as the Lev - Mar algorithm calculates Y - f (x). If you divide that result by the best linear fit (Y) and multiply by 100. This will give you your desired effect.

  I hope this helps you.

• Analysis of capacity Vcops algorithms

  Hello

  I would like to ask if anyone knows what are the specific algorithms used by Vcops to implement the predictive ability. Are they simple algorithms based on regression analysis for example or something more advanced, such as combined neural networks, etc.?

  Thank you!

  Its much more than just linear regression - there are at least 8 algorithms that work and are in competition to interepret data:

  http://www.VMware.com/files/PDF/vCenter/VMware-vCenter-operations-sophisticated-dynamic-thresholds-WP-en.PDF

• Forecasting functions - orders

  Is there a function I could use in a sql statement to predict orders for the same months in 2013 according to the planned 2012 data? Thank you.
  
  Full month
  01/01/2012 4 077
  01/02/2012 4 044
  01/03/2012 5 188
  01/04/2012 5 244
  01/05/2012 5 741
  01/06/2012 6 284
  01/07/2012 6 447
  01/08/2012 6 645
  01/09/2012 6 102
  01/10/2012 6 313
  01/11/2012 5 973
  01/12/2012 5 197
  01/01/2013 5 138

  Hello

  989618 wrote:
  Is there a function I could use in a sql statement to predict orders for the same months in 2013 according to the planned 2012 data?

  Of course, there are many functions.
  A common way to forecast includes regression functions, like REGR_INTERCEPT and REGR_SLOPE, like this:

  WITH   got_d         AS
  (
       SELECT     month
       ,     complete
       ,     month - DATE '2000-01-01'     AS d
       FROM     table_x
  --     WHERE     ...     -- If you want any filtering, put it here
  )
  ,     regr_results          AS
  (
       SELECT     REGR_SLOPE     (complete, d)     AS slope
       ,     REGR_INTERCEPT (complete, d)     AS intercept
       FROM     got_d
  )
  ,     all_months     AS
  (
       SELECT     ADD_MONTHS ( DATE '2011-12-01'
                    , LEVEL
                    )     AS month
       FROM    dual
       CONNECT BY     LEVEL     <= 24
  )
  SELECT       m.month
  ,       d.complete
  ,       (r.slope * (m.month - DATE '2000-01-01'))
             + r.intercept          AS estimated
  FROM              all_months     m
  LEFT OUTER JOIN  got_d          d  ON  d.month  = m.month
  CROSS JOIN       regr_results     r
  ORDER BY  m.month
  ;
  

  The 'magic' date of 1 January 2000 is an arbitrary point where we can count the days. You can use any DATE (before, during or after the period covered by the request), just be consistent in both places where you have to use it.

  Linear regression is to find a straight line such that none of the actual data points are far from the line; That's what the Group functions calculate.
  In the above query, we will extend this straight line in the main query, such that we display a few points on the line of the beach of the actual data.

  Thank you.

  Full month
  01/01/2012 4 077
  01/02/2012 4 044
  01/03/2012 5 188...

  If you would care to post CREATE TABLE and INSERT statements for your sample data, and then I could test this.
  See the FAQ forum {message identifier: = 9360002}

• algorithms of data processing

  I use the R client to connect to oracle. my local computer has 1 GB of RAM and server is superior to it. My data is huge and I do not want the functions of R as lm. But I don't know that makes data load on the client and then treat the client or my service running on the server?
  If 2' th is true (it's good for me) made any other algorithm to do R and the algorithms of ore?
  
  Published by: Nasiri Mahdi on 23 February 2013 01:16

  Nguyen,

  Both approaches are supported. It really depends on what you're trying to do. For linear regression, you can try ore.lm or ore.odmGLM. If you build models partitioned on a large data, you can try ore.groupApply lm inside. If you do not yet take a look at the examples that we have in our 'Learning R Series' [url http://www.oracle.com/technetwork/database/options/advanced-analytics/r-enterprise/index.html] here.

  Denis